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# Solução_Calculo_Stewart_6e

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F.<br />

TX.10<br />

<strong>Solução</strong> detalhada de todos os exercícios ímpares do livro: Cálculo, James <strong>Stewart</strong> - VOLUME I e VOLUME II<br />

"O que sabemos é uma gota, o que ignoramos é um oceano." Isaac Newton<br />

23/11/2010.TX

F.<br />

TX.10<br />

1 FUNCTIONS AND MODELS<br />

1.1 Four Ways to Represent a Function<br />

In exercises requiring estimations or approximations, your answers may vary slightly from the answers given here.<br />

1. (a) The point (−1, −2) is on the graph of f,sof(−1) = −2.<br />

(b) When x =2, y is about 2.8,sof(2) ≈ 2.8.<br />

(c) f(x) =2is equivalent to y =2.Wheny =2,wehavex = −3 and x =1.<br />

(d) Reasonable estimates for x when y =0are x = −2.5 and x =0.3.<br />

(e) The domain of f consists of all x-values on the graph of f. For this function, the domain is −3 ≤ x ≤ 3,or[−3, 3].<br />

The range of f consists of all y-values on the graph of f. For this function, the range is −2 ≤ y ≤ 3,or[−2, 3].<br />

(f ) As x increases from −1 to 3, y increases from −2 to 3. Thus,f is increasing on the interval [−1, 3].<br />

3. From Figure 1 in the text, the lowest point occurs at about (t, a) =(12, −85). The highest point occurs at about (17, 115).<br />

Thus, the range of the vertical ground acceleration is −85 ≤ a ≤ 115. Written in interval notation, we get [−85, 115].<br />

5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, the curve fails<br />

the Vertical Line Test.<br />

7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is [−3, 2] and the range<br />

is [−3, −2) ∪ [−1, 3].<br />

9. The person’s weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The person’s weight<br />

dropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30 years saw a gradual<br />

increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet, exercise, health problems.<br />

11. The water will cool down almost to freezing as the ice<br />

melts. Then, when the ice has melted, the water will<br />

slowly warm up to room temperature.<br />

13. Of course, this graph depends strongly on the<br />

geographical location!<br />

15. As the price increases, the amount sold decreases. 17.<br />

9

F.<br />

TX.10<br />

10 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

19. (a) (b) From the graph, we estimate the number of<br />

cell-phone subscribers worldwide to be about<br />

92 million in 1995 and 485 million in 1999.<br />

21. f(x) =3x 2 − x +2.<br />

f(2) = 3(2) 2 − 2+2=12− 2+2=12.<br />

f(−2) = 3(−2) 2 − (−2)+2=12+2+2=16.<br />

f(a) =3a 2 − a +2.<br />

f(−a) =3(−a) 2 − (−a)+2=3a 2 + a +2.<br />

f(a +1)=3(a +1) 2 − (a +1)+2=3(a 2 +2a +1)− a − 1+2=3a 2 +6a +3− a +1=3a 2 +5a +4.<br />

2f(a) =2· f(a) =2(3a 2 − a +2)=6a 2 − 2a +4.<br />

f(2a) =3(2a) 2 − (2a)+2=3(4a 2 ) − 2a +2=12a 2 − 2a +2.<br />

f(a 2 )=3(a 2 ) 2 − (a 2 )+2=3(a 4 ) − a 2 +2=3a 4 − a 2 +2.<br />

[f(a)] 2 = 3a 2 − a +2 2 = 3a 2 − a +2 3a 2 − a +2 <br />

=9a 4 − 3a 3 +6a 2 − 3a 3 + a 2 − 2a +6a 2 − 2a +4=9a 4 − 6a 3 +13a 2 − 4a +4.<br />

f(a + h) =3(a + h) 2 − (a + h)+2=3(a 2 +2ah + h 2 ) − a − h +2=3a 2 +6ah +3h 2 − a − h +2.<br />

23. f(x) =4+3x − x 2 ,sof(3 + h) =4+3(3+h) − (3 + h) 2 =4+9+3h − (9 + 6h + h 2 )=4− 3h − h 2 ,<br />

and<br />

f(3 + h) − f(3)<br />

h<br />

= (4 − 3h − h2 ) − 4<br />

h<br />

=<br />

h(−3 − h)<br />

h<br />

= −3 − h.<br />

25.<br />

f(x) − f(a)<br />

x − a<br />

=<br />

1<br />

x − 1 a − x<br />

a<br />

x − a = xa<br />

x − a =<br />

a − x −1(x − a)<br />

=<br />

xa(x − a) xa(x − a) = − 1<br />

ax<br />

27. f(x) =x/(3x − 1) is defined for all x except when 0=3x − 1 ⇔ x = 1 , so the domain<br />

3<br />

is <br />

x ∈ R | x 6= 1 3 = −∞,<br />

1<br />

3 ∪ 1 , ∞ 3<br />

.<br />

29. f(t) = √ t + 3√ t is defined when t ≥ 0. These values of t give real number results for √ t, whereas any value of t gives a real<br />

number result for 3√ t.Thedomainis[0, ∞).<br />

31. h(x) =1 √ 4<br />

x 2 − 5x is defined when x 2 − 5x >0 ⇔ x(x − 5) > 0. Note that x 2 − 5x 6= 0since that would result in<br />

division by zero. The expression x(x − 5) is positive if x5. (See Appendix A for methods for solving<br />

inequalities.) Thus, the domain is (−∞, 0) ∪ (5, ∞).

F.<br />

TX.10<br />

SECTION 1.1 FOUR WAYS TO REPRESENT A FUNCTION ¤ 11<br />

33. f(x) =5is defined for all real numbers, so the domain is R,or(−∞, ∞).<br />

The graph of f is a horizontal line with y-intercept 5.<br />

35. f(t) =t 2 − 6t is defined for all real numbers, so the domain is R,or<br />

(−∞, ∞). The graph of f is a parabola opening upward since the coefficient<br />

of t 2 is positive. To find the t-intercepts, let y =0and solve for t.<br />

0=t 2 − 6t = t(t − 6) ⇒ t =0and t =6.Thet-coordinate of the<br />

vertex is halfway between the t-intercepts, that is, at t =3.Since<br />

f(3) = 3 2 − 6 · 3=−9,thevertexis(3, −9).<br />

37. g(x) = √ x − 5 is defined when x − 5 ≥ 0 or x ≥ 5,sothedomainis[5, ∞).<br />

Since y = √ x − 5 ⇒ y 2 = x − 5 ⇒ x = y 2 +5,weseethatg is the<br />

tophalfofaparabola.<br />

39. G(x) =<br />

<br />

3x + |x|<br />

x if x ≥ 0<br />

.Since|x| =<br />

x<br />

−x if x0<br />

if x0<br />

if x0<br />

2 if x

F.<br />

TX.10<br />

12 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

47. We need to solve the given equation for y. x +(y − 1) 2 =0 ⇔ (y − 1) 2 = −x ⇔ y − 1=± √ −x ⇔<br />

y =1± √ −x. The expression with the positive radical represents the top half of the parabola, and the one with the negative<br />

radical represents the bottom half. Hence, we want f(x) =1− √ −x. Note that the domain is x ≤ 0.<br />

49. For 0 ≤ x ≤ 3, the graph is the line with slope −1 and y-intercept 3,thatis,y = −x +3.For3

F.<br />

TX.10<br />

SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 13<br />

61. f is an odd function because its graph is symmetric about the origin. g is an even function because its graph is symmetric with<br />

respect to the y-axis.<br />

63. (a) Because an even function is symmetric with respect to the y-axis, and the point (5, 3) is on the graph of this even function,<br />

the point (−5, 3) must also be on its graph.<br />

(b) Because an odd function is symmetric with respect to the origin, and the point (5, 3) is on the graph of this odd function,<br />

the point (−5, −3) must also be on its graph.<br />

65. f(x) = x<br />

x 2 +1 .<br />

f(−x) =<br />

−x<br />

(−x) 2 +1 =<br />

So f is an odd function.<br />

−x<br />

x 2 +1 = − x<br />

x 2 +1 = −f(x).<br />

67. f(x) = x<br />

x +1 ,sof(−x) =<br />

−x<br />

−x +1 =<br />

x<br />

x − 1 .<br />

Sincethisisneitherf(x) nor −f(x), the function f is<br />

neither even nor odd.<br />

69. f(x) =1+3x 2 − x 4 .<br />

f(−x) =1+3(−x) 2 − (−x) 4 =1+3x 2 − x 4 = f(x).<br />

So f is an even function.<br />

1.2 Mathematical Models: A Catalog of Essential Functions<br />

1. (a) f(x) = 5√ x is a root function with n =5.<br />

(b) g(x) = √ 1 − x 2 is an algebraic function because it is a root of a polynomial.<br />

(c) h(x) =x 9 + x 4 is a polynomial of degree 9.<br />

(d) r(x) = x2 +1<br />

is a rational function because it is a ratio of polynomials.<br />

x 3 + x<br />

(e) s(x) =tan2x is a trigonometric function.<br />

(f ) t(x) =log 10 x is a logarithmic function.<br />

3. We notice from the figure that g and h are even functions (symmetric with respect to the y-axis) and that f is an odd function<br />

(symmetric with respect to the origin). So (b) y = x 5 must be f. Sinceg is flatter than h near the origin, we must have<br />

(c) y = x 8 matched with g and (a) y = x 2 matched with h.

F.<br />

TX.10<br />

14 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

5. (a) An equation for the family of linear functions with slope 2<br />

is y = f(x) =2x + b,whereb is the y-intercept.<br />

(b) f(2) = 1 means that the point (2, 1) is on the graph of f. We can use the<br />

point-slope form of a line to obtain an equation for the family of linear<br />

functions through the point (2, 1). y − 1=m(x − 2), which is equivalent<br />

to y = mx +(1− 2m) in slope-intercept form.<br />

(c) To belong to both families, an equation must have slope m =2, so the equation in part (b), y = mx +(1− 2m),<br />

becomes y =2x − 3. Itistheonly function that belongs to both families.<br />

7. All members of the family of linear functions f(x) =c − x have graphs<br />

that are lines with slope −1. They-intercept is c.<br />

9. Since f(−1) = f(0) = f(2) = 0, f has zeros of −1, 0,and2,soanequationforf is f(x) =a[x − (−1)](x − 0)(x − 2),<br />

or f(x) =ax(x +1)(x − 2). Because f(1) = 6, we’ll substitute 1 for x and 6 for f(x).<br />

6=a(1)(2)(−1) ⇒ −2a =6 ⇒ a = −3, so an equation for f is f(x) =−3x(x +1)(x − 2).<br />

11. (a) D =200,soc =0.0417D(a +1)=0.0417(200)(a +1)=8.34a +8.34. The slope is 8.34, which represents the<br />

change in mg of the dosage for a child for each change of 1 year in age.<br />

(b) For a newborn, a =0,soc =8.34 mg.<br />

13. (a) (b) The slope of 9 means that F increases 9 degrees for each increase<br />

5 5<br />

of 1 ◦ C. (Equivalently, F increases by 9 when C increases by 5<br />

and F decreases by 9 when C decreases by 5.) The F -intercept of<br />

32 is the Fahrenheit temperature corresponding to a Celsius<br />

temperature of 0.

F.<br />

TX.10<br />

SECTION 1.2 MATHEMATICAL MODELS: A CATALOG OF ESSENTIAL FUNCTIONS ¤ 15<br />

15. (a) Using N in place of x and T in place of y,wefind the slope to be T 2 − T 1 80 − 70<br />

=<br />

N 2 − N 1 173 − 113 = 10<br />

equation is T − 80 = 1 (N − 173) ⇔ T − 80 = 1 N − 173 ⇔ T = 1 N + 307<br />

6 6 6 6 6<br />

60 = 1 . So a linear<br />

6<br />

307<br />

6<br />

=51.16 .<br />

(b) The slope of 1 means that the temperature in Fahrenheit degrees increases one-sixth as rapidly as the number of cricket<br />

6<br />

chirps per minute. Said differently, each increase of 6 cricket chirps per minute corresponds to an increase of 1 ◦ F.<br />

(c) When N =150, the temperature is given approximately by T = 1 307<br />

(150) + =76.16 ◦ F ≈ 76 ◦ F.<br />

6 6<br />

17. (a) We are given<br />

change in pressure<br />

10 feet change in depth = 4.34 =0.434. UsingP for pressure and d for depth with the point<br />

10<br />

(d, P )=(0, 15), we have the slope-intercept form of the line, P =0.434d +15.<br />

(b) When P = 100,then100 = 0.434d +15 ⇔ 0.434d =85 ⇔ d = 85<br />

0.434<br />

≈ 195.85 feet. Thus, the pressure is<br />

100 lb/in 2 at a depth of approximately 196 feet.<br />

19. (a) The data appear to be periodic and a sine or cosine function would make the best model. A model of the form<br />

f(x) =a cos(bx)+c seems appropriate.<br />

(b) The data appear to be decreasing in a linear fashion. A model of the form f(x) =mx + b seems appropriate.<br />

Some values are given to many decimal places. These are the results given by several computer algebra systems — rounding is left to the reader.<br />

21. (a)<br />

(b) Using the points (4000, 14.1) and (60,000, 8.2),weobtain<br />

8.2 − 14.1<br />

y − 14.1 = (x − 4000) or, equivalently,<br />

60,000 − 4000<br />

y ≈−0.000105357x +14.521429.<br />

A linear model does seem appropriate.<br />

(c) Using a computing device, we obtain the least squares regression line y = −0.0000997855x +13.950764.<br />

The following commands and screens illustrate how to find the least squares regression line on a TI-83 Plus.<br />

Enter the data into list one (L1) and list two (L2). Press<br />

to enter the editor.<br />

Find the regession line and store it in Y 1.Press .

F.<br />

TX.10<br />

16 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

Note from the last figure that the regression line has been stored in Y 1 andthatPlot1hasbeenturnedon(Plot1is<br />

highlighted). You can turn on Plot1 from the Y= menu by placing the cursor on Plot1 and pressing<br />

or by<br />

pressing .<br />

Now press<br />

to produce a graph of the data and the regression<br />

line. Note that choice 9 of the ZOOM menu automatically selects a window<br />

that displays all of the data.<br />

(d) When x =25,000, y ≈ 11.456; or about 11.5 per 100 population.<br />

(e) When x =80,000, y ≈ 5.968;orabouta6% chance.<br />

(f ) When x = 200,000, y is negative, so the model does not apply.<br />

23. (a)<br />

(b)<br />

A linear model does seem appropriate.<br />

Using a computing device, we obtain the least squares<br />

regression line y =0.089119747x − 158.2403249,<br />

where x is the year and y is the height in feet.<br />

(c) When x = 2000,themodelgivesy ≈ 20.00 ft. Note that the actual winning height for the 2000 Olympics is less than the<br />

winning height for 1996—so much for that prediction.<br />

(d) When x =2100, y ≈ 28.91 ft. This would be an increase of 9.49 ft from 1996 to 2100. Even though there was an increase<br />

of 8.59 ft from 1900 to 1996, it is unlikely that a similar increase will occur over the next 100 years.<br />

25. Using a computing device, we obtain the cubic<br />

function y = ax 3 + bx 2 + cx + d with<br />

a =0.0012937, b = −7.06142, c =12,823,<br />

and d = −7,743,770. Whenx = 1925,<br />

y ≈ 1914 (million).

F.<br />

TX.10<br />

SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 17<br />

1.3 New Functions from Old Functions<br />

1. (a) If the graph of f is shifted 3 units upward, its equation becomes y = f(x)+3.<br />

(b) If the graph of f is shifted 3 units downward, its equation becomes y = f(x) − 3.<br />

(c) If the graph of f is shifted 3 units to the right, its equation becomes y = f(x − 3).<br />

(d) If the graph of f is shifted 3 units to the left, its equation becomes y = f(x +3).<br />

(e) If the graph of f is reflected about the x-axis, its equation becomes y = −f(x).<br />

(f ) If the graph of f is reflected about the y-axis, its equation becomes y = f(−x).<br />

(g) If the graph of f isstretchedverticallybyafactorof3, its equation becomes y =3f(x).<br />

(h) If the graph of f is shrunk vertically by a factor of 3, its equation becomes y = 1 f(x). 3<br />

3. (a) (graph 3) The graph of f is shifted 4 units to the right and has equation y = f(x − 4).<br />

(b) (graph 1) The graph of f is shifted 3 units upward and has equation y = f(x)+3.<br />

(c) (graph 4) The graph of f is shrunk vertically by a factor of 3 and has equation y = 1 f(x). 3<br />

(d) (graph 5) The graph of f is shifted 4 units to the left and reflected about the x-axis. Its equation is y = −f(x +4).<br />

(e) (graph 2) The graph of f is shifted 6 units to the left and stretched vertically by a factor of 2. Its equation is<br />

y =2f(x +6).<br />

5. (a) To graph y = f(2x) we shrink the graph of f<br />

horizontally by a factor of 2.<br />

(b) To graph y = f 1<br />

2 x westretchthegraphoff<br />

horizontally by a factor of 2.<br />

The point (4, −1) on the graph of f corresponds to the<br />

point 1<br />

2 · 4, −1 =(2, −1).<br />

(c) To graph y = f(−x) we reflect the graph of f about<br />

the y-axis.<br />

The point (4, −1) on the graph of f corresponds to the<br />

point (2 · 4, −1) = (8, −1).<br />

(d) To graph y = −f(−x) we reflect the graph of f about<br />

the y-axis, then about the x-axis.<br />

The point (4, −1) on the graph of f corresponds to the<br />

point (−1 · 4, −1) = (−4, −1).<br />

The point (4, −1) on the graph of f corresponds to the<br />

point (−1 · 4, −1 · −1) = (−4, 1).

F.<br />

TX.10<br />

18 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

7. The graph of y = f(x) = √ 3x − x 2 has been shifted 4 units to the left, reflected about the x-axis, and shifted downward<br />

1 unit. Thus, a function describing the graph is<br />

y = −1 ·<br />

<br />

reflect<br />

f (x +4)<br />

<br />

shift<br />

4 units left<br />

− 1<br />

<br />

shift<br />

1 unit left<br />

This function can be written as<br />

y = −f(x +4)− 1=− 3(x +4)− (x +4) 2 − 1=− 3x +12− (x 2 +8x +16)− 1=− √ −x 2 − 5x − 4 − 1<br />

9. y = −x 3 : Start with the graph of y = x 3 and reflect<br />

gives the same result since substituting −x for x gives<br />

us y =(−x) 3 = −x 3 .<br />

11. y =(x +1) 2 : Start with the graph of y = x 2<br />

and shift 1 unit to the left.<br />

13. y =1+2cosx: Start with the graph of y =cosx, stretch vertically by a factor of 2,andthenshift1 unit upward.<br />

15. y =sin(x/2): Start with the graph of y =sinx and stretch horizontally by a factor of 2.

F.<br />

TX.10<br />

SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 19<br />

17. y = √ x +3: Start with the graph of<br />

y = √ x and shift 3 units to the left.<br />

19. y = 1 2 (x2 +8x) = 1 2 (x2 +8x +16)− 8= 1 2 (x +4)2 − 8: Start with the graph of y = x 2 , compress vertically by a<br />

factor of 2,shift4 units to the left, and then shift 8 units downward.<br />

0 0 0 0<br />

21. y =2/(x +1): Start with the graph of y =1/x,shift1 unit to the left, and then stretch vertically by a factor of 2.<br />

23. y = |sin x|: Start with the graph of y =sinx and reflect all the parts of the graph below the x-axis about the x-axis.<br />

25. This is just like the solution to Example 4 except the amplitude of the curve (the 30 ◦ NcurveinFigure9onJune21)is<br />

14 − 12 = 2. So the function is L(t) =12+2sin 2π<br />

365 (t − 80) . March 31 is the 90th day of the year, so the model gives<br />

L(90) ≈ 12.34 h. The daylight time (5:51 AM to 6:18 PM)is12 hours and 27 minutes, or 12.45 h. The model value differs<br />

from the actual value by 12.45−12.34<br />

12.45<br />

≈ 0.009,lessthan1%.<br />

27. (a) To obtain y = f(|x|), the portion of the graph of y = f(x) to the right of the y-axisisreflected about the y-axis.<br />

(b) y =sin|x|<br />

(c) y = |x|

F.<br />

TX.10<br />

20 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

29. f(x) =x 3 +2x 2 ; g(x) =3x 2 − 1. D = R for both f and g.<br />

(f + g)(x) =(x 3 +2x 2 )+(3x 2 − 1) = x 3 +5x 2 − 1, D = R.<br />

(f − g)(x) =(x 3 +2x 2 ) − (3x 2 − 1) = x 3 − x 2 +1, D = R.<br />

(fg)(x) =(x 3 +2x 2 )(3x 2 − 1) = 3x 5 +6x 4 − x 3 − 2x 2 , D = R.<br />

<br />

f<br />

(x) = x3 +2x 2<br />

g 3x 2 − 1 , D = x | x 6=± √ 1 <br />

since 3x 2 − 1 6= 0.<br />

3<br />

31. f(x) =x 2 − 1, D = R; g(x) =2x +1, D = R.<br />

(a) (f ◦ g)(x) =f(g(x)) = f(2x +1)=(2x +1) 2 − 1=(4x 2 +4x +1)− 1=4x 2 +4x, D = R.<br />

(b) (g ◦ f)(x) =g(f(x)) = g(x 2 − 1) = 2(x 2 − 1) + 1 = (2x 2 − 2) + 1 = 2x 2 − 1, D = R.<br />

(c) (f ◦ f)(x) =f(f(x)) = f(x 2 − 1) = (x 2 − 1) 2 − 1=(x 4 − 2x 2 +1)− 1=x 4 − 2x 2 , D = R.<br />

(d) (g ◦ g)(x) =g(g(x)) = g(2x +1)=2(2x +1)+1=(4x +2)+1=4x +3, D = R.<br />

33. f(x) =1− 3x; g(x) =cosx. D = R for both f and g, and hence for their composites.<br />

(a) (f ◦ g)(x) =f(g(x)) = f(cos x) =1− 3cosx.<br />

(b) (g ◦ f)(x) =g(f(x)) = g(1 − 3x) =cos(1− 3x).<br />

(c) (f ◦ f)(x) =f(f(x)) = f(1 − 3x) =1− 3(1 − 3x) =1− 3+9x =9x − 2.<br />

(d) (g ◦ g)(x) =g(g(x)) = g(cos x) =cos(cosx)<br />

[Note that this is not cos x · cos x.]<br />

35. f(x) =x + 1 +1<br />

, D = {x | x 6= 0}; g(x) =x , D = {x | x 6=−2}<br />

x x +2<br />

x +1<br />

(a) (f ◦ g)(x) =f(g(x)) = f = x +1<br />

x +2 x +2 + 1 = x +1<br />

x +1 x +2 + x +2<br />

x +1<br />

x +2<br />

=<br />

(x +1)(x +1)+(x +2)(x +2)<br />

(x +2)(x +1)<br />

=<br />

x 2 +2x +1 + x 2 +4x +4 <br />

(x +2)(x +1)<br />

Since g(x) is not defined for x = −2 and f(g(x)) is not defined for x = −2 and x = −1,<br />

the domain of (f ◦ g)(x) is D = {x | x 6=−2, −1}.<br />

<br />

(b) (g ◦ f)(x) =g(f(x)) = g x + 1 <br />

=<br />

x<br />

<br />

x + 1 <br />

+1<br />

x<br />

<br />

x + 1 =<br />

+2<br />

x<br />

x 2 +1+x<br />

x<br />

x 2 +1+2x<br />

x<br />

Since f(x) is not defined for x =0and g(f(x)) is not defined for x = −1,<br />

the domain of (g ◦ f)(x) is D = {x | x 6=−1, 0}.<br />

<br />

(c) (f ◦ f)(x)=f(f(x)) = f x + 1 <br />

= x + 1 <br />

+ 1<br />

x x x + 1 x<br />

= x(x) x 2 +1 +1 x 2 +1 + x(x)<br />

x(x 2 +1)<br />

= x4 +3x 2 +1<br />

, D = {x | x 6= 0}<br />

x(x 2 +1)<br />

= x + 1 x + 1<br />

x 2 +1<br />

x<br />

= x4 + x 2 + x 2 +1+x 2<br />

x(x 2 +1)<br />

= x2 + x +1<br />

x 2 +2x +1 = x2 + x +1<br />

(x +1) 2<br />

= x + 1 x + x<br />

x 2 +1<br />

= 2x2 +6x +5<br />

(x +2)(x +1)

F.<br />

TX.10<br />

x +1<br />

(d) (g ◦ g)(x) =g(g(x)) = g =<br />

x +2<br />

x +1<br />

x +2 +1<br />

=<br />

x +1<br />

x +2 +2<br />

SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS ¤ 21<br />

x +1+1(x +2)<br />

x +2<br />

x +1+2(x +2)<br />

x +2<br />

Since g(x) is not defined for x = −2 and g(g(x)) is not defined for x = − 5 3 ,<br />

the domain of (g ◦ g)(x) is D = x | x 6=−2, − 5 3<br />

<br />

.<br />

37. (f ◦ g ◦ h)(x) =f(g(h(x))) = f(g(x − 1)) = f(2(x − 1)) = 2(x − 1) + 1 = 2x − 1<br />

39. (f ◦ g ◦ h)(x) =f(g(h(x))) = f(g(x 3 +2))=f[(x 3 +2) 2 ]<br />

= f(x 6 +4x 3 +4)= (x 6 +4x 3 +4)− 3= √ x 6 +4x 3 +1<br />

=<br />

x +1+x +2 2x +3<br />

=<br />

x +1+2x +4 3x +5<br />

41. Let g(x) =x 2 +1and f(x) =x 10 .Then(f ◦ g)(x) =f(g(x)) = f(x 2 +1)=(x 2 +1) 10 = F (x).<br />

43. Let g(x) = 3√ x and f(x) = x<br />

1+x .Then(f ◦ g)(x) =f(g(x)) = f( 3√ x )= 3√ x<br />

1+ 3√ x = F (x).<br />

45. Let g(t) =cost and f(t) = √ t.Then(f ◦ g)(t) =f(g(t)) = f(cos t) = √ cos t = u(t).<br />

47. Let h(x) =x 2 , g(x) =3 x ,andf(x) =1− x. Then<br />

<br />

(f ◦ g ◦ h)(x) =f(g(h(x))) = f(g(x 2 )) = f 3 x2 =1− 3 x2 = H(x).<br />

49. Let h(x) = √ x, g(x) =secx, andf(x) =x 4 .Then<br />

(f ◦ g ◦ h)(x) =f(g(h(x))) = f(g( √ x )) = f(sec √ x )=(sec √ x ) 4 =sec 4 ( √ x )=H(x).<br />

51. (a) g(2) = 5, because the point (2, 5) is on the graph of g. Thus,f(g(2)) = f(5) = 4, because the point (5, 4) is on the<br />

graph of f.<br />

(b) g(f(0)) = g(0) = 3<br />

(c) (f ◦ g)(0) = f(g(0)) = f(3) = 0<br />

(d) (g ◦ f)(6) = g(f(6)) = g(6). This value is not defined, because there is no point on the graph of g that has<br />

x-coordinate 6.<br />

(e) (g ◦ g)(−2) = g(g(−2)) = g(1) = 4<br />

(f ) (f ◦ f)(4) = f(f(4)) = f(2) = −2<br />

53. (a) Using the relationship distance = rate · time with the radius r as the distance, we have r(t) =60t.<br />

(b) A = πr 2 ⇒ (A ◦ r)(t) =A(r(t)) = π(60t) 2 = 3600πt 2 . This formula gives us the extent of the rippled area<br />

(in cm 2 )atanytimet.<br />

55. (a) From the figure, we have a right triangle with legs 6 and d, and hypotenuse s.<br />

By the Pythagorean Theorem, d 2 +6 2 = s 2 ⇒ s = f(d) = √ d 2 +36.<br />

(b) Using d = rt,wegetd =(30km/hr)(t hr) =30t (in km). Thus,<br />

d = g(t) =30t.<br />

(c) (f ◦ g)(t) =f(g(t)) = f(30t) = (30t) 2 +36= √ 900t 2 +36. This function represents the distance between the<br />

lighthouse and the ship as a function of the time elapsed since noon.

F.<br />

TX.10<br />

22 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

57. (a)<br />

(b)<br />

H(t) =<br />

<br />

0 if t

F.<br />

TX.10<br />

1.4 Graphing Calculators and Computers<br />

SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS ¤ 23<br />

1. f(x) = √ x 3 − 5x 2<br />

(a) [−5, 5] by [−5, 5]<br />

(There is no graph shown.)<br />

(b) [0, 10] by [0, 2] (c) [0, 10] by [0, 10]<br />

The most appropriate graph is produced in viewing rectangle (c).<br />

3. Since the graph of f(x) =5+20x − x 2 is a<br />

parabola opening downward, an appropriate viewing<br />

rectangle should include the maximum point.<br />

5. f(x) = 4√ 81 − x 4 is defined when 81 − x 4 ≥ 0 ⇔<br />

x 4 ≤ 81 ⇔ |x| ≤ 3, so the domain of f is [−3, 3]. Also<br />

0 ≤ 4√ 81 − x 4 ≤ 4√ 81 = 3,sotherangeis[0, 3].<br />

7. The graph of f(x) =x 3 − 225x is symmetric with respect to the origin.<br />

Since f(x) =x 3 − 225x = x(x 2 − 225) = x(x + 15)(x − 15),there<br />

are x-intercepts at 0, −15,and15. f(20) = 3500.<br />

9. The period of g(x) = sin(1000x) is 2π ≈ 0.0063 and its range is<br />

1000<br />

[−1, 1]. Sincef(x) =sin 2 (1000x) is the square of g,itsrangeis<br />

[0, 1] and a viewing rectangle of [−0.01, 0.01] by [0, 1.1] seems<br />

appropriate.<br />

11. The domain of y = √ x is x ≥ 0,sothedomainoff(x) =sin √ x is [0, ∞)<br />

and the range is [−1, 1]. With a little trial-and-error experimentation, we find<br />

that an Xmax of 100 illustrates the general shape of f,soanappropriate<br />

viewing rectangle is [0, 100] by [−1.5, 1.5].

F.<br />

TX.10<br />

24 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

13. The first term, 10 sin x, has period 2π and range [−10, 10]. It will be the dominant term in any “large” graph of<br />

y =10sinx +sin100x, as shown in the first figure. The second term, sin 100x,hasperiod 2π = π and range [−1, 1].<br />

100 50<br />

Itcausesthebumpsinthefirst figure and will be the dominant term in any “small” graph, as shown in the view near the<br />

origin in the second figure.<br />

15. We must solve the given equation for y to obtain equations for the upper and<br />

lower halves of the ellipse.<br />

4x 2 +2y 2 =1 ⇔ 2y 2 =1− 4x 2 ⇔ y 2 = 1 − 4x2<br />

2<br />

<br />

1 − 4x<br />

2<br />

y = ±<br />

2<br />

⇔<br />

17. From the graph of y =3x 2 − 6x +1<br />

and y =0.23x − 2.25 in the viewing<br />

rectangle [−1, 3] by [−2.5, 1.5],itis<br />

difficult to see if the graphs intersect.<br />

If we zoom in on the fourth quadrant,<br />

we see the graphs do not intersect.<br />

19. From the graph of f(x) =x 3 − 9x 2 − 4, we see that there is one solution<br />

of the equation f(x) =0and it is slightly larger than 9. By zooming in or<br />

using a root or zero feature, we obtain x ≈ 9.05.<br />

21. We see that the graphs of f(x) =x 2 and g(x) =sinx intersect twice. One<br />

solution is x =0. The other solution of f = g is the x-coordinate of the<br />

point of intersection in the first quadrant. Using an intersect feature or<br />

zooming in, we find this value to be approximately 0.88. Alternatively, we<br />

could find that value by finding the positive zero of h(x) =x 2 − sin x.<br />

Note: After producing the graph on a TI-83 Plus, we can find the approximate value 0.88 by using the following keystrokes:<br />

. The “1” is just a guess for 0.88.

F.<br />

TX.10<br />

SECTION 1.4 GRAPHING CALCULATORS AND COMPUTERS ¤ 25<br />

23. g(x) =x 3 /10 is larger than f(x) =10x 2 whenever x>100.<br />

25. We see from the graphs of y = |sin x − x| and y =0.1 that there are<br />

two solutions to the equation |sin x − x| =0.1: x ≈−0.85 and<br />

x ≈ 0.85. The condition |sin x − x| < 0.1 holds for any x lying<br />

between these two values, that is, −0.85

F.<br />

TX.10<br />

26 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

33. y 2 = cx 3 + x 2 .Ifc0 (b) R (c) (0, ∞) (d) See Figures 4(c), 4(b), and 4(a), respectively.<br />

3. All of these graphs approach 0 as x →−∞, all of them pass through the point<br />

(0, 1), and all of them are increasing and approach ∞ as x →∞.Thelargerthe<br />

base, the faster the function increases for x>0, and the faster it approaches 0 as<br />

x →−∞.<br />

Note: The notation “x →∞” can be thought of as “x becomes large” at this point.<br />

More details on this notation are given in Chapter 2.<br />

5. The functions with bases greater than 1 (3 x and 10 x ) are increasing, while those<br />

with bases less than 1 <br />

1 x<br />

and <br />

1 x <br />

3<br />

10 are decreasing. The graph of 1<br />

x<br />

is the<br />

3<br />

reflection of that of 3 x about the y-axis, and the graph of 1<br />

10<br />

x<br />

is the reflection of<br />

that of 10 x about the y-axis. The graph of 10 x increases more quickly than that of<br />

3 x for x>0, and approaches 0 faster as x →−∞.<br />

7. We start with the graph of y =4 x (Figure 3) and then<br />

shift 3 units downward. This shift doesn’t affect the<br />

domain, but the range of y =4 x − 3 is (−3, ∞) .<br />

There is a horizontal asymptote of y = −3.<br />

y =4 x y =4 x − 3

F.<br />

TX.10<br />

SECTION 1.5 EXPONENTIAL FUNCTIONS ¤ 27<br />

9. We start with the graph of y =2 x (Figure 3),<br />

x-axis (or just rotate 180 ◦ to handle both<br />

reflections) to obtain the graph of y = −2 −x .<br />

In each graph, y =0is the horizontal<br />

asymptote.<br />

y =2 x y =2 −x y = −2 −x<br />

11. We start with the graph of y = e x (Figure 13) and reflect about the y-axis to get the graph of y = e −x . Then we compress the<br />

graphverticallybyafactorof2 to obtain the graph of y = 1 2 e−x andthenreflect about the x-axis to get the graph of<br />

y = − 1 2 e−x . Finally, we shift the graph upward one unit to get the graph of y =1− 1 2 e−x .<br />

13. (a) To find the equation of the graph that results from shifting the graph of y = e x 2 units downward, we subtract 2 from the<br />

original function to get y = e x − 2.<br />

(b) To find the equation of the graph that results from shifting the graph of y = e x 2 units to the right, we replace x with x − 2<br />

in the original function to get y = e (x−2) .<br />

(c) To find the equation of the graph that results from reflecting the graph of y = e x about the x-axis, we multiply the original<br />

function by −1 to get y = −e x .<br />

(d) To find the equation of the graph that results from reflecting the graph of y = e x about the y-axis, we replace x with −x in<br />

the original function to get y = e −x .<br />

(e) To find the equation of the graph that results from reflecting the graph of y = e x about the x-axis and then about the<br />

y-axis, we first multiply the original function by −1 (to get y = −e x ) and then replace x with −x in this equation to<br />

get y = −e −x .<br />

15. (a) The denominator 1+e x is never equal to zero because e x > 0, so the domain of f(x) =1/(1 + e x ) is R.<br />

(b) 1 − e x =0 ⇔ e x =1 ⇔ x =0, so the domain of f(x) =1/(1 − e x ) is (−∞, 0) ∪ (0, ∞).<br />

<br />

17. Use y = Ca x with the points (1, 6) and (3, 24). 6=Ca 1 C =<br />

6<br />

a<br />

6<br />

and 24 = Ca 3 ⇒ 24 = a 3<br />

a<br />

⇒<br />

4=a 2 ⇒ a =2 [since a>0] andC = 6 2 =3. The function is f(x) =3· 2x .<br />

19. If f(x) =5 x ,then<br />

f(x + h) − f(x)<br />

h<br />

= 5x+h − 5 x<br />

h<br />

= 5x 5 h − 5 x<br />

h<br />

= 5x 5 h − 1 <br />

h<br />

5<br />

=5 x h − 1<br />

.<br />

h

F.<br />

TX.10<br />

28 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

21. 2 ft =24in, f(24) = 24 2 in = 576 in =48ft. g(24) = 2 24 in =2 24 /(12 · 5280) mi ≈ 265 mi<br />

23. The graph of g finally surpasses that of f at x ≈ 35.8.<br />

25. (a) Fifteen hours represents 5 doubling periods (one doubling period is three hours). 100 · 2 5 = 3200<br />

(b) In t hours, there will be t/3 doubling periods. The initial population is 100,<br />

so the population y at time t is y = 100 · 2 t/3 .<br />

(c) t =20 ⇒ y = 100 · 2 20/3 ≈ 10,159<br />

(d) We graph y 1 = 100 · 2 x/3 and y 2 =50,000. The two curves intersect at<br />

x ≈ 26.9, so the population reaches 50,000 in about 26.9 hours.<br />

27. An exponential model is y = ab t ,wherea =3.154832569 × 10 −12<br />

and b =1.017764706. This model gives y(1993) ≈ 5498 million and<br />

y(2010) ≈ 7417 million.<br />

29. From the graph, it appears that f is an odd function (f is undefined for x =0).<br />

To prove this, we must show that f(−x) =−f(x).<br />

f(−x) = 1 − e1/(−x) 1 −<br />

1 − 1<br />

e(−1/x)<br />

=<br />

1+e1/(−x) 1+e = e 1/x<br />

(−1/x)<br />

1+ 1<br />

e 1/x<br />

= − 1 − e1/x<br />

= −f(x)<br />

1+e1/x so f is an odd function.<br />

· e1/x<br />

e = e1/x − 1<br />

1/x e 1/x +1<br />

1.6 Inverse Functions and Logarithms<br />

1. (a) See Definition 1.<br />

(b) It must pass the Horizontal Line Test.<br />

3. f is not one-to-one because 2 6= 6,butf(2) = 2.0 =f(6).<br />

5. No horizontal line intersects the graph of f more than once. Thus, by the Horizontal Line Test, f is one-to-one.

F.<br />

TX.10<br />

SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS ¤ 29<br />

7. The horizontal line y =0(the x-axis) intersects the graph of f in more than one point. Thus, by the Horizontal Line Test,<br />

f is not one-to-one.<br />

9. The graph of f(x) =x 2 − 2x is a parabola with axis of symmetry x = − b<br />

2a = − −2 =1.Pickanyx-values equidistant<br />

2(1)<br />

from 1 to find two equal function values. For example, f(0) = 0 and f(2) = 0,sof is not one-to-one.<br />

11. g(x) =1/x. x 1 6=x 2 ⇒ 1/x 1 6=1/x 2 ⇒ g (x 1 ) 6=g (x 2 ),sog is one-to-one.<br />

Geometric solution: The graph of g is the hyperbola shown in Figure 14 in Section 1.2. It passes the Horizontal Line Test,<br />

so g is one-to-one.<br />

13. A football will attain every height h up to its maximum height twice: once on the way up, and again on the way down. Thus,<br />

even if t 1 does not equal t 2 , f(t 1 ) may equal f(t 2 ),sof is not 1-1.<br />

15. Since f(2) = 9 and f is 1-1, we know that f −1 (9) = 2. Remember, if the point (2, 9) is on the graph of f, then the point<br />

(9, 2) is on the graph of f −1 .<br />

17. First, we must determine x such that g(x) =4. By inspection, we see that if x =0,theng(x) =4.Sinceg is 1-1 (g is an<br />

increasing function), it has an inverse, and g −1 (4) = 0.<br />

19. We solve C = 5 (F − 32) for F : 9 C = F − 32 ⇒ F = 9 9 5 5<br />

C +32. This gives us a formula for the inverse function, that<br />

is, the Fahrenheit temperature F as a function of the Celsius temperature C. F ≥−459.67 ⇒ 9 5 C +32≥−459.67 ⇒<br />

9<br />

5<br />

C ≥−491.67 ⇒ C ≥−273.15, the domain of the inverse function.<br />

21. f(x) = √ 10 − 3x ⇒ y = √ 10 − 3x (y ≥ 0) ⇒ y 2 =10− 3x ⇒ 3x =10− y 2 ⇒ x = − 1 3 y2 + 10<br />

3 .<br />

Interchange x and y: y = − 1 3 x2 + 10<br />

3 .Sof −1 (x) =− 1 3 x2 + 10<br />

3 . Note that the domain of f −1 is x ≥ 0.<br />

23. y = f(x) =e x3 ⇒ ln y = x 3 ⇒ x = 3√ ln y. Interchange x and y: y = 3√ ln x. Sof −1 (x) = 3√ ln x.<br />

25. y = f(x) =ln(x +3) ⇒ x +3=e y ⇒ x = e y − 3. Interchange x and y: y = e x − 3. Sof −1 (x) =e x − 3.<br />

27. y = f(x) =x 4 +1 ⇒ y − 1=x 4 ⇒ x = 4√ y − 1 (not ± since<br />

x ≥ 0). Interchange x and y: y = 4√ x − 1. Sof −1 (x) = 4√ x − 1. The<br />

graph of y = 4√ x − 1 is just the graph of y = 4√ x shifted right one unit.<br />

From the graph, we see that f and f −1 are reflections about the line y = x.<br />

29. Reflect the graph of f about the line y = x. The points (−1, −2), (1, −1),<br />

(2, 2),and(3, 3) on f are reflected to (−2, −1), (−1, 1), (2, 2),and(3, 3)<br />

on f −1 .

F.<br />

TX.10<br />

30 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

31. (a) It is defined as the inverse of the exponential function with base a,thatis,log a x = y ⇔ a y = x.<br />

(b) (0, ∞) (c) R (d) See Figure 11.<br />

33. (a) log 5 125 = 3 since 5 3 =125. (b) log 3<br />

1<br />

27 = −3 since 3−3 = 1 3 3 = 1 27 .<br />

35. (a) log 2 6 − log 2 15 + log 2 20 = log 2 ( 6<br />

15 )+log 2 20 [by Law 2]<br />

=log 2 ( 6 · 20) [by Law 1]<br />

15<br />

=log 2 8,and log 2 8=3since 2 3 =8.<br />

<br />

(b) log 3 100 − log 3 18 − log 3 50 = log 100<br />

<br />

3 18 − log3 50 = log 100<br />

<br />

3 18·50<br />

37. ln5+5ln3=ln5+ln3 5 [by Law 3]<br />

=ln(5· 3 5 ) [by Law 1]<br />

=ln1215<br />

=log 3 ( 1 9 ),and log 3 1<br />

9<br />

= −2 since 3 −2 = 1 9 .<br />

39. ln(1 + x 2 )+ 1 2 ln x − ln sin x =ln(1+x2 )+lnx 1/2 − ln sin x =ln[(1+x 2 ) √ x ] − ln sin x =ln (1 + x2 ) √ x<br />

sin x<br />

41. To graph these functions, we use log 1.5 x = ln x<br />

ln 1.5 and log 50 x = ln x<br />

ln 50 .<br />

These graphs all approach −∞ as x → 0 + , and they all pass through the<br />

point (1, 0). Also, they are all increasing, and all approach ∞ as x →∞.<br />

The functions with larger bases increase extremely slowly, and the ones with<br />

smaller bases do so somewhat more quickly. The functions with large bases<br />

approach the y-axis more closely as x → 0 + .<br />

43. 3 ft =36in, so we need x such that log 2 x =36 ⇔ x =2 36 =68,719,476,736. Inmiles,thisis<br />

68,719,476,736 in · 1 ft<br />

12 in · 1 mi<br />

≈ 1,084,587.7 mi.<br />

5280 ft<br />

45. (a) Shift the graph of y =log 10 x five units to the left to<br />

obtain the graph of y =log 10 (x +5).Notethevertical<br />

asymptote of x = −5.<br />

(b) Reflect the graph of y =lnx about the x-axis to obtain<br />

the graph of y = − ln x.<br />

y =log 10 x y =log 10 (x +5)<br />

47. (a) 2lnx =1 ⇒ ln x = 1 2<br />

⇒ x = e 1/2 = √ e<br />

(b) e −x =5 ⇒ −x =ln5 ⇒ x = − ln 5<br />

y =lnx<br />

y = − ln x<br />

49. (a) 2 x−5 =3 ⇔ log 2 3=x − 5 ⇔ x =5+log 2 3.<br />

Or: 2 x−5 =3 ⇔ ln 2 x−5 =ln3 ⇔ (x − 5) ln 2 = ln 3 ⇔ x − 5= ln 3<br />

ln 2<br />

⇔ x =5+ ln 3<br />

ln 2

F.<br />

TX.10<br />

SECTION 1.6 INVERSE FUNCTIONS AND LOGARITHMS ¤ 31<br />

(b) ln x +ln(x − 1) = ln(x(x − 1)) = 1 ⇔ x(x − 1) = e 1 ⇔ x 2 − x − e =0. The quadratic formula (with a =1,<br />

b = −1,andc = −e)givesx = 1 2<br />

<br />

1 ±<br />

√ 1+4e<br />

<br />

, but we reject the negative root since the natural logarithm is not<br />

defined for xe −1 ⇒ x>e −1 ⇒ x ∈ (1/e, ∞)<br />

53. (a) For f(x) = √ 3 − e 2x ,wemusthave3 − e 2x ≥ 0 ⇒ e 2x ≤ 3 ⇒ 2x ≤ ln 3 ⇒ x ≤ 1 2<br />

ln 3. Thus, the domain<br />

of f is (−∞, 1 2<br />

ln 3].<br />

(b) y = f(x) = √ 3 − e 2x [note that y ≥ 0] ⇒ y 2 =3− e 2x ⇒ e 2x =3− y 2 ⇒ 2x =ln(3− y 2 ) ⇒<br />

x = 1 2 ln(3 − y2 ). Interchange x and y: y = 1 2 ln(3 − x2 ).Sof −1 (x) = 1 2 ln(3 − x2 ). For the domain of f −1 ,wemust<br />

have 3 − x 2 > 0 ⇒ x 2 < 3 ⇒ |x| < √ 3 ⇒ − √ 3

F.<br />

TX.10<br />

32 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

63. (a) In general, tan(arctan x) =x for any real number x. Thus,tan(arctan 10) = 10.<br />

(b) sin −1 sin 7π 3<br />

<br />

=sin<br />

−1 sin π 3<br />

<br />

=sin<br />

−1 √ 3<br />

2 = π 3 since sin π 3 = √ 3<br />

2 and π 3 is in − π 2 , π 2<br />

<br />

.<br />

[Recall that 7π 3<br />

= π 3<br />

+2π and the sine function is periodic with period 2π.]<br />

65. Let y =sin −1 x.Then− π 2 ≤ y ≤ π 2<br />

⇒ cos y ≥ 0,socos(sin −1 x)=cosy = 1 − sin 2 y = √ 1 − x 2 .<br />

67. Let y =tan −1 x.Thentan y = x, so from the triangle we see that<br />

sin(tan −1 x)=siny =<br />

x<br />

√<br />

1+x<br />

2 .<br />

69. The graph of sin −1 x is the reflection of the graph of<br />

sin x about the line y = x.<br />

71. g(x) =sin −1 (3x +1).<br />

Domain (g) ={x | −1 ≤ 3x +1≤ 1} = {x | −2 ≤ 3x ≤ 0} = x | − 2 3 ≤ x ≤ 0 = − 2 3 , 0 .<br />

Range (g) = y | − π 2 ≤ y ≤ π 2<br />

=<br />

−<br />

π<br />

2 , π 2<br />

.<br />

73. (a) If the point (x, y) is on the graph of y = f(x), then the point (x − c, y) is that point shifted c units to the left. Since f is<br />

1-1, the point (y, x) is on the graph of y = f −1 (x) and the point corresponding to (x − c, y) on the graph of f is<br />

(y, x − c) on the graph of f −1 . Thus, the curve’s reflection is shifted down the same number of units as the curve itself is<br />

shiftedtotheleft.Soanexpressionfortheinversefunctionisg −1 (x) =f −1 (x) − c.<br />

(b) If we compress (or stretch) a curve horizontally, the curve’s reflection in the line y = x is compressed (or stretched)<br />

vertically by the same factor. Using this geometric principle, we see that the inverse of h(x) =f(cx) canbeexpressedas<br />

h −1 (x) =(1/c) f −1 (x).

F.<br />

TX.10<br />

CHAPTER 1 REVIEW ¤ 33<br />

1 Review<br />

1. (a) A function f is a rule that assigns to each element x in a set A exactly one element, called f(x),inasetB. ThesetA is<br />

called the domain of the function. The range of f is the set of all possible values of f(x) as x varies throughout the<br />

domain.<br />

(b) If f is a function with domain A,thenitsgraph is the set of ordered pairs {(x, f(x)) | x ∈ A}.<br />

(c) Use the Vertical Line Test on page 16.<br />

2. The four ways to represent a function are: verbally, numerically, visually, and algebraically. An example of each is given<br />

below.<br />

Verbally: An assignment of students to chairs in a classroom (a description in words)<br />

Numerically: A tax table that assigns an amount of tax to an income (a table of values)<br />

Visually: A graphical history of the Dow Jones average (a graph)<br />

Algebraically: A relationship between distance, rate, and time: d = rt (an explicit formula)<br />

3. (a) An even function f satisfies f(−x) =f(x) for every number x in its domain. It is symmetric with respect to the y-axis.<br />

(b) An odd function g satisfies g(−x) =−g(x) for every number x in its domain. It is symmetric with respect to the origin.<br />

4. A function f is called increasing on an interval I if f(x 1)

F.<br />

TX.10<br />

34 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

(c) (d) (e)<br />

(f ) (g) (h)<br />

9. (a) The domain of f + g is the intersection of the domain of f and the domain of g;thatis,A ∩ B.<br />

(b) The domain of fg is also A ∩ B.<br />

(c) The domain of f/g must exclude values of x that make g equal to 0;thatis,{x ∈ A ∩ B | g(x) 6= 0}.<br />

10. Given two functions f and g,thecomposite function f ◦ g is defined by (f ◦ g)(x) =f(g (x)). The domain of f ◦ g is the<br />

set of all x in the domain of g such that g(x) is in the domain of f.<br />

11. (a) If the graph of f is shifted 2 units upward, its equation becomes y = f(x)+2.<br />

(b) If the graph of f is shifted 2 units downward, its equation becomes y = f(x) − 2.<br />

(c) If the graph of f is shifted 2 units to the right, its equation becomes y = f(x − 2).<br />

(d) If the graph of f is shifted 2 units to the left, its equation becomes y = f(x +2).<br />

(e) If the graph of f is reflected about the x-axis, its equation becomes y = −f(x).<br />

(f ) If the graph of f is reflected about the y-axis, its equation becomes y = f(−x).<br />

(g) If the graph of f isstretchedverticallybyafactorof2, its equation becomes y =2f(x).<br />

(h) If the graph of f is shrunk vertically by a factor of 2, its equation becomes y = 1 2 f(x).<br />

(i) If the graph of f is stretched horizontally by a factor of 2, its equation becomes y = f 1<br />

2 x .<br />

(j) If the graph of f is shrunk horizontally by a factor of 2, its equation becomes y = f(2x).<br />

12. (a) A function f is called a one-to-one function if it never takes on the same value twice; that is, if f(x 1 ) 6=f(x 2 ) whenever<br />

x 1 6=x 2 .(Or,f is 1-1 if each output corresponds to only one input.)<br />

once.<br />

Use the Horizontal Line Test: A function is one-to-one if and only if no horizontal line intersects its graph more than<br />

(b) If f is a one-to-one function with domain A and range B,thenitsinverse function f −1 has domain B and range A and is<br />

defined by<br />

f −1 (y) =x ⇔ f(x) =y<br />

for any y in B. The graph of f −1 is obtained by reflecting the graph of f about the line y = x.

F.<br />

TX.10<br />

CHAPTER 1 REVIEW ¤ 35<br />

13. (a)Theinversesinefunctionf(x) =sin −1 x is defined as follows:<br />

sin −1 x = y ⇔ sin y = x and − π 2 ≤ y ≤ π 2<br />

Its domain is −1 ≤ x ≤ 1 and its range is − π 2 ≤ y ≤ π 2 .<br />

(b) The inverse cosine function f(x) =cos −1 x is defined as follows:<br />

cos −1 x = y ⇔ cos y = x and 0 ≤ y ≤ π<br />

Its domain is −1 ≤ x ≤ 1 and its range is 0 ≤ y ≤ π.<br />

(c) The inverse tangent function f(x) =tan −1 x is defined as follows:<br />

tan −1 x = y ⇔ tan y = x and − π 2

F.<br />

TX.10<br />

36 ¤ CHAPTER 1 FUNCTIONS AND MODELS<br />

3. f(x) =x 2 − 2x +3,sof(a + h) =(a + h) 2 − 2(a + h)+3=a 2 +2ah + h 2 − 2a − 2h +3,and<br />

f(a + h) − f(a)<br />

h<br />

= (a2 +2ah + h 2 − 2a − 2h +3)− (a 2 − 2a +3)<br />

h<br />

=<br />

h(2a + h − 2)<br />

h<br />

=2a + h − 2.<br />

5. f(x) =2/(3x − 1). Domain: 3x − 1 6= 0 ⇒ 3x 6= 1 ⇒ x 6= 1 . D = <br />

−∞, 1 3 3 ∪ 1 , ∞ 3<br />

Range: all reals except 0 (y =0is the horizontal asymptote for f.) R =(−∞, 0) ∪ (0, ∞)<br />

7. h(x) =ln(x +6). Domain: x +6> 0 ⇒ x>−6. D =(−6, ∞)<br />

Range: x +6> 0,soln(x +6)takes on all real numbers and, hence, the range is R.<br />

R =(−∞, ∞)<br />

9. (a)Toobtainthegraphofy = f(x)+8,weshiftthegraphofy = f(x) up 8 units.<br />

(b)Toobtainthegraphofy = f(x +8),weshiftthegraphofy = f(x) left 8 units.<br />

(c) To obtain the graph of y =1+2f(x), we stretch the graph of y = f(x) vertically by a factor of 2, and then shift the<br />

resulting graph 1 unit upward.<br />

(d)Toobtainthegraphofy = f(x − 2) − 2, we shift the graph of y = f(x) right 2 units (for the “−2” insidethe<br />

parentheses), and then shift the resulting graph 2 units downward.<br />

(e)Toobtainthegraphofy = −f(x),wereflect the graph of y = f(x) about the x-axis.<br />

(f)Toobtainthegraphofy = f −1 (x),wereflect the graph of y = f(x) about the line y = x (assuming f is one–to-one).<br />

11. y = − sin 2x: Start with the graph of y =sinx, compress horizontally by a factor of 2,andreflect about the x-axis.<br />

13. y = 1 (1 + 2 ex ):<br />

Startwiththegraphofy = e x ,<br />

shift 1 unit upward, and compress<br />

vertically by a factor of 2.<br />

15. f(x) = 1<br />

x +2 :<br />

Startwiththegraphoff(x) =1/x<br />

and shift 2 units to the left.

F.<br />

TX.10<br />

CHAPTER 1 REVIEW ¤ 37<br />

17. (a) The terms of f are a mixture of odd and even powers of x,sof is neither even nor odd.<br />

(b) The terms of f are all odd powers of x,sof is odd.<br />

(c) f(−x) =e −(−x)2 = e −x2 = f(x),sof is even.<br />

(d) f(−x) =1+sin(−x) =1− sin x. Nowf(−x) 6=f(x) and f(−x) 6=−f(x),sof is neither even nor odd.<br />

19. f(x) =lnx, D =(0, ∞); g(x) =x 2 − 9, D = R.<br />

(a) (f ◦ g)(x) =f(g(x)) = f(x 2 − 9) = ln(x 2 − 9).<br />

Domain: x 2 − 9 > 0 ⇒ x 2 > 9 ⇒ |x| > 3 ⇒ x ∈ (−∞, −3) ∪ (3, ∞)<br />

(b) (g ◦ f)(x) =g(f(x)) = g(ln x) =(lnx) 2 − 9. Domain: x>0,or(0, ∞)<br />

(c) (f ◦ f)(x) =f(f(x)) = f(ln x) =ln(lnx). Domain: ln x>0 ⇒ x>e 0 =1,or(1, ∞)<br />

(d) (g ◦ g)(x) =g(g(x)) = g(x 2 − 9) = (x 2 − 9) 2 − 9. Domain: x ∈ R,or(−∞, ∞)<br />

21. Many models appear to be plausible. Your choice depends on whether you<br />

think medical advances will keep increasing life expectancy, or if there is<br />

bound to be a natural leveling-off of life expectancy. A linear model,<br />

y =0.2493x − 423.4818, gives us an estimate of 77.6 years for the<br />

year 2010.<br />

23. We need to know the value of x such that f(x) =2x +lnx =2.Sincex =1gives us y =2, f −1 (2) = 1.<br />

25. (a) e 2ln3 = e ln 3 2<br />

=3 2 =9<br />

(b) log 10 25 + log 10 4=log 10 (25 · 4) = log 10 100 = log 10 10 2 =2<br />

(c) tan <br />

arcsin 1 2 =tan<br />

π<br />

= √ 1 6 3<br />

(d) Let θ =cos −1 4 5 ,socos θ = 4 5 .Thensin cos −1 4 5<br />

<br />

=sinθ =<br />

√<br />

1 − cos2 θ =<br />

<br />

1 − <br />

4 2 9<br />

5<br />

= = 3 . 25 5<br />

27. (a) The population would reach 900 in about 4.4 years.<br />

100,000<br />

(b) P =<br />

100 + 900e ⇒ 100P −t +900Pe−t = 100,000 ⇒ 900Pe −t = 100,000 − 100P ⇒<br />

<br />

<br />

e −t 100,000 − 100P<br />

1000 − P<br />

1000 − P<br />

= ⇒ −t =ln<br />

⇒ t = − ln<br />

,orln<br />

900P<br />

9P<br />

9P<br />

required for the population to reach a given number P .<br />

9 · 900<br />

(c) P =900 ⇒ t =ln<br />

=ln81≈ 4.4 years, as in part (a).<br />

1000 − 900<br />

9P<br />

1000 − P<br />

<br />

;thisisthetime

F.<br />

TX.10

F.<br />

TX.10<br />

PRINCIPLES OF PROBLEM SOLVING<br />

1. By using the area formula for a triangle, 1 (base)(height), in two ways, we see that<br />

2<br />

1<br />

(4) (y) = 1 (h)(a),soa = 4y 2 2<br />

h .Since42 + y 2 = h 2 , y = √ h 2 − 16,and<br />

a = 4√ h 2 − 16<br />

.<br />

h<br />

3. |2x − 1| =<br />

<br />

2x − 1 if x ≥<br />

1<br />

2<br />

1 − 2x if x< 1 2<br />

and |x +5| =<br />

<br />

x +5<br />

−x − 5<br />

if x ≥−5<br />

if x

F.<br />

TX.10<br />

40 ¤ CHAPTER 1 PRINCIPLES OF PROBLEM SOLVING<br />

9. |x| + |y| ≤ 1. The boundary of the region has equation |x| + |y| =1. In quadrants<br />

I,II,III,andIV,thisbecomesthelinesx + y =1, −x + y =1, −x − y =1,and<br />

x − y =1respectively.<br />

11. (log 2 3)(log 3 4)(log 4 5) ···(log 31 32) =<br />

ln 3<br />

ln 2<br />

ln 4<br />

ln 3<br />

ln 5<br />

···<br />

ln 4<br />

ln 32 ln 32 ln 25<br />

= =<br />

ln 31 ln 2 ln 2 = 5ln2<br />

ln 2 =5<br />

13. ln x 2 − 2x − 2 ≤ 0 ⇒ x 2 − 2x − 2 ≤ e 0 =1 ⇒ x 2 − 2x − 3 ≤ 0 ⇒ (x − 3)(x +1)≤ 0 ⇒ x ∈ [−1, 3].<br />

Since the argument must be positive, x 2 − 2x − 2 > 0 ⇒ x − 1 − √ 3 x − 1+ √ 3 > 0 ⇒<br />

x ∈ −∞, 1 − √ 3 ∪ 1+ √ 3, ∞ . The intersection of these intervals is −1, 1 − √ 3 ∪ 1+ √ 3, 3 .<br />

15. Let d be the distance traveled on each half of the trip. Let t 1 and t 2 be the times taken for the first and second halves of the trip.<br />

For the first half of the trip we have t 1 = d/30 and for the second half we have t 2 = d/60. Thus, the average speed for the<br />

entire trip is<br />

is 40 mi/h.<br />

total distance<br />

total time<br />

= 2d<br />

t 1 + t 2<br />

=<br />

2d<br />

d<br />

30 + d 60<br />

17. Let S n be the statement that 7 n − 1 is divisible by 6.<br />

• S 1 is true because 7 1 − 1=6is divisible by 6.<br />

· 60<br />

60 = 120d<br />

2d + d = 120d =40. The average speed for the entire trip<br />

3d<br />

• Assume S k is true, that is, 7 k − 1 is divisible by 6. Inotherwords,7 k − 1=6m for some positive integer m. Then<br />

7 k+1 − 1=7 k · 7 − 1=(6m +1)· 7 − 1=42m +6=6(7m +1), which is divisible by 6,soS k+1 is true.<br />

• Therefore, by mathematical induction, 7 n − 1 is divisible by 6 for every positive integer n.<br />

19. f 0 (x) =x 2 and f n+1 (x) =f 0 (f n (x)) for n =0, 1, 2,....<br />

f 1 (x) =f 0 (f 0 (x)) = f 0<br />

<br />

x<br />

2 = x 22 = x 4 , f 2 (x) =f 0 (f 1 (x)) = f 0 (x 4 )=(x 4 ) 2 = x 8 ,<br />

f 3(x) =f 0(f 2(x)) = f 0(x 8 )=(x 8 ) 2 = x 16 , .... Thus, a general formula is f n(x) =x 2n+1 .

F.<br />

TX.10<br />

2 LIMITS AND DERIVATIVES<br />

2.1 The Tangent and Velocity Problems<br />

(b) Using the values of t that correspond to the points<br />

1. (a) Using P (15, 250), we construct the following table:<br />

30 (30, 0)<br />

0−250<br />

30−15 15<br />

closest to P (t =10and t =20), we have<br />

t Q slope = m PQ<br />

5 (5, 694)<br />

694−250<br />

= − 444 −38.8+(−27.8)<br />

5−15 10<br />

= −33.3<br />

2<br />

444−250<br />

10 (10, 444)<br />

= − 194 = −38.8<br />

10−15 5<br />

111−250<br />

20 (20, 111)<br />

20−15<br />

= − 139<br />

5<br />

= −27.8<br />

28−250<br />

25 (25, 28)<br />

25−15 10<br />

(c) From the graph, we can estimate the slope of the<br />

tangent line at P to be −300<br />

9<br />

= −33.3.<br />

3. (a)<br />

x Q m PQ<br />

(i) 0.5 (0.5, 0.333333) 0.333333<br />

(ii) 0.9 (0.9, 0.473684) 0.263158<br />

(iii) 0.99 (0.99, 0.497487) 0.251256<br />

(iv) 0.999 (0.999, 0.499750) 0.250125<br />

(v) 1.5 (1.5, 0.6) 0.2<br />

(vi) 1.1 (1.1, 0.523810) 0.238095<br />

(vii) 1.01 (1.01, 0.502488) 0.248756<br />

(viii) 1.001 (1.001, 0.500250) 0.249875<br />

(b) The slope appears to be 1 4 .<br />

(c) y − 1 2 = 1 4 (x − 1) or y = 1 4 x + 1 4 .<br />

5. (a) y = y(t) =40t − 16t 2 .Att =2, y =40(2)− 16(2) 2 =16. The average velocity between times 2 and 2+h is<br />

<br />

y(2 + h) − y(2)<br />

40(2 + h) − 16(2 + h)<br />

2<br />

− 16 −24h − 16h2<br />

v ave = =<br />

= = −24 − 16h, ifh 6= 0.<br />

(2 + h) − 2<br />

h<br />

h<br />

(i) [2, 2.5]: h =0.5, v ave = −32 ft/s<br />

(ii) [2, 2.1]: h =0.1, v ave = −25.6 ft/s<br />

(iii) [2, 2.05]: h =0.05, v ave = −24.8 ft/s<br />

(iv) [2, 2.01]: h =0.01, v ave = −24.16 ft/s<br />

(b) The instantaneous velocity when t =2(h approaches 0)is−24 ft/s.<br />

41

F.<br />

42 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

7. (a) (i) On the interval [1, 3], v ave =<br />

(ii) On the interval [2, 3], v ave =<br />

(iii) On the interval [3, 5], v ave =<br />

(iv) On the interval [3, 4], v ave =<br />

s(3) − s(1)<br />

3 − 1<br />

s(3) − s(2)<br />

3 − 2<br />

s(5) − s(3)<br />

5 − 3<br />

s(4) − s(3)<br />

4 − 3<br />

=<br />

=<br />

=<br />

=<br />

10.7 − 1.4<br />

2<br />

10.7 − 5.1<br />

1<br />

25.8 − 10.7<br />

2<br />

17.7 − 10.7<br />

1<br />

= 9.3<br />

2<br />

=5.6 m/s.<br />

= 15.1<br />

2<br />

=7m/s.<br />

=4.65 m/s.<br />

=7.55 m/s.<br />

(b)<br />

Using the points (2, 4) and (5, 23) from the approximate tangent<br />

line, the instantaneous velocity at t =3is about 23 − 4<br />

5 − 2<br />

≈ 6.3 m/s.<br />

9. (a) For the curve y =sin(10π/x) and the point P (1, 0):<br />

x Q m PQ<br />

2 (2, 0) 0<br />

1.5 (1.5, 0.8660) 1.7321<br />

1.4 (1.4, −0.4339) −1.0847<br />

1.3 (1.3, −0.8230) −2.7433<br />

1.2 (1.2, 0.8660) 4.3301<br />

1.1 (1.1, −0.2817) −2.8173<br />

x Q m PQ<br />

0.5 (0.5, 0) 0<br />

0.6 (0.6, 0.8660) −2.1651<br />

0.7 (0.7, 0.7818) −2.6061<br />

0.8 (0.8, 1) −5<br />

0.9 (0.9, −0.3420) 3.4202<br />

As x approaches 1, the slopes do not appear to be approaching any particular value.<br />

(b)<br />

We see that problems with estimation are caused by the frequent<br />

oscillations of the graph. The tangent is so steep at P that we need to<br />

take x-values much closer to 1 in order to get accurate estimates of<br />

its slope.<br />

(c) If we choose x =1.001, then the point Q is (1.001, −0.0314) and m PQ ≈−31.3794. Ifx =0.999, thenQ is<br />

(0.999, 0.0314) and m PQ = −31.4422. The average of these slopes is −31.4108. So we estimate that the slope of the<br />

tangent line at P is about −31.4.

F.<br />

TX.10<br />

SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 43<br />

2.2 The Limit of a Function<br />

1. As x approaches 2, f(x) approaches 5. [Or,thevaluesoff(x) can be made as close to 5 as we like by taking x sufficiently<br />

close to 2 (but x 6= 2).] Yes, the graph could have a hole at (2, 5) and be definedsuchthatf(2) = 3.<br />

3. (a) lim f(x) =∞ means that the values of f(x) canbemadearbitrarilylarge(aslargeasweplease)bytakingx<br />

x→−3<br />

(b)<br />

sufficiently close to −3 (but not equal to −3).<br />

lim f(x) =−∞ means that the values of f(x) can be made arbitrarily large negative by taking x sufficiently close to 4<br />

x→4 +<br />

through values larger than 4.<br />

5. (a) f(x) approaches 2 as x approaches 1 from the left, so lim f(x) =2.<br />

x→1− (b) f(x) approaches 3 as x approaches 1 from the right, so lim f(x) =3.<br />

x→1 +<br />

(c) lim<br />

x→1<br />

f(x) does not exist because the limits in part (a) and part (b) are not equal.<br />

(d) f(x) approaches 4 as x approaches 5 from the left and from the right, so lim<br />

x→5<br />

f(x) =4.<br />

(e) f(5) is not defined, so it doesn’t exist.<br />

7. (a) lim g(t) =−1<br />

t→0− (b) lim g(t) =−2<br />

+<br />

(c) lim<br />

t→0<br />

g(t) does not exist because the limits in part (a) and part (b) are not equal.<br />

(d) lim g(t) =2<br />

t→2− t→0<br />

(e) lim g(t) =0<br />

+<br />

(f ) lim<br />

t→2<br />

g(t) does not exist because the limits in part (d) and part (e) are not equal.<br />

(g) g(2) = 1<br />

t→2<br />

(h) lim<br />

t→4<br />

g(t) =3<br />

9. (a) lim f(x) =−∞ (b) lim f(x) =∞<br />

x→−7 x→−3<br />

(d)<br />

lim f(x) =−∞<br />

x→6− (e) lim f(x) =∞<br />

+<br />

x→6<br />

(f ) The equations of the vertical asymptotes are x = −7, x = −3, x =0,andx =6.<br />

(c) lim f(x) =∞<br />

x→0<br />

11. (a) lim f(x) =1<br />

x→0− (b)<br />

lim f(x) =0<br />

x→0 +<br />

(c) lim<br />

x→0<br />

f(x) does not exist because the limits<br />

in part (a) and part (b) are not equal.<br />

13. lim f(x) =2, lim<br />

x→1− x→1<br />

f(x) =−2, f(1) = 2 15. lim f(x) =4, lim<br />

+ +<br />

x→3<br />

f(3) = 3, f(−2) = 1<br />

f(x) =2, lim f(x) =2,<br />

x→3− x→−2

F.<br />

44 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

17. For f(x) = x2 − 2x<br />

x 2 − x − 2 :<br />

x<br />

f(x)<br />

2.5 0.714286<br />

2.1 0.677419<br />

2.05 0.672131<br />

2.01 0.667774<br />

2.005 0.667221<br />

2.001 0.666778<br />

x<br />

f(x)<br />

1.9 0.655172<br />

1.95 0.661017<br />

1.99 0.665552<br />

1.995 0.666110<br />

1.999 0.666556<br />

x 2 − 2x<br />

It appears that lim<br />

x→2 x 2 − x − 2 =0.¯6 = 2 . 3<br />

TX.10<br />

19. For f(x) = ex − 1 − x<br />

x 2 :<br />

x<br />

f(x)<br />

1 0.718282<br />

0.5 0.594885<br />

0.1 0.517092<br />

0.05 0.508439<br />

0.01 0.501671<br />

x<br />

f(x)<br />

−1 0.367879<br />

−0.5 0.426123<br />

−0.1 0.483742<br />

−0.05 0.491770<br />

−0.01 0.498337<br />

It appears that lim<br />

x→0<br />

e x − 1 − x<br />

x 2 =0.5 = 1 2 .<br />

√ x +4− 2<br />

21. For f(x) =<br />

:<br />

x<br />

x<br />

f(x)<br />

1 0.236068<br />

0.5 0.242641<br />

0.1 0.248457<br />

0.05 0.249224<br />

0.01 0.249844<br />

It appears that lim<br />

x→0<br />

√ x +4− 2<br />

x<br />

23. For f(x) = x6 − 1<br />

x 10 − 1 :<br />

x f(x)<br />

x f(x)<br />

−1 0.267949<br />

0.5 0.985337<br />

−0.5 0.258343<br />

0.9 0.719397<br />

−0.1 0.251582<br />

0.95 0.660186<br />

−0.05 0.250786<br />

0.99 0.612018<br />

−0.01 0.250156<br />

0.999 0.601200<br />

It appears that lim<br />

x→1<br />

x 6 − 1<br />

x 10 − 1 =0.6 = 3 5 .<br />

x<br />

f(x)<br />

1.5 0.183369<br />

1.1 0.484119<br />

1.05 0.540783<br />

1.01 0.588022<br />

1.001 0.598800<br />

25. lim<br />

x→−3 + x +2<br />

x +3 = −∞ since the numerator is negative and the denominator approaches 0 from the positive side as x →−3+ .<br />

2 − x<br />

27. lim = ∞ since the numerator is positive and the denominator approaches 0 through positive values as x → 1.<br />

x→1 (x − 1)<br />

2<br />

29. Let t = x 2 − 9. Thenasx → 3 + , t → 0 + ,and lim ln(x2 − 9) = lim ln t = −∞ by (3).<br />

x→3 + t→0 +<br />

31. lim<br />

x→2π − x csc x =<br />

lim<br />

x→2π −<br />

x<br />

= −∞ since the numerator is positive and the denominator approaches 0 through negative<br />

sin x<br />

values as x → 2π − .<br />

33. (a) f(x) = 1<br />

x 3 − 1 .<br />

lim f(x) =−∞ and lim x→1− x→1 + 0.99 −33.7<br />

x f(x)<br />

0.5 −1.14<br />

From these calculations, it seems that<br />

0.9 −3.69<br />

0.999 −333.7<br />

0.9999 −3333.7<br />

0.99999 −33,333.7<br />

x f(x)<br />

1.5 0.42<br />

1.1 3.02<br />

1.01 33.0<br />

1.001 333.0<br />

1.0001 3333.0<br />

1.00001 33,333.3

F.<br />

TX.10<br />

SECTION 2.2 THE LIMIT OF A FUNCTION ¤ 45<br />

(b) If x is slightly smaller than 1,thenx 3 − 1 will be a negative number close to 0,andthereciprocalofx 3 − 1,thatis,f(x),<br />

will be a negative number with large absolute value. So lim f(x) =−∞.<br />

x→1− If x is slightly larger than 1,thenx 3 − 1 will be a small positive number, and its reciprocal, f(x), will be a large positive<br />

number. So lim f(x) =∞.<br />

x→1 +<br />

(c) It appears from the graph of f that<br />

lim<br />

x→1<br />

f(x) =−∞ and lim f(x) =∞.<br />

− +<br />

x→1<br />

35. (a) Let h(x) =(1+x) 1/x .<br />

(b)<br />

x<br />

h(x)<br />

−0.001 2.71964<br />

−0.0001 2.71842<br />

−0.00001 2.71830<br />

−0.000001 2.71828<br />

0.000001 2.71828<br />

0.00001 2.71827<br />

0.0001 2.71815<br />

0.001 2.71692<br />

It appears that lim<br />

x→0<br />

(1 + x) 1/x ≈ 2.71828, which is approximately e.<br />

In Section 3.6 we will see that the value of the limit is exactly e.<br />

37. For f(x) =x 2 − (2 x /1000):<br />

(a)<br />

x f(x)<br />

1 0.998000<br />

0.8 0.638259<br />

0.6 0.358484<br />

0.4 0.158680<br />

0.2 0.038851<br />

0.1 0.008928<br />

0.05 0.001465<br />

It appears that lim<br />

x→0<br />

f(x) =0.<br />

(b)<br />

x<br />

f(x)<br />

0.04 0.000572<br />

0.02 −0.000614<br />

0.01 −0.000907<br />

0.005 −0.000978<br />

0.003 −0.000993<br />

0.001 −0.001000<br />

It appears that lim<br />

x→0<br />

f(x) =−0.001.

F.<br />

46 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

39. No matter how many times we zoom in toward the origin, the graphs of f(x) =sin(π/x) appear to consist of almost-vertical<br />

lines. This indicates more and more frequent oscillations as x → 0.<br />

41. There appear to be vertical asymptotes of the curve y =tan(2sinx) at x ≈ ±0.90<br />

and x ≈ ±2.24. Tofind the exact equations of these asymptotes, we note that the<br />

graph of the tangent function has vertical asymptotes at x = π + πn. Thus,we<br />

2<br />

must have 2sinx = π + πn, or equivalently, sin x = π + π n.Since<br />

2 4 2<br />

−1 ≤ sin x ≤ 1,wemusthavesin x = ± π and so x = ± 4 sin−1 π (corresponding<br />

4<br />

to x ≈ ±0.90). Just as 150 ◦ is the reference angle for 30 ◦ , π − sin −1 π 4<br />

is the<br />

reference angle for sin −1 π .Sox = ± <br />

4<br />

π − sin −1 π 4 are also equations of<br />

vertical asymptotes (corresponding to x ≈ ±2.24).<br />

2.3 Calculating Limits Using the Limit Laws<br />

1. (a) lim<br />

x→2<br />

[f(x)+5g(x)] = lim<br />

x→2<br />

f(x) + lim<br />

x→2<br />

[5g(x)] [Limit Law 1]<br />

=lim<br />

x→2<br />

f(x)+5lim<br />

x→2<br />

g(x) [Limit Law 3]<br />

=4+5(−2) = −6<br />

3<br />

(b) lim [g(x)] 3 = lim g(x) [Limit Law 6]<br />

x→2 x→2<br />

=(−2) 3 = −8<br />

(c) lim<br />

x→2<br />

<br />

f(x)=<br />

<br />

lim<br />

x→2<br />

f(x) [Limit Law 11]<br />

= √ 4=2

F.<br />

TX.10<br />

SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 47<br />

3f(x)<br />

lim [3f(x)]<br />

(d) lim<br />

x→2 g(x) = x→2<br />

lim g(x) [Limit Law 5]<br />

x→2<br />

=<br />

3lim<br />

x→2<br />

f(x)<br />

lim<br />

x→2 g(x) [Limit Law 3]<br />

= 3(4)<br />

−2 = −6<br />

g(x)<br />

(e) Because the limit of the denominator is 0, we can’t use Limit Law 5. The given limit, lim , does not exist because the<br />

x→2 h(x)<br />

denominator approaches 0 while the numerator approaches a nonzero number.<br />

g(x) h(x)<br />

lim [g(x) h(x)]<br />

x→2<br />

(f ) lim =<br />

x→2 f(x) lim f(x) [Limit Law 5]<br />

x→2<br />

=<br />

lim g(x) · lim h(x)<br />

x→2 x→2<br />

lim f(x) [Limit Law 4]<br />

x→2<br />

= −2 · 0<br />

4<br />

=0<br />

3. lim<br />

x→−2 (3x4 +2x 2 − x +1)= lim<br />

x→−2 3x4 + lim<br />

x→−2 2x2 − lim x + lim 1 [Limit Laws 1 and 2]<br />

x→−2 x→−2<br />

= 3 lim<br />

x→−2 x4 +2 lim<br />

x→−2 x2 − lim<br />

x→−2 x + lim<br />

x→−2 1 [3]<br />

=3(−2) 4 +2(−2) 2 − (−2) + (1)<br />

[9,8,and7]<br />

=48+8+2+1=59<br />

5. lim (1 + 3√ x )(2− 6x 2 + x 3 ) = lim (1 + 3√ x ) · lim(2 − 6x 2 + x 3 ) [Limit Law 4]<br />

x→8 x→8 x→8<br />

<br />

<br />

<br />

= lim 1 + lim 3√ x · lim 2 − 6lim<br />

x→8 x→8<br />

x→8 x→8 x2 +limx 3 [1,2,and3]<br />

x→8<br />

7. lim<br />

x→1<br />

<br />

3<br />

1+3x<br />

=<br />

1+4x 2 +3x 4<br />

=<br />

<br />

lim<br />

x→1<br />

<br />

= 1+ 3√ 8 · 2<br />

− 6 · 8 2 +8 3 [7, 10, 9]<br />

= (3)(130) = 390<br />

3<br />

1+3x<br />

[6]<br />

1+4x 2 +3x 4 3<br />

[5]<br />

lim (1 + 3x)<br />

x→1<br />

lim (1 +<br />

x→1 4x2 +3x 4 )<br />

<br />

lim 1+3limx<br />

3<br />

x→1 x→1<br />

=<br />

lim 1+4lim<br />

[2, 1, and 3]<br />

x→1 x→1 x2 +3limx 4<br />

x→1<br />

<br />

3 3 3<br />

1 + 3(1)<br />

4 1<br />

=<br />

= = = 1 [7, 8, and 9]<br />

1 + 4(1) 2 +3(1) 4 8 2 8<br />

9. lim<br />

x→4 − √<br />

16 − x2 = lim<br />

x→4 − (16 − x2 ) [11]<br />

= lim 16 − lim x2 [2]<br />

x→4− x→4 −<br />

= 16 − (4) 2 =0 [7 and 9]

F.<br />

48 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

x 2 + x − 6 (x +3)(x − 2)<br />

11. lim<br />

=lim<br />

=lim(x +3)=2+3=5<br />

x→2 x − 2 x→2 x − 2<br />

x→2<br />

x 2 − x +6<br />

13. lim<br />

does not exist since x − 2 → 0 but x 2 − x +6→ 8 as x → 2.<br />

x→2 x − 2<br />

15. lim<br />

t→−3<br />

t 2 − 9<br />

2t 2 +7t +3 = lim<br />

t→−3<br />

(t +3)(t − 3)<br />

(2t +1)(t +3) = lim<br />

t→−3<br />

t − 3<br />

2t +1 = −3 − 3<br />

2(−3) + 1 = −6<br />

−5 = 6 5<br />

(4 + h) 2 − 16 (16 + 8h + h 2 ) − 16 8h + h 2 h(8 + h)<br />

17. lim<br />

=lim<br />

=lim =lim =lim(8 + h) =8+0=8<br />

h→0 h<br />

h→0 h<br />

h→0 h h→0 h h→0<br />

19. Bytheformulaforthesumofcubes,wehave<br />

lim<br />

x→−2<br />

x +2<br />

x 3 +8 = lim<br />

x→−2<br />

x +2<br />

(x +2)(x 2 − 2x +4) = lim<br />

x→−2<br />

1<br />

x 2 − 2x +4 = 1<br />

4+4+4 = 1<br />

12 .<br />

√ √ <br />

9 − t 3+ t 3 − t<br />

21. lim<br />

t→9 3 − √ t =lim<br />

t→9 3 − √ √ √<br />

=lim 3+ t =3+ 9=6<br />

t<br />

t→9<br />

√ √ √ x +2− 3 x +2− 3 x +2+3<br />

(x +2)− 9<br />

23. lim<br />

=lim<br />

· √ = lim<br />

x→7 x − 7 x→7 x − 7 x +2+3 x→7 (x − 7) √ x +2+3 <br />

25. lim<br />

x→−4<br />

27. lim<br />

x→16<br />

29. lim<br />

t→0<br />

<br />

=lim<br />

x→7<br />

x − 7<br />

(x − 7) √ x +2+3 = lim<br />

x→7<br />

1<br />

4 + 1 x +4<br />

x<br />

4+x = lim 4x<br />

x→−4 4+x = lim<br />

x→−4<br />

4 − √ x<br />

16x − x 2 = lim<br />

x→16<br />

1<br />

t √ 1+t − 1 <br />

t<br />

x +4<br />

4x(4 + x) = lim<br />

x→−4<br />

(4 − √ x )(4 + √ x )<br />

(16x − x 2 )(4 + √ x ) = lim<br />

x→16<br />

1<br />

√ x +2+3<br />

=<br />

1<br />

√<br />

9+3<br />

= 1 6<br />

1<br />

4x = 1<br />

4(−4) = − 1 16<br />

16 − x<br />

x(16 − x)(4 + √ x )<br />

1<br />

= lim<br />

x→16 x(4 + √ x ) = 1<br />

16 4+ √ 16 = 1<br />

16(8) = 1<br />

128<br />

1 − √ √ √ <br />

1+t 1 − 1+t 1+ 1+t<br />

=lim<br />

t→0 t √ = lim<br />

1+t t→0 t √ t +1 1+ √ 1+t =lim<br />

t→0<br />

=lim<br />

t→0<br />

−1<br />

√ 1+t<br />

1+<br />

√ 1+t<br />

=<br />

−1<br />

√ 1+0<br />

1+<br />

√ 1+0<br />

= − 1 2<br />

−t<br />

t √ 1+t 1+ √ 1+t <br />

31. (a)<br />

lim<br />

x→0<br />

x<br />

√ 1+3x − 1<br />

≈ 2 3<br />

(b)<br />

x<br />

f(x)<br />

−0.001 0.6661663<br />

−0.0001 0.6666167<br />

−0.00001 0.6666617<br />

−0.000001 0.6666662<br />

0.000001 0.6666672<br />

0.00001 0.6666717<br />

0.0001 0.6667167<br />

0.001 0.6671663<br />

The limit appears to be 2 3 .

F.<br />

(c) lim<br />

x→0<br />

<br />

TX.10 SECTION 2.3 CALCULATING LIMITS USING THE LIMIT LAWS ¤ 49<br />

√ <br />

x 1+3x +1 x √ 1+3x +1 <br />

x √ 1+3x +1 <br />

√ · √ =lim<br />

=lim<br />

1+3x − 1 1+3x +1 x→0 (1 + 3x) − 1 x→0 3x<br />

= 1 3 lim √ <br />

1+3x +1<br />

x→0<br />

= 1 3<br />

lim<br />

= 1 3<br />

lim<br />

x→0<br />

(1 + 3x)+lim<br />

x→0<br />

1<br />

x→0<br />

1+3lim<br />

x→0<br />

x +1<br />

= 1 √ <br />

1+3· 0+1<br />

3<br />

<br />

<br />

[Limit Law 3]<br />

[1 and 11]<br />

[1,3,and7]<br />

[7 and 8]<br />

= 1 3 (1 + 1) = 2 3<br />

33. Let f(x) =−x 2 , g(x) =x 2 cos 20πx and h(x) =x 2 .Then<br />

−1 ≤ cos 20πx ≤ 1 ⇒ −x 2 ≤ x 2 cos 20πx ≤ x 2 ⇒ f(x) ≤ g(x) ≤ h(x).<br />

So since lim<br />

x→0<br />

f(x) = lim<br />

x→0<br />

h(x) =0, by the Squeeze Theorem we have<br />

lim g(x) =0.<br />

x→0<br />

35. We have lim<br />

x→4<br />

(4x − 9) = 4(4) − 9=7and lim<br />

x→4<br />

<br />

x 2 − 4x +7 =4 2 − 4(4) + 7 = 7. Since4x − 9 ≤ f(x) ≤ x 2 − 4x +7<br />

for x ≥ 0, lim<br />

x→4<br />

f(x) =7by the Squeeze Theorem.<br />

<br />

37. −1 ≤ cos(2/x) ≤ 1 ⇒ −x 4 ≤ x 4 cos(2/x) ≤ x 4 .Sincelim<br />

−x<br />

4<br />

=0and lim x 4 =0,wehave<br />

x→0 x→0<br />

<br />

lim x 4 cos(2/x) =0by the Squeeze Theorem.<br />

x→0<br />

39. |x − 3| =<br />

41.<br />

<br />

x − 3 if x − 3 ≥ 0<br />

−(x − 3) if x − 3 < 0 = <br />

x − 3 if x ≥ 3<br />

3 − x if x

F.<br />

50 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

1<br />

43. Since |x| = −x for x0, lim sgn x = lim 1=1.<br />

x→0 + x→0 +<br />

(ii) Since sgn x = −1 for x

F.<br />

TX.10<br />

SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ¤ 51<br />

<br />

f(x) − 8<br />

f(x) − 8<br />

55. lim [f(x) − 8] = lim<br />

· (x − 1) =lim · lim (x − 1) = 10 · 0=0.<br />

x→1 x→1 x − 1<br />

x→1 x − 1 x→1<br />

Thus, lim<br />

x→1<br />

f(x) = lim<br />

x→1<br />

{[f(x) − 8] + 8} =lim<br />

x→1<br />

[f(x) − 8] + lim<br />

x→1<br />

8=0+8=8.<br />

f(x) − 8<br />

f(x) − 8<br />

Note: The value of lim does not affect the answer since it’s multiplied by 0. What’s important is that lim<br />

x→1 x − 1<br />

x→1 x − 1<br />

exists.<br />

57. Observe that 0 ≤ f(x) ≤ x 2 for all x,and lim<br />

x→0<br />

0=0= lim<br />

x→0<br />

x 2 . So, by the Squeeze Theorem, lim<br />

x→0<br />

f(x) =0.<br />

59. Let f(x) =H(x) and g(x) =1− H(x),whereH is the Heaviside function definedinExercise1.3.57.<br />

Thus, either f or g is 0 for any value of x. Thenlim<br />

x→0<br />

f(x) and lim<br />

x→0<br />

g(x) do not exist, but lim<br />

x→0<br />

[f(x)g(x)] = lim<br />

x→0<br />

0=0.<br />

61. Since the denominator approaches 0 as x →−2, the limit will exist only if the numerator also approaches<br />

<br />

0 as x →−2. In order for this to happen, we need lim 3x 2 + ax + a +3 =0<br />

x→−2<br />

⇔<br />

3(−2) 2 + a(−2) + a +3=0 ⇔ 12 − 2a + a +3=0 ⇔ a =15.Witha =15, the limit becomes<br />

3x 2 +15x +18 3(x +2)(x +3)<br />

lim<br />

= lim<br />

x→−2 x 2 + x − 2 x→−2 (x − 1)(x +2) = lim 3(x +3)<br />

= 3(−2+3) = 3<br />

x→−2 x − 1 −2 − 1 −3 = −1.<br />

2.4 The Precise Definition of a Limit<br />

1. On the left side of x =2,weneed|x − 2| < 10<br />

− 2 = 4 . On the right side, we need |x − 2| < 10<br />

− 2 = 4 . For both of<br />

7 7 3 3<br />

these conditions to be satisfied at once, we need the more restrictive of the two to hold, that is, |x − 2| < 4 . So we can choose<br />

7<br />

δ = 4 7<br />

, or any smaller positive number.<br />

3. The leftmost question mark is the solution of √ x =1.6 and the rightmost, √ x =2.4. Sothevaluesare1.6 2 =2.56 and<br />

2.4 2 =5.76. On the left side, we need |x − 4| < |2.56 − 4| =1.44. On the right side, we need |x − 4| < |5.76 − 4| =1.76.<br />

To satisfy both conditions, we need the more restrictive condition to hold — namely, |x − 4| < 1.44. Thus, we can choose<br />

δ =1.44, or any smaller positive number.<br />

5. From the graph, we find that tan x =0.8 when x ≈ 0.675,so<br />

π<br />

− 4 δ1 ≈ 0.675 ⇒ δ1 ≈ π 4<br />

− 0.675 ≈ 0.1106. Also,tan x =1.2<br />

when x ≈ 0.876,so π 4 + δ 2 ≈ 0.876 ⇒ δ 2 =0.876 − π 4 ≈ 0.0906.<br />

Thus, we choose δ =0.0906 (or any smaller positive number) since this is<br />

the smaller of δ 1 and δ 2 .

F.<br />

52 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

7. For ε =1,thedefinition of a limit requires that we find δ such that 4+x − 3x 3 − 2 < 1 ⇔ 1 < 4+x − 3x 3 < 3<br />

whenever 0 < |x − 1|

F.<br />

TX.10<br />

SECTION 2.4 THE PRECISE DEFINITION OF A LIMIT ¤ 53<br />

15. Given ε>0, we need δ>0 such that if 0 < |x − 1|

F.<br />

54 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

29. Given ε>0,weneedδ>0 such that if 0 < |x − 2|

F.<br />

TX.10<br />

SECTION 2.5 CONTINUITY ¤ 55<br />

39. Suppose that lim f(x) =L. Givenε = 1 ,thereexistsδ>0 such that 0 < |x|

F.<br />

56 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

<br />

<br />

4<br />

11. lim f(x) = lim x +2x<br />

3 4 = lim x +2 lim<br />

x→−1 x→−1<br />

x→−1 x→−1 x3 = −1+2(−1) 34 =(−3) 4 =81=f(−1).<br />

By the definition of continuity, f is continuous at a = −1.<br />

13. For a>2,wehave<br />

2x +3<br />

lim (2x +3)<br />

lim f(x)= lim<br />

x→a x→a x − 2 = x→a<br />

lim (x − 2) [Limit Law 5]<br />

x→a<br />

2 lim x + lim 3<br />

x→a x→a<br />

=<br />

lim x − lim 2<br />

x→a x→a<br />

=<br />

2a +3<br />

a − 2<br />

= f(a)<br />

TX.10<br />

[1,2,and3]<br />

[7 and 8]<br />

Thus, f is continuous at x = a for every a in (2, ∞);thatis,f is continuous on (2, ∞).<br />

15. f(x) =ln|x − 2| is discontinuous at 2 since f(2) = ln 0 is not defined.<br />

17. f(x) =<br />

<br />

e<br />

x<br />

if x

F.<br />

TX.10<br />

SECTION 2.5 CONTINUITY ¤ 57<br />

25. By Theorem 7, the exponential function e −5t and the trigonometric function cos 2πt are continuous on (−∞, ∞).<br />

By part 4 of Theorem 4, L(t) =e −5t cos 2πt is continuous on (−∞, ∞).<br />

27. By Theorem 5, the polynomial t 4 − 1 is continuous on (−∞, ∞). By Theorem 7, ln x is continuous on its domain, (0, ∞).<br />

By Theorem 9, ln t 4 − 1 is continuous on its domain, which is<br />

<br />

t | t 4 − 1 > 0 = t | t 4 > 1 = {t ||t| > 1} =(−∞, −1) ∪ (1, ∞)<br />

29. The function y =<br />

1<br />

is discontinuous at x =0because the<br />

1+e1/x left- and right-hand limits at x =0are different.<br />

31. Because we are dealing with root functions, 5+ √ x is continuous on [0, ∞), √ x +5is continuous on [−5, ∞), sothe<br />

quotient f(x) = 5+√ x<br />

√ 5+x<br />

is continuous on [0, ∞). Sincef is continuous at x =4, lim<br />

x→4<br />

f(x) =f(4) = 7 3 .<br />

33. Because x 2 − x is continuous on R, the composite function f(x) =e x2 −x is continuous on R, so<br />

lim<br />

x→1 f(x) =f(1) = e1 − 1 = e 0 =1.<br />

<br />

x<br />

2<br />

if x

F.<br />

58 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

⎧<br />

x +2 ⎪⎨<br />

if x1<br />

TX.10<br />

f is continuous on (−∞, 0) and (1, ∞) sinceoneachoftheseintervals<br />

it is a polynomial; it is continuous on (0, 1) since it is an exponential.<br />

Now lim f(x) = lim +2)=2and lim f(x) = lim<br />

x→0− x→0−(x ex =1,sof is discontinuous at 0. Sincef(0) = 1, f is<br />

x→0 + x→0 +<br />

continuous from the right at 0. Also lim f(x) = lim ex = e and lim f(x) = lim − x) =1,sof is discontinuous<br />

x→1− x→1 − x→1 + x→1 +(2<br />

at 1. Sincef(1) = e, f is continuous from the left at 1.<br />

41. f(x) =<br />

<br />

cx 2 +2x if x

F.<br />

TX.10<br />

SECTION 2.5 CONTINUITY ¤ 59<br />

51. (a) f(x) =cosx − x 3 is continuous on the interval [0, 1], f(0) = 1 > 0,andf(1) = cos 1 − 1 ≈−0.46 < 0. Since<br />

1 > 0 > −0.46, there is a number c in (0, 1) such that f(c) =0by the Intermediate Value Theorem. Thus, there is a root<br />

of the equation cos x − x 3 =0,orcos x = x 3 , in the interval (0, 1).<br />

(b) f(0.86) ≈ 0.016 > 0 and f(0.87) ≈−0.014 < 0, so there is a root between 0.86 and 0.87, that is, in the interval<br />

(0.86, 0.87).<br />

53. (a) Let f(x) =100e −x/100 − 0.01x 2 . Then f(0) = 100 > 0 and<br />

f(100) = 100e −1 − 100 ≈−63.2 < 0. So by the Intermediate<br />

Value Theorem, there is a number c in (0, 100) such that f(c) =0.<br />

This implies that 100e −c/100 =0.01c 2 .<br />

(b) Using the intersect feature of the graphing device, we find that the<br />

root of the equation is x =70.347, correct to three decimal places.<br />

55. (⇒)Iff is continuous at a,thenbyTheorem8withg(h) =a + h,wehave<br />

<br />

<br />

lim f(a + h) =f lim (a + h) = f(a).<br />

h→0 h→0<br />

(⇐)Letε>0. Sincelim<br />

h→0<br />

f(a + h) =f(a),thereexistsδ>0 such that 0 < |h|

F.<br />

60 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

lim<br />

x→0 (−x4 )=0and lim x 4 =0, the Squeeze Theorem gives us lim(x 4 sin(1/x)) = 0,whichequalsf(0). Thus,f is<br />

x→0 x→0<br />

continuous at 0 and, hence, on (−∞, ∞).<br />

65. Define u(t) to be the monk’s distance from the monastery, as a function of time, on the first day, and define d(t) to be his<br />

distance from the monastery, as a function of time, on the second day. Let D be the distance from the monastery to the top of<br />

the mountain. From the given information we know that u(0) = 0, u(12) = D, d(0) = D and d(12) = 0. Now consider the<br />

function u − d, which is clearly continuous. We calculate that (u − d)(0) = −D and (u − d)(12) = D.Sobythe<br />

Intermediate Value Theorem, there must be some time t 0 between 0 and 12 such that (u − d)(t 0)=0 ⇔ u(t 0)=d(t 0).<br />

So at time t 0 after 7:00 AM, the monk will be at the same place on both days.<br />

2.6 Limits at Infinity; Horizontal Asymptotes<br />

1. (a) As x becomes large, the values of f(x) approach 5.<br />

(b) As x becomes large negative, the values of f(x) approach 3.<br />

3. (a) lim f(x) =∞ (b) lim f(x) =∞ (c) lim f(x) =−∞<br />

x→2 − +<br />

x→−1<br />

(d) lim f(x) =1 (e) lim f(x) =2 (f ) Vertical: x = −1, x =2; Horizontal: y =1, y =2<br />

x→∞ x→−∞<br />

x→−1<br />

5. f(0) = 0, f(1) = 1,<br />

lim f(x) =0,<br />

x→∞<br />

f is odd<br />

7. lim f(x) =−∞, lim f(x) =∞, 9. f(0) = 3, lim f(x) =4,<br />

x→2 x→∞ x→0− lim f(x) =0, lim f(x) =∞, lim f(x) =2,<br />

x→−∞ x→0 +<br />

x→0 +<br />

lim f(x) =−∞ lim f(x) =−∞, lim f(x) =−∞,<br />

x→0 − x→−∞ x→4− lim f(x) =∞, lim<br />

x→4 +<br />

f(x) =3<br />

x→∞<br />

11. If f(x) =x 2 /2 x , then a calculator gives f(0) = 0, f(1) = 0.5, f(2) = 1, f(3) = 1.125, f(4) = 1, f(5) = 0.78125,<br />

f(6) = 0.5625, f(7) = 0.3828125, f(8) = 0.25, f(9) = 0.158203125, f(10) = 0.09765625, f(20) ≈ 0.00038147,<br />

f(50) ≈ 2.2204 × 10 −12 , f(100) ≈ 7.8886 × 10 −27 .<br />

<br />

It appears that lim x 2 /2 x =0.<br />

x→∞

F.<br />

TX.10<br />

SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 61<br />

13. lim<br />

x→∞<br />

3x 2 − x +4<br />

2x 2 +5x − 8 = lim<br />

x→∞<br />

(3x 2 − x +4)/x 2<br />

(2x 2 +5x − 8)/x 2 [divide both the numerator and denominator by x 2<br />

(the highest power of x thatappears in the denominator)]<br />

=<br />

lim (3 − 1/x<br />

x→∞ +4/x2 )<br />

lim (2 + 5/x −<br />

x→∞ 8/x2 )<br />

lim 3 − lim (1/x)+ lim<br />

x→∞ x→∞ x→∞ (4/x2 )<br />

=<br />

lim 2 + lim (5/x) − lim<br />

x→∞ x→∞ x→∞ (8/x2 )<br />

=<br />

3 − lim (1/x)+4 lim<br />

x→∞ x→∞ (1/x2 )<br />

2+5 lim(1/x) − 8 lim<br />

x→∞ x→∞ (1/x2 )<br />

= 3 − 0+4(0)<br />

2+5(0)− 8(0)<br />

[Limit Law 5]<br />

[Limit Laws 1 and 2]<br />

[Limit Laws 7 and 3]<br />

[Theorem 5 of Section 2.5]<br />

= 3 2<br />

1<br />

15. lim<br />

x→∞ 2x +3 = lim 1/x<br />

lim (1/x)<br />

lim<br />

x→∞ (2x +3)/x = x→∞<br />

=<br />

lim (2 + 3/x)<br />

1 − x − x 2<br />

17. lim<br />

x→−∞ 2x 2 − 7<br />

x→∞<br />

x→∞ (1/x)<br />

lim 2 + 3 lim (1/x) = 0<br />

x→∞ x→∞<br />

(1 − x − x 2 )/x 2 lim<br />

x→−∞ (1/x2 − 1/x − 1)<br />

= lim<br />

=<br />

x→−∞ (2x 2 − 7)/x 2 lim (2 −<br />

x→−∞ 7/x2 )<br />

lim<br />

x→−∞ (1/x2 ) − lim (1/x) − lim<br />

x→−∞<br />

=<br />

lim 2 − 7 lim<br />

x→−∞ x→−∞ (1/x2 )<br />

x→−∞ 1<br />

= 0 − 0 − 1<br />

2 − 7(0) = −1 2<br />

2 + 3(0) = 0 2 =0<br />

19. Divide both the numerator and denominator by x 3 (the highest power of x that occurs in the denominator).<br />

lim<br />

x→∞<br />

x 3 +5x<br />

2x 3 − x 2 +4 = lim<br />

x→∞<br />

=<br />

x 3 +5x<br />

x 3<br />

2x 3 − x 2 +4<br />

x 3<br />

lim 1+5 lim<br />

x→∞ x→∞<br />

lim 2 − lim<br />

x→∞ x→∞<br />

= lim<br />

x→∞<br />

1<br />

x 2<br />

1<br />

+4 lim<br />

x x→∞<br />

1+ 5 x 2<br />

2 − 1 x + 4 x 3 =<br />

lim<br />

1+ 5 <br />

x<br />

2 2 − 1 x + 4 <br />

x 3<br />

x→∞<br />

lim<br />

x→∞<br />

= 1 + 5(0)<br />

1 2 − 0+4(0) = 1 2<br />

x 3<br />

21. First, multiply the factors in the denominator. Then divide both the numerator and denominator by u 4 .<br />

lim<br />

u→∞<br />

4u 4 +5<br />

(u 2 − 2)(2u 2 − 1) = lim<br />

u→∞<br />

=<br />

lim<br />

u→∞<br />

4u 4 +5<br />

2u 4 − 5u 2 +2 = lim<br />

u→∞<br />

lim<br />

4+ 5 <br />

u<br />

2 4 − 5 u + 2 =<br />

2 u 4<br />

u→∞<br />

4u 4 +5<br />

u 4<br />

2u 4 − 5u 2 +2<br />

u 4<br />

= lim<br />

u→∞<br />

lim 4+5 lim<br />

u→∞ u→∞<br />

lim 2 − 5 lim<br />

u→∞ u→∞<br />

1<br />

u 4<br />

1<br />

+2 lim<br />

u2 u→∞<br />

4+ 5 u 4<br />

2 − 5 u 2 + 2 u 4<br />

=<br />

1<br />

u 4<br />

4+5(0)<br />

2 − 5(0) + 2(0) = 4 2 =2

F.<br />

62 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

√ √<br />

9x6 − x 9x6 − x/x 3<br />

23. lim<br />

= lim<br />

=<br />

x→∞ x 3 +1 x→∞ (x 3 +1)/x 3<br />

=<br />

<br />

lim 9 − 1/x<br />

5<br />

x→∞<br />

lim<br />

x→∞<br />

= √ 9 − 0=3<br />

<br />

lim<br />

1+ lim<br />

x→∞ (1/x3 ) =<br />

<br />

lim (9x6 − x)/x 6<br />

x→∞<br />

lim (1 +<br />

x→∞ 1/x3 )<br />

x→∞ 9 − lim<br />

x→∞ (1/x5 )<br />

1+0<br />

[since x 3 = √ x 6 for x>0]<br />

√<br />

25. lim 9x2 + x − 3x = lim<br />

x→∞<br />

x→∞<br />

= lim<br />

x→∞<br />

= lim<br />

x→∞<br />

√<br />

27. lim x2 + ax − √ x 2 + bx = lim<br />

x→∞<br />

x→∞<br />

√<br />

9x2 + x − 3x √ 9x 2 + x +3x <br />

√<br />

9x2 + x +3x<br />

9x 2 + x − 9x 2<br />

√<br />

9x2 + x +3x = lim<br />

x→∞<br />

x/x<br />

<br />

9x2 /x 2 + x/x 2 +3x/x = lim<br />

x→∞<br />

= lim<br />

x→∞<br />

= lim<br />

x→∞<br />

= lim<br />

x→∞<br />

x<br />

√<br />

9x2 + x +3x · 1/x<br />

1/x<br />

1<br />

<br />

9+1/x +3<br />

=<br />

√<br />

9x2 + x 2<br />

− (3x)<br />

2<br />

√<br />

9x2 + x +3x<br />

√<br />

x2 + ax − √ x 2 + bx √ x 2 + ax + √ x 2 + bx <br />

√<br />

x2 + ax + √ x 2 + bx<br />

(x 2 + ax) − (x 2 + bx)<br />

√<br />

x2 + ax + √ x 2 + bx = lim<br />

x→∞<br />

a − b<br />

<br />

1+a/x +<br />

<br />

1+b/x<br />

=<br />

1<br />

√<br />

9+3<br />

= 1<br />

3+3 = 1 6<br />

[(a − b)x]/x<br />

√<br />

x2 + ax + √ x 2 + bx / √ x 2<br />

a − b<br />

√ 1+0+<br />

√ 1+0<br />

= a − b<br />

2<br />

x + x 3 + x 5<br />

29. lim<br />

x→∞ 1 − x 2 + x = lim (x + x 3 + x 5 )/x 4<br />

[divide by the highest power of x in the denominator]<br />

4 x→∞ (1 − x 2 + x 4 )/x 4<br />

= lim<br />

x→∞<br />

1/x 3 +1/x + x<br />

1/x 4 − 1/x 2 +1 = ∞<br />

because (1/x 3 +1/x + x) →∞and (1/x 4 − 1/x 2 +1)→ 1 as x →∞.<br />

31. lim<br />

x→−∞ (x4 + x 5 )= lim<br />

x→−∞ x5 ( 1 +1) [factor out the largest power of x] = −∞ because x x5 →−∞and 1/x +1→ 1<br />

as x →−∞.<br />

<br />

Or: lim x 4 + x 5 = lim<br />

x→−∞<br />

x→−∞ x4 (1 + x) =−∞.<br />

33. lim<br />

x→∞<br />

1 − e x<br />

1+2e x = lim<br />

x→∞<br />

(1 − e x )/e x<br />

(1 + 2e x )/e = lim 1/e x − 1<br />

x x→∞ 1/e x +2 = 0 − 1<br />

0+2 = −1 2<br />

35. Since −1 ≤ cos x ≤ 1 and e −2x > 0, wehave−e −2x ≤ e −2x cos x ≤ e −2x .Weknowthat lim<br />

<br />

lim<br />

e<br />

−2x<br />

=0,sobytheSqueezeTheorem, lim<br />

x→∞<br />

x→∞ (e−2x cos x) =0.<br />

x→∞ (−e−2x )=0and<br />

37. (a)<br />

x<br />

f(x)<br />

−10,000 −0.4999625<br />

−100,000 −0.4999962<br />

−1,000,000 −0.4999996<br />

From the graph of f(x) = √ x 2 + x +1+x,weestimate<br />

the value of<br />

lim<br />

x→−∞<br />

f(x) to be −0.5.<br />

(b)<br />

Fromthetable,weestimatethelimit<br />

to be −0.5.

F.<br />

(c)<br />

lim<br />

x→−∞<br />

√<br />

x2 + x +1+x =<br />

lim<br />

x→−∞<br />

= lim<br />

x→−∞<br />

=<br />

√<br />

x2 + x +1+x √ <br />

x 2 + x +1− x<br />

√ = lim<br />

x2 + x +1− x x→−∞<br />

(x +1)(1/x)<br />

√<br />

x2 + x +1− x (1/x) =<br />

1+0<br />

− √ 1+0+0− 1 = − 1 2<br />

lim<br />

x→−∞<br />

1+(1/x)<br />

− 1+(1/x)+(1/x 2 ) − 1<br />

Note that for x

F.<br />

64 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

45. From the graph, it appears y =1is a horizontal asymptote.<br />

3x 3 +500x 2<br />

3x 3 +500x 2<br />

lim<br />

x→∞ x 3 + 500x 2 +100x +2000 = lim<br />

x 3<br />

3+(500/x)<br />

= lim<br />

x→∞ x 3 +500x 2 + 100x + 2000 x→∞ 1+(500/x) + (100/x 2 )+(2000/x 3 )<br />

x 3<br />

3+0<br />

=<br />

=3, soy =3is a horizontal asymptote.<br />

1+0+0+0<br />

The discrepancy can be explained by the choice of the viewing window. Try<br />

[−100,000, 100,000] by [−1, 4] to get a graph that lends credibility to our<br />

calculation that y =3is a horizontal asymptote.<br />

47. Let’s look for a rational function.<br />

(1) lim f(x) =0 ⇒ degree of numerator < degree of denominator<br />

x→±∞<br />

(2) lim f(x) =−∞ ⇒ there is a factor of x 2 in the denominator (not just x, since that would produce a sign<br />

x→0<br />

change at x =0), and the function is negative near x =0.<br />

(3) lim f(x) =∞ and lim f(x) =−∞ ⇒ vertical asymptote at x =3;thereisafactorof(x − 3) in the<br />

x→3− x→3 +<br />

denominator.<br />

(4) f(2) = 0 ⇒ 2 is an x-intercept; there is at least one factor of (x − 2) in the numerator.<br />

Combining all of this information and putting in a negative sign to give us the desired left- and right-hand limits gives us<br />

f(x) =<br />

2 − x as one possibility.<br />

x 2 (x − 3)<br />

49. y = f(x) =x 4 − x 6 = x 4 (1 − x 2 )=x 4 (1 + x)(1 − x). They-intercept is<br />

f(0) = 0. Thex-intercepts are 0, −1,and1 [found by solving f(x) =0for x].<br />

Since x 4 > 0 for x 6= 0, f doesn’t change sign at x =0. The function does change<br />

sign at x = −1 and x =1.Asx → ±∞, f(x) =x 4 (1 − x 2 ) approaches −∞<br />

because x 4 →∞and (1 − x 2 ) →−∞.<br />

51. y = f(x) =(3− x)(1 + x) 2 (1 − x) 4 .They-intercept is f(0) = 3(1) 2 (1) 4 =3.<br />

The x-intercepts are 3, −1,and1. There is a sign change at 3,butnotat−1 and 1.<br />

When x is large positive, 3 − x is negative and the other factors are positive, so<br />

lim f(x) =−∞. Whenx is large negative, 3 − x is positive, so<br />

x→∞<br />

lim f(x) =∞.<br />

x→−∞<br />

53. (a) Since −1 ≤ sin x ≤ 1 for all x, − 1 x ≤ sin x<br />

x<br />

≤ 1 for x>0. Asx →∞, −1/x → 0 and 1/x → 0, so by the Squeeze<br />

x<br />

sin x<br />

Theorem, (sin x)/x → 0. Thus, lim<br />

x→∞ x =0.

F.<br />

TX.10<br />

SECTION 2.6 LIMITS AT INFINITY; HORIZONTAL ASYMPTOTES ¤ 65<br />

(b) From part (a), the horizontal asymptote is y =0. The function<br />

y =(sinx)/x crosses the horizontal asymptote whenever sin x =0;<br />

that is, at x = πn for every integer n. Thus, the graph crosses the<br />

asymptote an infinite number of times.<br />

55. Divide the numerator and the denominator by the highest power of x in Q(x).<br />

(a) If deg Pdeg Q, then the numerator → ±∞ but the denominator doesn’t, so lim [P (x)/Q(x)] = ±∞<br />

x→∞<br />

57. lim<br />

x→∞<br />

(depending on the ratio of the leading coefficients of P and Q).<br />

5 √ x<br />

√ · 1/√ x<br />

x − 1 1/ √ x = lim<br />

x→∞<br />

5<br />

<br />

1 − (1/x)<br />

=<br />

5<br />

√ 1 − 0<br />

=5and<br />

10e x − 21<br />

lim<br />

· 1/ex<br />

x→∞ 2e x 1/e = lim 10 − (21/e x )<br />

= 10 − 0 =5.Since 10ex − 21<br />

F.<br />

66 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

65. (a) 1/x 2 < 0.0001 ⇔ x 2 > 1/0.0001 = 10 000 ⇔ x>100 (x >0)<br />

(b) If ε>0 is given, then 1/x 2 1/ε ⇔ x>1/ √ ε.LetN =1/ √ ε.<br />

Then x>N ⇒ x> √ 1 ⇒<br />

ε 1 <br />

x − 0 = 1 1<br />

M.Nowe x >M ⇔ x>ln M, sotake<br />

N =max(1, ln M). (ThisensuresthatN>0.) Then x>N=max(1, ln M) ⇒ e x > max(e, M) ≥ M,<br />

so lim<br />

x→∞ ex = ∞.<br />

71. Suppose that lim f(x) =L. Then for every ε>0 there is a corresponding positive number N such that |f(x) − L| N.Ift =1/x,thenx>N ⇔ 0 < 1/x < 1/N ⇔ 0 0 (namely 1/N ) such that |f(1/t) − L|

F.<br />

TX.10<br />

SECTION 2.7 DERIVATIVES AND RATES OF CHANGE ¤ 67<br />

(c)<br />

The graph of y =2x +1is tangent to the graph of y =4x − x 2 at the<br />

point (1, 3). Now zoom in toward the point (1, 3) until the parabola and<br />

the tangent line are indistiguishable.<br />

5. Using (1) with f(x) = x − 1 and P (3, 2),<br />

x − 2<br />

x − 1<br />

f(x) − f(a)<br />

m = lim<br />

=lim<br />

x − 2 − 2<br />

x→a x − a x→3 x − 3<br />

=lim<br />

x→3<br />

3 − x<br />

(x − 2)(x − 3) =lim<br />

x→3<br />

x − 1 − 2(x − 2)<br />

=lim<br />

x − 2<br />

x→3 x − 3<br />

−1<br />

x − 2 = −1<br />

1 = −1<br />

Tangent line: y − 2=−1(x − 3) ⇔ y − 2=−x +3 ⇔ y = −x +5<br />

√ √<br />

x − 1 ( √ x − 1)( √ x +1)<br />

7. Using (1), m =lim =lim<br />

x→1 x − 1 x→1 (x − 1)( √ x − 1<br />

= lim<br />

x +1) x→1 (x − 1)( √ x +1) =lim 1<br />

√ = 1<br />

x→1 x +1 2 .<br />

Tangent line: y − 1= 1 2 (x − 1) ⇔ y = 1 2 x + 1 2<br />

9. (a) Using (2) with y = f(x) =3+4x 2 − 2x 3 ,<br />

f(a + h) − f(a) 3+4(a + h) 2 − 2(a + h) 3 − (3 + 4a 2 − 2a 3 )<br />

m =lim<br />

= lim<br />

h→0 h<br />

h→0 h<br />

=lim<br />

h→0<br />

3+4(a 2 +2ah + h 2 ) − 2(a 3 +3a 2 h +3ah 2 + h 3 ) − 3 − 4a 2 +2a 3<br />

h<br />

=lim<br />

h→0<br />

3+4a 2 +8ah +4h 2 − 2a 3 − 6a 2 h − 6ah 2 − 2h 3 − 3 − 4a 2 +2a 3<br />

h<br />

8ah +4h 2 − 6a 2 h − 6ah 2 − 2h 3 h(8a +4h − 6a 2 − 6ah − 2h 2 )<br />

=lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

=lim<br />

h→0<br />

(8a +4h − 6a 2 − 6ah − 2h 2 )=8a − 6a 2<br />

(b) At (1, 5): m =8(1)− 6(1) 2 =2,soanequationofthetangentline<br />

is y − 5=2(x − 1) ⇔ y =2x +3.<br />

(c)<br />

At (2, 3): m =8(2)− 6(2) 2 = −8, so an equation of the tangent<br />

line is y − 3=−8(x − 2) ⇔ y = −8x +19.<br />

11. (a) The particle is moving to the right when s is increasing; that is, on the intervals (0, 1) and (4, 6). The particle is moving to<br />

the left when s is decreasing; that is, on the interval (2, 3). The particle is standing still when s is constant; that is, on the<br />

intervals (1, 2) and (3, 4).

F.<br />

68 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

(b) The velocity of the particle is equal to the slope of the tangent line of the<br />

graph. Note that there is no slope at the corner points on the graph. On the<br />

interval (0, 1), the slope is 3 − 0 =3.Ontheinterval(2, 3), the slope is<br />

1 − 0<br />

1 − 3<br />

3 − 1<br />

= −2. On the interval (4, 6), the slope is<br />

3 − 2 6 − 4 =1.<br />

13. Let s(t) =40t − 16t 2 .<br />

<br />

s(t) − s(2)<br />

40t − 16t<br />

2<br />

− 16 −16t 2 +40t − 16 −8 2t 2 − 5t +2 <br />

v(2) = lim<br />

=lim<br />

=lim<br />

=lim<br />

t→2 t − 2 t→2 t − 2<br />

t→2 t − 2<br />

t→2 t − 2<br />

−8(t − 2)(2t − 1)<br />

=lim<br />

= −8lim(2t − 1) = −8(3) = −24<br />

t→2 t − 2<br />

t→2<br />

Thus, the instantaneous velocity when t =2is −24 ft/s.<br />

1<br />

s(a + h) − s(a) (a + h) − 1 15. v(a)= lim<br />

= lim<br />

2 a 2<br />

h→0 h<br />

h→0 h<br />

a 2 − (a + h) 2<br />

a<br />

=lim<br />

2 (a + h) 2<br />

h→0 h<br />

= lim<br />

h→0<br />

a 2 − (a 2 +2ah + h 2 )<br />

ha 2 (a + h) 2<br />

−(2ah + h 2 )<br />

=lim<br />

h→0 ha 2 (a + h) =lim −h(2a + h)<br />

2 h→0 ha 2 (a + h) = lim −(2a + h)<br />

2 h→0 a 2 (a + h) = −2a<br />

2 a 2 · a = −2<br />

2 a m/s 3<br />

So v (1) = −2<br />

−2<br />

= −2 m/s, v(2) =<br />

13 2 = − 1 −2<br />

m/s, and v(3) = 3 4 3 = − 2 3 27 m/s.<br />

17. g 0 (0) istheonlynegativevalue.Theslopeatx =4is smaller than the slope at x =2and both are smaller than the slope at<br />

x = −2. Thus, g 0 (0) < 0

F.<br />

TX.10<br />

SECTION 2.7 DERIVATIVES AND RATES OF CHANGE ¤ 69<br />

23. (a) Using Definition 2 with F (x) =5x/(1 + x 2 ) and the point (2, 2),wehave (b)<br />

5(2 + h)<br />

F 0 F (2 + h) − F (2) 1+(2+h) − 2<br />

(2) = lim<br />

=lim<br />

2<br />

h→0 h<br />

h→0 h<br />

5h +10<br />

= lim<br />

h 2 +4h +5 − 2 5h +10− 2(h 2 +4h +5)<br />

=lim<br />

h 2 +4h +5<br />

h→0 h<br />

h→0 h<br />

−2h 2 − 3h<br />

= lim<br />

h→0 h(h 2 +4h +5) =lim h(−2h − 3)<br />

h→0 h(h 2 +4h +5) =lim −2h − 3<br />

h→0 h 2 +4h +5 = −3<br />

5<br />

So an equation of the tangent line at (2, 2) is y − 2=− 3 (x − 2) or y = − 3 x + 16 . 5 5 5<br />

25. Use Definition 2 with f(x) =3− 2x +4x 2 .<br />

f 0 f(a + h) − f(a) [3 − 2(a + h)+4(a + h) 2 ] − (3 − 2a +4a 2 )<br />

(a) = lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

= lim<br />

h→0<br />

(3 − 2a − 2h +4a 2 +8ah +4h 2 ) − (3 − 2a +4a 2 )<br />

h<br />

−2h +8ah +4h 2 h(−2+8a +4h)<br />

= lim<br />

=lim<br />

=lim(−2+8a +4h) =−2+8a<br />

h→0 h<br />

h→0 h<br />

h→0<br />

27. Use Definition 2 with f(t) =(2t +1)/(t +3).<br />

2(a + h)+1<br />

f 0 f(a + h) − f(a) (a + h)+3<br />

(a) = lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

−<br />

2a +1<br />

a +3<br />

= lim<br />

h→0<br />

(2a 2 +6a +2ah +6h + a +3)− (2a 2 +2ah +6a + a + h +3)<br />

h(a + h +3)(a +3)<br />

= lim<br />

h→0<br />

5h<br />

h(a + h +3)(a +3) =lim<br />

h→0<br />

29. Use Definition 2 with f(x) =1/ √ x +2.<br />

f 0 f(a + h) − f(a)<br />

(a) = lim<br />

=lim<br />

h→0 h<br />

√ √ a +2− a + h +2<br />

= lim<br />

h→0 h √ a + h +2 √ a +2 ·<br />

= lim<br />

h→0<br />

5<br />

(a + h +3)(a +3) = 5<br />

(a +3) 2<br />

= lim<br />

h→0<br />

(2a +2h +1)(a +3)− (2a +1)(a + h +3)<br />

h(a + h +3)(a +3)<br />

√ √<br />

1<br />

1<br />

− √ a +2− a + h +2<br />

√ √ (a + h)+2 a +2<br />

a + h +2 a +2<br />

= lim<br />

h→0 h<br />

h→0 h<br />

√ √ a +2+ a + h +2<br />

(a +2)− (a + h +2)<br />

√ √ = lim<br />

a +2+ a + h +2 h→0 h √ a + h +2 √ a +2 √ a +2+ √ a + h +2 <br />

−h<br />

h √ a + h +2 √ a +2 √ a +2+ √ a + h +2 =lim<br />

h→0<br />

−1<br />

1<br />

= √ 2 √ = −<br />

a +2 2 a +2 2(a +2) 3/2<br />

Note that the answers to Exercises 31 – 36 are not unique.<br />

(1 + h) 10 − 1<br />

31. By Definition 2, lim<br />

= f 0 (1),wheref(x) =x 10 and a =1.<br />

h→0 h<br />

(1 + h) 10 − 1<br />

Or: By Definition 2, lim<br />

= f 0 (0),wheref(x) =(1+x) 10 and a =0.<br />

h→0 h<br />

−1<br />

√ a + h +2<br />

√ a +2<br />

√ a +2+<br />

√ a + h +2

F.<br />

70 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

33. By Equation 3, lim<br />

x→5<br />

2 x − 32<br />

x − 5 = f 0 (5),wheref(x) =2 x and a =5.<br />

cos(π + h)+1<br />

35. By Definition 2, lim<br />

= f 0 (π),wheref(x) =cosx and a = π.<br />

h→0 h<br />

cos(π + h)+1<br />

Or: ByDefinition 2, lim<br />

= f 0 (0),wheref(x) =cos(π + x) and a =0.<br />

h→0 h<br />

37. v(5) = f 0 f(5 + h) − f(5) [100 + 50(5 + h) − 4.9(5 + h) 2 ] − [100 + 50(5) − 4.9(5) 2 ]<br />

(5) = lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

(100 + 250 + 50h − 4.9h 2 − 49h − 122.5) − (100 + 250 − 122.5) −4.9h 2 + h<br />

=lim<br />

= lim<br />

h→0 h<br />

h→0 h<br />

h(−4.9h +1)<br />

=lim<br />

=lim(−4.9h +1)=1m/s<br />

h→0 h<br />

h→0<br />

The speed when t =5is |1| =1m/s.<br />

39. Thesketchshowsthegraphforaroomtemperatureof72 ◦ and a refrigerator<br />

temperature of 38 ◦ . The initial rate of change is greater in magnitude than the<br />

rate of change after an hour.<br />

41. (a) (i) [2000, 2002]:<br />

(ii) [2000, 2001]:<br />

(iii) [1999, 2000]:<br />

P (2002) − P (2000)<br />

2002 − 2000<br />

P (2001) − P (2000)<br />

2001 − 2000<br />

P (2000) − P (1999)<br />

2000 − 1999<br />

(b) Using the values from (ii) and (iii), we have<br />

=<br />

=<br />

=<br />

77 − 55<br />

2<br />

68 − 55<br />

1<br />

55 − 39<br />

1<br />

13 + 16<br />

2<br />

= 22<br />

2 =11percent/year<br />

=13percent/year<br />

=16percent/year<br />

=14.5 percent/year.<br />

(c) Estimating A as (1999, 40) and B as (2001, 70), the slope at 2000 is<br />

70 − 40<br />

2001 − 1999 = 30<br />

2 =15percent/year.<br />

43. (a) (i) ∆C<br />

∆x<br />

(ii) ∆C<br />

∆x<br />

(b)<br />

=<br />

C(105) − C(100)<br />

105 − 100<br />

=<br />

6601.25 − 6500<br />

5<br />

6520.05 − 6500<br />

1<br />

=\$20.25/unit.<br />

C(101) − C(100)<br />

= = =\$20.05/unit.<br />

101 − 100<br />

5000 + 10(100 + h)+0.05(100 + h)<br />

2<br />

− 6500<br />

=<br />

=<br />

h<br />

=20+0.05h, h 6= 0<br />

C(100 + h) − C(100)<br />

h<br />

20h +0.05h2<br />

h<br />

C(100 + h) − C(100)<br />

So the instantaneous rate of change is lim<br />

= lim (20 + 0.05h) =\$20/unit.<br />

h→0 h<br />

h→0<br />

45. (a) f 0 (x) is the rate of change of the production cost with respect to the number of ounces of gold produced. Its units are<br />

dollars per ounce.

F.<br />

TX.10<br />

SECTION 2.8 THEDERIVATIVEASAFUNCTION ¤ 71<br />

(b) After 800 ounces of gold have been produced, the rate at which the production cost is increasing is \$17/ounce. So the cost<br />

of producing the 800th (or 801st) ounce is about \$17.<br />

(c) In the short term, the values of f 0 (x) will decrease because more efficient use is made of start-up costs as x increases. But<br />

eventually f 0 (x) might increase due to large-scale operations.<br />

47. T 0 (10) is the rate at which the temperature is changing at 10:00 AM.ToestimatethevalueofT 0 (10), we will average the<br />

difference quotients obtained using the times t =8and t =12.LetA =<br />

B =<br />

T (12) − T (10)<br />

12 − 10<br />

=<br />

88 − 81<br />

2<br />

T (8) − T (10)<br />

8 − 10<br />

=<br />

72 − 81<br />

−2<br />

=4.5 and<br />

=3.5. ThenT 0 T (t) − T (10)<br />

(10) = lim<br />

≈ A + B = 4.5+3.5 =4 ◦ F/h.<br />

t→10 t − 10 2 2<br />

49. (a) S 0 (T ) is the rate at which the oxygen solubility changes with respect to the water temperature. Its units are (mg/L)/ ◦ C.<br />

(b) For T =16 ◦ C, it appears that the tangent line to the curve goes through the points (0, 14) and (32, 6). So<br />

S 0 (16) ≈ 6 − 14<br />

32 − 0 = − 8<br />

32 = −0.25 (mg/L)/◦ C. This means that as the temperature increases past 16 ◦ C, the oxygen<br />

solubility is decreasing at a rate of 0.25 (mg/L)/ ◦ C.<br />

51. Since f(x) =x sin(1/x) when x 6= 0and f(0) = 0, wehave<br />

f 0 f(0 + h) − f(0) h sin(1/h) − 0<br />

(0) = lim<br />

=lim<br />

=limsin(1/h). This limit does not exist since sin(1/h) takes the<br />

h→0 h<br />

h→0 h<br />

h→0<br />

values −1 and 1 on any interval containing 0. (Compare with Example 4 in Section 2.2.)<br />

2.8 The Derivative as a Function<br />

1. It appears that f is an odd function, so f 0 will be an even<br />

function—that is, f 0 (−a) =f 0 (a).<br />

(a) f 0 (−3) ≈ 1.5<br />

(b) f 0 (−2) ≈ 1<br />

(c) f 0 (−1) ≈ 0<br />

(d) f 0 (0) ≈−4<br />

(e) f 0 (1) ≈ 0<br />

(f ) f 0 (2) ≈ 1<br />

(g) f 0 (3) ≈ 1.5<br />

3. (a) 0 = II, since from left to right, the slopes of the tangents to graph (a) start out negative, become 0, then positive, then 0,then<br />

negative again. The actual function values in graph II follow the same pattern.<br />

(b) 0 = IV, since from left to right, the slopes of the tangents to graph (b) start out at a fixed positive quantity, then suddenly<br />

become negative, then positive again. The discontinuities in graph IV indicate sudden changes in the slopes of the tangents.<br />

(c) 0 = I, since the slopes of the tangents to graph (c) are negative for x0,asarethefunctionvaluesof<br />

graph I.<br />

(d) 0 = III, since from left to right, the slopes of the tangents to graph (d) are positive, then 0, then negative, then 0, then<br />

positive, then 0, then negative again, and the function values in graph III follow the same pattern.

F.<br />

72 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

Hints for Exercises 4 –11: First plot x-intercepts on the graph of f 0 for any horizontal tangents on the graph of f . Look for any corners on the graph<br />

of f— there will be a discontinuity on the graph of f 0 . On any interval where f has a tangent with positive (or negative) slope, the graph of f 0 will be<br />

positive (or negative). If the graph of the function is linear, the graph of f 0 will be a horizontal line.<br />

5. 7.<br />

9. 11.<br />

13. It appears that there are horizontal tangents on the graph of M for t = 1963<br />

and t = 1971. Thus, there are zeros for those values of t on the graph of<br />

M 0 . The derivative is negative for the years 1963 to 1971.<br />

15.<br />

The slope at 0 appears to be 1 and the slope at 1<br />

appears to be 2.7. Asx decreases, the slope gets<br />

closer to 0. Since the graphs are so similar, we might<br />

guess that f 0 (x) =e x .

F.<br />

TX.10<br />

SECTION 2.8 THEDERIVATIVEASAFUNCTION ¤ 73<br />

17. (a) By zooming in, we estimate that f 0 (0) = 0, f 0 1<br />

2<br />

<br />

=1, f 0 (1) = 2,<br />

and f 0 (2) = 4.<br />

(b) By symmetry, f 0 (−x) =−f 0 (x). Sof 0 − 1 2<br />

= −1, f 0 (−1) = −2,<br />

and f 0 (−2) = −4.<br />

(c) It appears that f 0 (x) is twice the value of x, so we guess that f 0 (x) =2x.<br />

(d) f 0 f(x + h) − f(x) (x + h) 2 − x 2<br />

(x) =lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

x 2 +2hx + h 2 − x 2 2hx + h 2 h(2x + h)<br />

=lim<br />

=lim =lim<br />

=lim(2x + h) =2x<br />

h→0 h<br />

h→0 h h→0 h<br />

h→0<br />

1<br />

19. f 0 f(x + h) − f(x)<br />

(x)= lim<br />

= lim<br />

(x + h) − 1<br />

2 3 − 1 x − <br />

1<br />

2 3<br />

h→0 h<br />

h→0 h<br />

=lim<br />

1<br />

h 2<br />

h→0<br />

h =lim<br />

h→0<br />

1<br />

2 = 1 2<br />

Domain of f = domain of f 0 = R.<br />

<br />

21. f 0 f(t + h) − f(t)<br />

5(t + h) − 9(t + h)<br />

2<br />

− (5t − 9t 2 )<br />

(t) = lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

1<br />

= lim<br />

x + 1 h − 1 − 1 x + 1 2 2 3 2 3<br />

h→0 h<br />

5t +5h − 9(t 2 +2th + h 2 ) − 5t +9t 2 5t +5h − 9t 2 − 18th − 9h 2 − 5t +9t 2<br />

= lim<br />

= lim<br />

h→0 h<br />

h→0 h<br />

5h − 18th − 9h 2 h(5 − 18t − 9h)<br />

= lim<br />

= lim<br />

=lim(5 − 18t − 9h) =5− 18t<br />

h→0 h<br />

h→0 h<br />

h→0<br />

Domain of f = domain of f 0 = R.<br />

<br />

23. f 0 f(x + h) − f(x) (x + h) 3 − 3(x + h)+5 − (x 3 − 3x +5)<br />

(x) = lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

x 3 +3x 2 h +3xh 2 + h 3 − 3x − 3h +5 − x 3 − 3x +5 <br />

3x 2 h +3xh 2 + h 3 − 3h<br />

= lim<br />

= lim<br />

h→0 h<br />

h→0 h<br />

h 3x 2 +3xh + h 2 − 3 <br />

= lim<br />

=lim 3x 2 +3xh + h 2 − 3 =3x 2 − 3<br />

h→0 h<br />

h→0<br />

Domain of f = domain of f 0 = R.<br />

√ √ <br />

25. g 0 g(x + h) − g(x) 1+2(x + h) − 1+2x 1+2(x + h)+ 1+2x<br />

(x) =lim<br />

= lim<br />

√<br />

h→0 h<br />

h→0 h<br />

1+2(x + h)+ 1+2x<br />

(1 + 2x +2h) − (1 + 2x)<br />

2<br />

2<br />

=lim <br />

h→0<br />

1+2(x √ =lim√ √ =<br />

h<br />

+ h)+ 1+2x h→0 1+2x +2h + 1+2x 2 √ 1+2x = 1<br />

√ 1+2x<br />

Domain of g = − 1 2 , ∞ , domain of g 0 = − 1 2 , ∞ .

F.<br />

74 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

4(t + h)<br />

27. G 0 G(t + h) − G(t) (t + h)+1 − 4t 4(t + h)(t +1)− 4t(t + h +1)<br />

t +1<br />

(t + h +1)(t +1)<br />

(t) =lim<br />

=lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

h→0 h<br />

<br />

4t 2 +4ht +4t +4h − 4t 2 +4ht +4t <br />

4h<br />

=lim<br />

= lim<br />

h→0 h(t + h +1)(t +1)<br />

h→0 h(t + h +1)(t +1)<br />

=lim<br />

h→0<br />

4<br />

(t + h +1)(t +1) = 4<br />

(t +1) 2<br />

Domain of G = domain of G 0 =(−∞, −1) ∪ (−1, ∞).<br />

<br />

29. f 0 f(x + h) − f(x) (x + h) 4 − x 4 x 4 +4x 3 h +6x 2 h 2 +4xh 3 + h 4 − x 4<br />

(x) =lim<br />

=lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

h→0 h<br />

4x 3 h +6x 2 h 2 +4xh 3 + h 4 <br />

=lim<br />

=lim 4x 3 +6x 2 h +4xh 2 + h 3 =4x 3<br />

h→0 h<br />

h→0<br />

Domain of f = domain of f 0 = R.<br />

31. (a) f 0 f(x + h) − f(x) [(x + h) 4 +2(x + h)] − (x 4 +2x)<br />

(x)= lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

=lim<br />

h→0<br />

x 4 +4x 3 h +6x 2 h 2 +4xh 3 + h 4 +2x +2h − x 4 − 2x<br />

h<br />

4x 3 h +6x 2 h 2 +4xh 3 + h 4 +2h h(4x 3 +6x 2 h +4xh 2 + h 3 +2)<br />

=lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

=lim<br />

h→0<br />

(4x 3 +6x 2 h +4xh 2 + h 3 +2)=4x 3 +2<br />

(b) Notice that f 0 (x) =0when f has a horizontal tangent, f 0 (x) is<br />

positive when the tangents have positive slope, and f 0 (x) is<br />

negative when the tangents have negative slope.<br />

33. (a) U 0 (t) is the rate at which the unemployment rate is changing with respect to time. Its units are percent per year.<br />

(b) To find U 0 U(t + h) − U(t) U(t + h) − U(t)<br />

(t),weuse lim<br />

≈ for small values of h.<br />

h→0 h<br />

h<br />

For 1993: U 0 (1993) ≈<br />

U(1994) − U(1993)<br />

1994 − 1993<br />

=<br />

6.1 − 6.9<br />

1<br />

= −0.80<br />

For 1994: We estimate U 0 (1994) by using h = −1 and h =1, and then average the two results to obtain a final estimate.<br />

h = −1 ⇒ U 0 (1994) ≈<br />

h =1 ⇒ U 0 (1994) ≈<br />

U(1993) − U(1994)<br />

1993 − 1994<br />

U(1995) − U(1994)<br />

1995 − 1994<br />

=<br />

=<br />

6.9 − 6.1<br />

−1<br />

5.6 − 6.1<br />

1<br />

So we estimate that U 0 (1994) ≈ 1 [(−0.80) + (−0.50)] = −0.65.<br />

2<br />

= −0.80;<br />

= −0.50.<br />

t 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002<br />

U 0 (t) −0.80 −0.65 −0.35 −0.35 −0.45 −0.35 −0.25 0.25 0.90 1.10

F.<br />

TX.10<br />

SECTION 2.8 THEDERIVATIVEASAFUNCTION ¤ 75<br />

35. f is not differentiable at x = −4, because the graph has a corner there, and at x =0, because there is a discontinuity there.<br />

37. f is not differentiable at x = −1, because the graph has a vertical tangent there, and at x =4, because the graph has a corner<br />

there.<br />

39. As we zoom in toward (−1, 0), the curve appears more and more like a<br />

straight line, so f(x) =x + |x| is differentiable at x = −1. But no<br />

matter how much we zoom in toward the origin, the curve doesn’t straighten<br />

out—we can’t eliminate the sharp point (a cusp). So f is not differentiable<br />

at x =0.<br />

41. a = f, b = f 0 , c = f 00 . We can see this because where a has a horizontal tangent, b =0,andwhereb has a horizontal tangent,<br />

c =0. We can immediately see that c can be neither f nor f 0 , since at the points where c has a horizontal tangent, neither a<br />

nor b is equal to 0.<br />

43. We can immediately see that a is the graph of the acceleration function, since at the points where a has a horizontal tangent,<br />

neither c nor b is equal to 0. Next, we note that a =0at the point where b has a horizontal tangent, so b must be the graph of<br />

the velocity function, and hence, b 0 = a. We conclude that c is the graph of the position function.<br />

<br />

45. f 0 f(x + h) − f(x)<br />

1+4(x + h) − (x + h)<br />

2<br />

− (1 + 4x − x 2 )<br />

(x) = lim<br />

= lim<br />

h→0 h<br />

h→0 h<br />

(1 + 4x +4h − x 2 − 2xh − h 2 ) − (1 + 4x − x 2 ) 4h − 2xh − h 2<br />

=lim<br />

= lim<br />

= lim (4 − 2x − h) =4− 2x<br />

h→0 h<br />

h→0 h<br />

h→0<br />

f 00 f 0 (x + h) − f 0 (x) [4 − 2(x + h)] − (4 − 2x) −2h<br />

(x) = lim<br />

= lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

h→0 h<br />

= lim (−2) = −2<br />

h→0<br />

We see from the graph that our answers are reasonable because the graph of<br />

f 0 is that of a linear function and the graph of f 00 is that of a constant<br />

function.

F.<br />

76 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

<br />

47. f 0 f(x + h) − f(x) 2(x + h) 2 − (x + h) 3 − (2x 2 − x 3 )<br />

(x) = lim<br />

= lim<br />

h→0 h<br />

h→0 h<br />

h(4x +2h − 3x 2 − 3xh − h 2 )<br />

=lim<br />

=lim(4x +2h − 3x 2 − 3xh − h 2 )=4x − 3x 2<br />

h→0 h<br />

h→0<br />

<br />

f 00 f 0 (x + h) − f 0 (x)<br />

4(x + h) − 3(x + h)<br />

2<br />

− (4x − 3x 2 ) h(4 − 6x − 3h)<br />

(x) = lim<br />

= lim<br />

= lim<br />

h→0 h<br />

h→0 h<br />

h→0 h<br />

=lim(4 − 6x − 3h) =4− 6x<br />

h→0<br />

f 000 f 00 (x + h) − f 00 (x) [4 − 6(x + h)] − (4 − 6x) −6h<br />

(x) = lim<br />

=lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

h→0 h<br />

=lim (−6) = −6<br />

h→0<br />

f (4) f 000 (x + h) − f 000 (x) −6 − (−6) 0<br />

(x) = lim<br />

=lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

h→0 h =lim(0)<br />

= 0<br />

h→0<br />

The graphs are consistent with the geometric interpretations of the<br />

derivatives because f 0 has zeros where f has a local minimum and a local<br />

maximum, f 00 has a zero where f 0 has a local maximum, and f 000 is a<br />

constant function equal to the slope of f 00 .<br />

49. (a)Notethatwehavefactoredx − a as the difference of two cubes in the third step.<br />

f 0 f(x) − f(a) x 1/3 − a 1/3<br />

x 1/3 − a 1/3<br />

(a) =lim<br />

=lim<br />

= lim<br />

x→a x − a x→a x − a x→a (x 1/3 − a 1/3 )(x 2/3 + x 1/3 a 1/3 + a 2/3 )<br />

1<br />

=lim<br />

x→a x 2/3 + x 1/3 a 1/3 + a = 1<br />

2/3 3a or 1 2/3 3 a−2/3<br />

(b) f 0 f(0 + h) − f(0)<br />

√ 3<br />

h − 0<br />

(0) = lim<br />

= lim = lim<br />

h→0 h<br />

h→0 h<br />

exist, and therefore f 0 (0) does not exist.<br />

(c) lim<br />

x→0<br />

|f 0 (x)| =lim<br />

51. f(x) =|x − 6| =<br />

x→0<br />

1<br />

TX.10<br />

1<br />

h→0 h<br />

x − 6 = lim<br />

x→6<br />

|x − 6| .<br />

2/3<br />

. This function increases without bound, so the limit does not<br />

= ∞ and f is continuous at x =0(root function), so f has a vertical tangent at x =0.<br />

3x2/3 <br />

x − 6 if x − 6 ≥ 6 x − 6 if x ≥ 6<br />

−(x − 6) if x − 6 < 0 = 6 − x if x6<br />

However, a formula for f 0 is f 0 (x) =<br />

−1 if x

F.<br />

TX.10<br />

SECTION 2.8 THEDERIVATIVEASAFUNCTION ¤ 77<br />

53. (a) f(x) =x |x| =<br />

<br />

x<br />

2<br />

if x ≥ 0<br />

−x 2 if x0.<br />

[See Exercise 2.8.17(d).] Similarly, since f(x) =−x 2 for<br />

x

F.<br />

78 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

2 Review<br />

1. (a) lim<br />

x→a<br />

f(x) =L: SeeDefinition2.2.1andFigures1and2inSection2.2.<br />

(b)<br />

(c)<br />

lim f(x) =L: See the paragraph after Definition 2.2.2 and Figure 9(b) in Section 2.2.<br />

x→a +<br />

lim f(x) =L: SeeDefinition 2.2.2 and Figure 9(a) in Section 2.2.<br />

x→a− (d) lim<br />

x→a<br />

f(x) =∞: SeeDefinition 2.2.4 and Figure 12 in Section 2.2.<br />

(e) lim f(x) =L: SeeDefinition 2.6.1 and Figure 2 in Section 2.6.<br />

x→∞<br />

2. In general, the limit of a function fails to exist when the function does not approach a fixed number. For each of the following<br />

functions, the limit fails to exist at x =2.<br />

The left- and right-hand<br />

limits are not equal.<br />

There is an<br />

infinite discontinuity.<br />

There are an infinite<br />

number of oscillations.<br />

3. (a) – (g) See the statements of Limit Laws 1– 6 and 11 in Section 2.3.<br />

4. See Theorem 3 in Section 2.3.<br />

5. (a) See Definition 2.2.6 and Figures 12–14 in Section 2.2.<br />

(b) See Definition 2.6.3 and Figures 3 and 4 in Section 2.6.<br />

6. (a) y = x 4 :Noasymptote (b)y =sinx: No asymptote<br />

(c) y =tanx: Vertical asymptotes x = π + πn, n an integer (d) y 2 =tan−1 x: Horizontal asymptotes y = ± π 2<br />

(e) y = e x : Horizontal asymptote y =0<br />

<br />

<br />

lim<br />

x→−∞ ex =0<br />

(g) y =1/x: Vertical asymptote x =0,<br />

horizontal asymptote y =0<br />

(f ) y =lnx: Vertical asymptote x =0<br />

<br />

lim<br />

x→0 + ln x = −∞ <br />

(h) y = √ x: No asymptote<br />

7. (a) A function f is continuous at a number a if f(x) approaches f(a) as x approaches a;thatis, lim<br />

x→a<br />

f(x) =f(a).<br />

(b) A function f is continuous on the interval (−∞, ∞) if f is continuous at every real number a. The graph of such a<br />

function has no breaks and every vertical line crosses it.

F.<br />

TX.10<br />

CHAPTER 2 REVIEW ¤ 79<br />

8. See Theorem 2.5.10.<br />

9. See Definition 2.7.1.<br />

10. See the paragraph containing Formula 3 in Section 2.7.<br />

11. (a) The average rate of change of y with respect to x over the interval [x 1 ,x 2 ] is<br />

f(x2) − f(x1)<br />

x 2 − x 1<br />

.<br />

f(x 2) − f(x 1)<br />

(b) The instantaneous rate of change of y with respect to x at x = x 1 is lim<br />

.<br />

x 2 →x 1 x 2 − x 1<br />

12. See Definition 2.7.2. The pages following the definition discuss interpretations of f 0 (a) as the slope of a tangent line to the<br />

graph of f at x = a and as an instantaneous rate of change of f(x) with respect to x when x = a.<br />

13. See the paragraphs before and after Example 6 in Section 2.8.<br />

14. (a) A function f is differentiable at a number a if its derivative f 0 exists<br />

(c)<br />

at x = a;thatis,iff 0 (a) exists.<br />

(b) See Theorem 2.8.4. This theorem also tells us that if f is not<br />

continuous at a,thenf is not differentiable at a.<br />

15. See the discussion and Figure 7 on page 159.<br />

1. False. Limit Law 2 applies only if the individual limits exist (these don’t).<br />

3. True. Limit Law 5 applies.<br />

x(x − 5) sin(x − 5)<br />

5. False. Consider lim or lim .Thefirst limit exists and is equal to 5. By Example 3 in Section 2.2,<br />

x→5 x − 5 x→5 x − 5<br />

we know that the latter limit exists (and it is equal to 1).<br />

7. True. A polynomial is continuous everywhere, so lim<br />

x→b<br />

p(x) exists and is equal to p(b).<br />

9. True. See Figure 8 in Section 2.6.<br />

11. False. Consider f(x) =<br />

<br />

1/(x − 1) if x 6= 1<br />

2 if x =1<br />

13. True. Use Theorem 2.5.8 with a =2, b =5,andg(x) =4x 2 − 11. Notethatf(4) = 3 is not needed.<br />

15. True, by the definition of a limit with ε =1.<br />

17. False. See the note after Theorem 4 in Section 2.8.<br />

19. False.<br />

<br />

d 2 2<br />

y<br />

dy<br />

dx is the second derivative while is the first derivative squared. For example, if y = x,<br />

2 dx<br />

then d 2 2<br />

y dy<br />

dx =0,but =1.<br />

2 dx

F.<br />

80 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

1. (a) (i) lim f(x) =3 (ii) lim f(x) =0<br />

x→2 + x→−3 +<br />

(iii)<br />

lim f(x) does not exist since the left and right limits are not equal. (The left limit is −2.)<br />

x→−3<br />

(iv) lim<br />

x→4<br />

f(x) =2<br />

(v) lim f(x) =∞ (vi) lim f(x) =−∞<br />

x→0 −<br />

(vii) lim f(x) =4 (viii) lim f(x) =−1<br />

x→∞ x→−∞<br />

(b) The equations of the horizontal asymptotes are y = −1 and y =4.<br />

(c) The equations of the vertical asymptotes are x =0and x =2.<br />

(d) f is discontinuous at x = −3, 0, 2,and4. The discontinuities are jump, infinite, infinite, and removable, respectively.<br />

x→2<br />

3. Since the exponential function is continuous, lim<br />

x→1<br />

e x3 −x = e 1−1 = e 0 =1.<br />

x 2 − 9<br />

5. lim<br />

x→−3 x 2 +2x − 3 = lim (x +3)(x − 3)<br />

x→−3 (x +3)(x − 1) = lim x − 3<br />

x→−3 x − 1 = −3 − 3<br />

−3 − 1 = −6<br />

−4 = 3 2<br />

<br />

(h − 1) 3 +1 h 3 − 3h 2 +3h − 1 +1 h 3 − 3h 2 +3h <br />

7. lim<br />

= lim<br />

= lim<br />

= lim h 2 − 3h +3 =3<br />

h→0 h<br />

h→0 h<br />

h→0 h<br />

h→0<br />

Another solution: Factor the numerator as a sum of two cubes and then simplify.<br />

(h − 1) 3 +1 (h − 1) 3 +1 3 [(h − 1) + 1] (h − 1) 2 − 1(h − 1) + 1 2<br />

lim<br />

= lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

h→0 h<br />

9. lim<br />

r→9<br />

= lim<br />

h→0<br />

(h − 1) 2 − h +2 =1− 0+2=3<br />

√ r<br />

(r − 9) 4 = ∞ since (r − 9)4 → 0 as r → 9 and<br />

√ r<br />

> 0 for r 6= 9.<br />

(r − 9)<br />

4<br />

u 4 − 1<br />

11. lim<br />

u→1 u 3 +5u 2 − 6u = lim (u 2 +1)(u 2 − 1)<br />

u→1 u(u 2 +5u − 6) =lim (u 2 +1)(u +1)(u − 1) (u 2 +1)(u +1)<br />

= lim<br />

= 2(2)<br />

u→1 u(u +6)(u − 1) u→1 u(u +6) 1(7) = 4 7<br />

13. Since x is positive, √ x 2 = |x| = x. Thus,<br />

√ √<br />

x2 − 9<br />

lim<br />

x→∞ 2x − 6<br />

= lim<br />

x2 − 9/ √ √<br />

x 2 1 − 9/x<br />

x→∞ (2x − 6)/x = lim<br />

2 1 − 0<br />

x→∞ 2 − 6/x = 2 − 0 = 1 2<br />

15. Let t =sinx. Thenasx → π − , sin x → 0 + ,sot → 0 + . Thus, lim ln(sin x) = lim ln t = −∞.<br />

x→π− t→0 +<br />

√<br />

17. lim x2 +4x +1− x √ √ <br />

x2 +4x +1− x x2 +4x +1+x (x 2 +4x +1)− x 2<br />

= lim<br />

· √ = lim √<br />

x→∞<br />

x→∞ 1<br />

x2 +4x +1+x x→∞ x2 +4x +1+x<br />

(4x +1)/x<br />

= lim<br />

x→∞ ( √ x 2 +4x +1+x)/x<br />

<br />

divide by x = √ <br />

x 2 for x>0<br />

4+1/x<br />

= lim <br />

x→∞ 1+4/x +1/x2 +1 = 4+0<br />

√ = 4 1+0+0+1 2 =2

F.<br />

TX.10<br />

CHAPTER 2 REVIEW ¤ 81<br />

19. Let t =1/x. Thenasx → 0 + , t →∞,and lim<br />

x→0 + tan−1 (1/x) = lim<br />

t→∞<br />

tan −1 t = π 2 .<br />

21. From the graph of y = cos 2 x /x 2 , it appears that y =0is the horizontal<br />

asymptote and x =0is the vertical asymptote. Now 0 ≤ (cos x) 2 ≤ 1<br />

0<br />

x ≤ cos2 x<br />

≤ 1 ⇒ 0 ≤ cos2 x<br />

≤ 1 .But lim<br />

2 x 2 x 2 x 2 x 0=0and<br />

2 x→±∞<br />

lim<br />

x→±∞<br />

1<br />

x<br />

2<br />

=0, so by the Squeeze Theorem, lim<br />

x→±∞<br />

cos 2 x<br />

x 2 =0.<br />

⇒<br />

cos 2 x<br />

Thus, y =0is the horizontal asymptote. lim = ∞ because cos 2 x → 1 and x 2 → 0 as x → 0, sox =0is the<br />

x→0 x 2<br />

vertical asymptote.<br />

23. Since 2x − 1 ≤ f(x) ≤ x 2 for 0 0 such that if 0 < |x − 2|

F.<br />

82 ¤ CHAPTER 2 LIMITS AND DERIVATIVES<br />

TX.10<br />

35. (a) The slope of the tangent line at (2, 1) is<br />

f(x) − f(2) 9 − 2x 2 − 1 8 − 2x 2<br />

lim<br />

=lim<br />

= lim<br />

x→2 x − 2 x→2 x − 2 x→2 x − 2<br />

= lim −2(x 2 − 4) −2(x − 2)(x +2)<br />

= lim<br />

x→2 x − 2 x→2 x − 2<br />

=lim<br />

x→2<br />

[−2(x +2)]=−2 · 4=−8<br />

(b) An equation of this tangent line is y − 1=−8(x − 2) or y = −8x +17.<br />

37. (a) s = s(t) =1+2t + t 2 /4. The average velocity over the time interval [1, 1+h] is<br />

v ave =<br />

s(1 + h) − s(1)<br />

(1 + h) − 1<br />

= 1+2(1+h)+(1+h)2 4 − 13/4<br />

h<br />

=<br />

10h + h2<br />

4h<br />

= 10 + h<br />

4<br />

So for the following intervals the average velocities are:<br />

(i) [1, 3]: h =2, v ave =(10+2)/4 =3m/s<br />

(iii) [1, 1.5]: h =0.5, v ave =(10+0.5)/4 =2.625 m/s<br />

(ii) [1, 2]: h =1, v ave =(10+1)/4 =2.75 m/s<br />

(iv) [1, 1.1]: h =0.1, v ave =(10+0.1)/4 =2.525 m/s<br />

s(1 + h) − s(1) 10 + h<br />

(b) When t =1, the instantaneous velocity is lim<br />

= lim = 10<br />

h→0 h<br />

h→0 4 4<br />

=2.5 m/s.<br />

39. (a) f 0 f(x) − f(2) x 3 − 2x − 4<br />

(2) = lim<br />

=lim<br />

x→2 x − 2 x→2 x − 2<br />

(x − 2) x 2 +2x +2 <br />

= lim<br />

x→2 x − 2<br />

= lim<br />

x→2<br />

x 2 +2x +2 =10<br />

(c)<br />

(b) y − 4=10(x − 2) or y =10x − 16<br />

41. (a) f 0 (r) is the rate at which the total cost changes with respect to the interest rate. Its units are dollars/(percent per year).<br />

(b) The total cost of paying off the loan is increasing by \$1200/(percent per year) as the interest rate reaches 10%. Soifthe<br />

interest rate goes up from 10% to 11%, the cost goes up approximately \$1200.<br />

(c) As r increases, C increases. So f 0 (r) will always be positive.<br />

43.

F.<br />

TX.10<br />

CHAPTER 2 REVIEW ¤ 83<br />

√<br />

45. (a) f 0 f(x + h) − f(x) 3 − 5(x + h) − 3 − 5x<br />

(x) =lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

[3 − 5(x + h)] − (3 − 5x)<br />

=lim <br />

h→0<br />

3 √ = lim<br />

h − 5(x + h)+ 3 − 5x h→0<br />

<br />

3 − 5(x + h)+<br />

√ 3 − 5x<br />

<br />

3 − 5(x + h)+<br />

√ 3 − 5x<br />

−5<br />

<br />

3 − 5(x + h)+<br />

√ 3 − 5x<br />

=<br />

−5<br />

2 √ 3 − 5x<br />

(b) Domain of f: (the radicand must be nonnegative) 3 − 5x ≥ 0<br />

5x ≤ 3 ⇒ x ∈ <br />

−∞, 3 5<br />

⇒<br />

Domain of f 0 :exclude 3 because it makes the denominator zero;<br />

5<br />

x ∈ <br />

−∞, 3 5<br />

(c) Our answer to part (a) is reasonable because f 0 (x) is always negative and f<br />

is always decreasing.<br />

47. f is not differentiable: at x = −4 because f is not continuous, at x = −1 because f has a corner, at x =2because f is not<br />

continuous, and at x =5because f has a vertical tangent.<br />

49. C 0 (1990) is the rate at which the total value of US currency in circulation is changing in billions of dollars per year. To<br />

estimate the value of C 0 (1990), we will average the difference quotients obtained using the times t =1985and t = 1995.<br />

Let A =<br />

B =<br />

C(1985) − C(1990)<br />

1985 − 1990<br />

C(1995) − C(1990)<br />

1995 − 1990<br />

C 0 (1990) =<br />

=<br />

=<br />

187.3 − 271.9<br />

−5<br />

409.3 − 271.9<br />

5<br />

= −84.6<br />

−5<br />

= 137.4<br />

5<br />

=16.92 and<br />

=27.48. Then<br />

C(t) − C(1990)<br />

lim<br />

≈ A + B 16.92 + 27.48<br />

= = 44.4 =22.2 billion dollars/year.<br />

t→1990 t − 1990 2<br />

2 2<br />

51. |f(x)| ≤ g(x) ⇔ −g(x) ≤ f(x) ≤ g(x) and lim<br />

x→a<br />

g(x) = 0 = lim<br />

x→a<br />

−g(x).<br />

Thus, by the Squeeze Theorem, lim<br />

x→a<br />

f(x) =0.

F.<br />

TX.10

F.<br />

TX.10<br />

PROBLEMS PLUS<br />

1. Let t = 6√ x,sox = t 6 .Thent → 1 as x → 1,so<br />

lim<br />

x→1<br />

3√ x − 1<br />

√ x − 1<br />

= lim<br />

t→1<br />

t 2 − 1<br />

t 3 − 1 =lim<br />

t→1<br />

(t − 1)(t +1)<br />

(t − 1) (t 2 + t +1) =lim<br />

t→1<br />

t +1<br />

t 2 + t +1 = 1+1<br />

1 2 +1+1 = 2 3 .<br />

Another method: Multiply both the numerator and the denominator by ( √ <br />

x +1)<br />

3√<br />

x2 + 3√ <br />

x +1 .<br />

3. For − 1 0,so|2x − 1| = −(2x − 1) and |2x +1| =2x +1.<br />

2 2<br />

|2x − 1| − |2x +1| −(2x − 1) − (2x +1) −4x<br />

Therefore, lim<br />

=lim<br />

=lim<br />

x→0 x<br />

x→0 x<br />

x→0 x<br />

= lim (−4) = −4.<br />

x→0<br />

5. Since [x] ≤ x 0,thenG(a +180 ◦ )=T (a +360 ◦ ) − T (a + 180 ◦ )=T (a) − T (a +180 ◦ )=−G(a) < 0.<br />

Also, G is continuous since temperature varies continuously. So, by the Intermediate Value Theorem, G has a zero on the<br />

interval [a, a +180 ◦ ].IfG(a) < 0, then a similar argument applies.<br />

85

F.<br />

86 ¤ CHAPTER 2 PROBLEMS PLUS<br />

TX.10<br />

(b) Yes. The same argument applies.<br />

(c) The same argument applies for quantities that vary continuously, such as barometric pressure. But one could argue that<br />

altitude above sea level is sometimes discontinuous, so the result might not always hold for that quantity.<br />

13. (a) Put x =0and y =0in the equation: f(0 + 0) = f(0) + f(0) + 0 2 · 0+0· 0 2 ⇒ f(0) = 2f(0).<br />

Subtracting f(0) from each side of this equation gives f(0) = 0.<br />

<br />

(b) f 0 f(0 + h) − f(0) f(0) + f(h)+0 2 h +0h 2 − f(0) f(h)<br />

(0) = lim<br />

=lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

h→0 h<br />

= lim f(x)<br />

x→0 x =1<br />

<br />

(c) f 0 f(x + h) − f(x) f(x)+f(h)+x 2 h + xh 2 − f(x)<br />

(x) =lim<br />

=lim<br />

h→0 h<br />

h→0 h<br />

<br />

f(h)<br />

=lim<br />

h→0 h<br />

+ x2 + xh =1+x 2<br />

=lim<br />

h→0<br />

f(h)+x 2 h + xh 2<br />

h

F.<br />

TX.10<br />

3 DIFFERENTIATION RULES<br />

3.1 Derivatives of Polynomials and Exponential Functions<br />

e h − 1<br />

1. (a) e is the number such that lim =1.<br />

h→0 h<br />

(b)<br />

x<br />

2.7 x − 1<br />

x<br />

−0.001 0.9928<br />

−0.0001 0.9932<br />

0.001 0.9937<br />

0.0001 0.9933<br />

x<br />

2.8 x − 1<br />

x<br />

−0.001 1.0291<br />

−0.0001 1.0296<br />

0.001 1.0301<br />

0.0001 1.0297<br />

From the tables (to two decimal places),<br />

2.7 h − 1<br />

2.8 h − 1<br />

lim =0.99 and lim =1.03.<br />

h→0 h<br />

h→0 h<br />

Since 0.99 < 1 < 1.03, 2.7

F.<br />

88 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

27. We first expand using the Binomial Theorem (see Reference Page 1).<br />

H(x) =(x + x −1 ) 3 = x 3 +3x 2 x −1 +3x(x −1 ) 2 +(x −1 ) 3 = x 3 +3x +3x −1 + x −3<br />

⇒<br />

H 0 (x) =3x 2 +3+3(−1x −2 )+(−3x −4 )=3x 2 +3− 3x −2 − 3x −4<br />

29. u = 5√ t +4 √ <br />

t 5 = t 1/5 +4t 5/2 ⇒ u 0 = 1 5 t−4/5 5<br />

+4<br />

2 t3/2 = 1 5 t−4/5 +10t 3/2 or 1/<br />

5 5√ <br />

t 4 +10 √ t 3<br />

31. z = A<br />

y + 10 Bey = Ay −10 + Be y ⇒ z 0 = −10Ay −11 + Be y = − 10A<br />

y + 11 Bey<br />

33. y = 4√ x = x 1/4 ⇒ y 0 = 1 4 x−3/4 = 1<br />

4 4√ x . At(1, 1), 3 y0 = 1 4<br />

and an equation of the tangent line is<br />

y − 1= 1 4 (x − 1) or y = 1 4 x + 3 4 .<br />

35. y = x 4 +2e x ⇒ y 0 =4x 3 +2e x . At (0, 2), y 0 =2and an equation of the tangent line is y − 2=2(x − 0)<br />

or y =2x +2. The slope of the normal line is − 1 (the negative reciprocal of 2) and an equation of the normal line is<br />

2<br />

y − 2=− 1 (x − 0) or y = − 1 x +2.<br />

2 2<br />

37. y =3x 2 − x 3 ⇒ y 0 =6x − 3x 2 .<br />

At (1, 2), y 0 =6− 3=3, so an equation of the tangent line is<br />

y − 2=3(x − 1) or y =3x − 1.<br />

39. f(x) =e x − 5x ⇒ f 0 (x) =e x − 5.<br />

Notice that f 0 (x) =0when f has a horizontal tangent, f 0 is positive<br />

when f is increasing, and f 0 is negative when f is decreasing.<br />

41. f(x) =3x 15 − 5x 3 +3 ⇒ f 0 (x) =45x 14 − 15x 2 .<br />

Notice that f 0 (x) =0when f has a horizontal tangent, f 0 is positive<br />

when f is increasing, and f 0 is negative when f is decreasing.

F.<br />

SECTION TX.10 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 89<br />

43. (a) (b) From the graph in part (a), it appears that f 0 is zero at x 1 ≈−1.25, x 2 ≈ 0.5,<br />

and x 3 ≈ 3. The slopes are negative (so f 0 is negative) on (−∞,x 1) and<br />

(x 2,x 3). The slopes are positive (so f 0 is positive) on (x 1,x 2) and (x 3, ∞).<br />

(c) f(x) =x 4 − 3x 3 − 6x 2 +7x +30<br />

⇒<br />

f 0 (x) =4x 3 − 9x 2 − 12x +7<br />

45. f(x) =x 4 − 3x 3 +16x ⇒ f 0 (x) =4x 3 − 9x 2 +16 ⇒ f 00 (x) =12x 2 − 18x<br />

47. f(x) =2x − 5x 3/4 ⇒ f 0 (x) =2− 15<br />

4 x−1/4 ⇒ f 00 (x) = 15<br />

16 x−5/4<br />

Note that f 0 is negative when f is decreasing and positive when f is<br />

increasing. f 00 is always positive since f 0 is always increasing.<br />

49. (a) s = t 3 − 3t ⇒ v(t) =s 0 (t) =3t 2 − 3 ⇒ a(t) =v 0 (t) =6t<br />

(b) a(2) = 6(2) = 12 m/s 2<br />

(c) v(t) =3t 2 − 3=0when t 2 =1,thatis,t =1and a(1) = 6 m/s 2 .<br />

51. The curve y =2x 3 +3x 2 − 12x +1has a horizontal tangent when y 0 =6x 2 +6x − 12 = 0 ⇔ 6(x 2 + x − 2) = 0 ⇔<br />

6(x +2)(x − 1) = 0 ⇔ x = −2 or x =1. The points on the curve are (−2, 21) and (1, −6).<br />

53. y =6x 3 +5x − 3 ⇒ m = y 0 =18x 2 +5,butx 2 ≥ 0 for all x,som ≥ 5 for all x.<br />

55. The slope of the line 12x − y =1(or y =12x − 1)is12, so the slope of both lines tangent to the curve is 12.<br />

y =1+x 3 ⇒ y 0 =3x 2 . Thus, 3x 2 =12 ⇒ x 2 =4 ⇒ x = ±2, which are the x-coordinates at which the tangent<br />

lines have slope 12. The points on the curve are (2, 9) and (−2, −7), so the tangent line equations are y − 9=12(x − 2)<br />

or y =12x − 15 and y + 7 = 12(x +2)or y =12x +17.

F.<br />

90 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

57. The slope of y = x 2 − 5x +4is given by m = y 0 =2x − 5. The slope of x − 3y =5 ⇔ y = 1 3 x − 5 3 is 1 3 ,<br />

so the desired normal line must have slope 1 , and hence, the tangent line to the parabola must have slope −3. This occurs if<br />

3<br />

2x − 5=−3 ⇒ 2x =2 ⇒ x =1.Whenx =1, y =1 2 − 5(1) + 4 = 0, and an equation of the normal line is<br />

y − 0= 1 3 (x − 1) or y = 1 3 x − 1 3 .<br />

59. Let a, a 2 be a point on the parabola at which the tangent line passes through the<br />

point (0, −4). The tangent line has slope 2a and equation y − (−4) = 2a(x − 0)<br />

⇔<br />

y =2ax − 4. Since a, a 2 also lies on the line, a 2 =2a(a) − 4,ora 2 =4.So<br />

a = ±2 and the points are (2, 4) and (−2, 4).<br />

1<br />

61. f 0 f(x + h) − f(x)<br />

(x) = lim<br />

= lim<br />

x + h − 1 x<br />

h→0 h<br />

h→0 h<br />

= lim<br />

h→0<br />

x − (x + h)<br />

hx(x + h) =lim<br />

h→0<br />

−h<br />

hx(x + h) =lim<br />

h→0<br />

63. Let P (x) =ax 2 + bx + c. ThenP 0 (x) =2ax + b and P 00 (x) =2a. P 00 (2) = 2 ⇒ 2a =2 ⇒ a =1.<br />

P 0 (2) = 3 ⇒ 2(1)(2) + b =3 ⇒ 4+b =3 ⇒ b = −1.<br />

P (2) = 5 ⇒ 1(2) 2 +(−1)(2) + c =5 ⇒ 2+c =5 ⇒ c =3.SoP (x) =x 2 − x +3.<br />

−1<br />

x(x + h) = − 1 x 2<br />

65. y = f(x) =ax 3 + bx 2 + cx + d ⇒ f 0 (x) =3ax 2 +2bx + c. Thepoint(−2, 6) is on f, sof(−2) = 6 ⇒<br />

−8a +4b − 2c + d =6 (1). The point (2, 0) is on f, sof(2) = 0 ⇒ 8a +4b +2c + d =0 (2). Since there are<br />

horizontal tangents at (−2, 6) and (2, 0), f 0 (±2) = 0. f 0 (−2) = 0 ⇒ 12a − 4b + c =0 (3) and f 0 (2) = 0 ⇒<br />

12a +4b + c =0 (4). Subtracting equation (3) from (4) gives 8b =0 ⇒ b =0. Adding (1) and (2) gives 8b +2d =6,<br />

so d =3since b =0.From(3) we have c = −12a,so(2) becomes 8a +4(0)+2(−12a)+3=0 ⇒ 3=16a ⇒<br />

a = 3 <br />

3<br />

16 16 = −<br />

9<br />

3<br />

and the desired cubic function is y =<br />

4 16 x3 − 9 x +3. 4<br />

67. f(x) =2− x if x ≤ 1 and f(x) =x 2 − 2x +2if x>1. Now we compute the right- and left-hand derivatives defined in<br />

Exercise 2.8.54:<br />

f 0 −(1) =<br />

f(1 + h) − f(1) 2 − (1 + h) − 1 −h<br />

lim<br />

= lim<br />

= lim<br />

h→0 − h<br />

h→0 − h<br />

h→0 − h = lim −1 =−1 and<br />

h→0− f+(1) 0 f(1 + h) − f(1) (1 + h) 2 − 2(1 + h)+2− 1 h 2<br />

= lim<br />

= lim<br />

= lim<br />

h→0 + h<br />

h→0 + h<br />

h→0 + h = lim h =0.<br />

h→0 +<br />

Thus, f 0 (1) does not exist since f−(1) 0 6=f+(1),sof<br />

0<br />

is not differentiable at 1. Butf 0 (x) =−1 for x1.

F.<br />

SECTION TX.10 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 91<br />

69. (a) Note that x 2 − 9 < 0 for x 2 < 9 ⇔ |x| < 3 ⇔ −3

F.<br />

92 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

79. y = x 2 ⇒ y 0 =2x, so the slope of a tangent line at the point (a, a 2 ) is y 0 =2a and the slope of a normal line is −1/(2a),<br />

for a 6= 0. The slope of the normal line through the points (a, a 2 ) and (0,c) is a2 − c − c<br />

a − 0 ,soa2 = − 1<br />

a 2a<br />

⇒<br />

a 2 − c = − 1 ⇒ a 2 = c − 1 . The last equation has two solutions if c> 1 , one solution if c = 1 , and no solution if<br />

2 2 2 2<br />

c< 1 2 . Since the y-axis is normal to y = x2 regardless of the value of c (thisisthecasefora =0), we have three normal lines<br />

if c> 1 2 and one normal line if c ≤ 1 2 .<br />

3.2 The Product and Quotient Rules<br />

1. Product Rule: y =(x 2 +1)(x 3 +1) ⇒<br />

y 0 =(x 2 +1)(3x 2 )+(x 3 + 1)(2x) =3x 4 +3x 2 +2x 4 +2x =5x 4 +3x 2 +2x.<br />

Multiplying first: y =(x 2 +1)(x 3 +1)=x 5 + x 3 + x 2 +1 ⇒ y 0 =5x 4 +3x 2 +2x (equivalent).<br />

3. By the Product Rule, f(x) =(x 3 +2x)e x ⇒<br />

f 0 (x)=(x 3 +2x)(e x ) 0 + e x (x 3 +2x) 0 =(x 3 +2x)e x + e x (3x 2 +2)<br />

= e x [(x 3 +2x)+(3x 2 +2)]=e x (x 3 +3x 2 +2x +2)<br />

x 2 d<br />

5. By the Quotient Rule, y = ex<br />

⇒ y 0 = dx (ex ) − e x d<br />

dx (x2 )<br />

= x2 (e x ) − e x (2x)<br />

= xex (x − 2)<br />

= ex (x − 2)<br />

.<br />

x 2 (x 2 ) 2 x 4 x 4 x 3<br />

The notations<br />

PR<br />

⇒<br />

and<br />

QR<br />

⇒<br />

indicate the use of the Product and Quotient Rules, respectively.<br />

7. g(x) = 3x − 1<br />

2x +1<br />

QR<br />

⇒ g 0 (2x + 1)(3) − (3x − 1)(2) 6x +3− 6x +2 5<br />

(x) = = =<br />

(2x +1) 2 (2x +1) 2 (2x +1) 2<br />

9. V (x) =(2x 3 +3)(x 4 − 2x)<br />

PR<br />

⇒<br />

V 0 (x) =(2x 3 + 3)(4x 3 − 2) + (x 4 − 2x)(6x 2 )=(8x 6 +8x 3 − 6) + (6x 6 − 12x 3 )=14x 6 − 4x 3 − 6<br />

1<br />

11. F (y) =<br />

y − 3 <br />

(y +5y 3 )= y −2 − 3y −4 y +5y 3 PR<br />

⇒<br />

2 y 4<br />

F 0 (y) =(y −2 − 3y −4 )(1 + 15y 2 )+(y +5y 3 )(−2y −3 +12y −5 )<br />

=(y −2 +15− 3y −4 − 45y −2 )+(−2y −2 +12y −4 − 10 + 60y −2 )<br />

=5+14y −2 +9y −4 or 5+14/y 2 +9/y 4<br />

13. y = x3<br />

1 − x 2 QR<br />

⇒<br />

y 0 = (1 − x2 )(3x 2 ) − x 3 (−2x)<br />

= x2 (3 − 3x 2 +2x 2 )<br />

= x2 (3 − x 2 )<br />

(1 − x 2 ) 2 (1 − x 2 ) 2 (1 − x 2 ) 2

F.<br />

TX.10<br />

SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 93<br />

15. y =<br />

t 2 +2<br />

t 4 − 3t 2 +1<br />

QR<br />

⇒<br />

y 0 = (t4 − 3t 2 +1)(2t) − (t 2 +2)(4t 3 − 6t)<br />

= 2t[(t4 − 3t 2 +1)− (t 2 + 2)(2t 2 − 3)]<br />

(t 4 − 3t 2 +1) 2 (t 4 − 3t 2 +1) 2<br />

= 2t(t4 − 3t 2 +1− 2t 4 − 4t 2 +3t 2 +6)<br />

= 2t(−t4 − 4t 2 +7)<br />

(t 4 − 3t 2 +1) 2 (t 4 − 3t 2 +1) 2<br />

17. y =(r 2 − 2r)e r PR<br />

⇒ y 0 =(r 2 − 2r)(e r )+e r (2r − 2) = e r (r 2 − 2r +2r − 2) = e r (r 2 − 2)<br />

19. y = v3 − 2v √ v<br />

v<br />

= v 2 − 2 √ v = v 2 − 2v 1/2 ⇒ y 0 =2v − 2 1<br />

2<br />

<br />

v −1/2 =2v − v −1/2 .<br />

We can change the form of the answer as follows: 2v − v −1/2 =2v − 1 √ v<br />

= 2v √ v − 1<br />

√ v<br />

= 2v3/2 − 1<br />

√ v<br />

21. f(t) = 2t<br />

2+ √ t<br />

<br />

QR<br />

⇒<br />

(2 + t 1/2 1<br />

)(2) − 2t<br />

f 0 (t) =<br />

(2 + √ t ) 2 = 4+2t1/2 − t 1/2<br />

(2 + √ t ) 2 = 4+t1/2<br />

(2 + √ t ) 2 or 4+ √ t<br />

(2 + √ t ) 2<br />

2 t−1/2 <br />

23. f(x) =<br />

A<br />

B + Ce x<br />

QR<br />

⇒ f 0 (x) = (B + Cex ) · 0 − A(Ce x )<br />

= − ACex<br />

(B + Ce x ) 2 (B + Ce x ) 2<br />

x<br />

25. f(x) =<br />

x + c/x ⇒ f 0 (x) = (x + c/x)(1) − x(1 − c/x2 ) x + c/x − x + c/x<br />

<br />

x + c 2<br />

= x 2 2<br />

= 2c/x<br />

+ c (x 2 + c) 2<br />

x<br />

x<br />

x 2<br />

· x2<br />

x = 2cx<br />

2 (x 2 + c) 2<br />

27. f(x) =x 4 e x ⇒ f 0 (x) =x 4 e x + e x · 4x 3 = x 4 +4x 3 e x or x 3 e x (x +4) ⇒<br />

f 00 (x)=(x 4 +4x 3 )e x + e x (4x 3 +12x 2 )=(x 4 +4x 3 +4x 3 +12x 2 )e x<br />

=(x 4 +8x 3 +12x 2 )e x<br />

or x 2 e x (x +2)(x +6) <br />

29. f(x) = x2<br />

1+2x ⇒ f 0 (x) = (1 + 2x)(2x) − x2 (2)<br />

(1 + 2x) 2 = 2x +4x2 − 2x 2<br />

(1 + 2x) 2 = 2x2 +2x<br />

(1 + 2x) 2 ⇒<br />

f 00 (x)= (1 + 2x)2 (4x +2)− (2x 2 +2x)(1 + 4x +4x 2 ) 0<br />

= 2(1 + 2x)2 (2x +1)− 2x(x +1)(4+8x)<br />

[(1 + 2x) 2 ] 2 (1 + 2x) 4<br />

= 2(1 + 2x)[(1 + 2x)2 − 4x(x +1)]<br />

= 2(1 + 4x +4x2 − 4x 2 − 4x) 2<br />

=<br />

(1 + 2x) 4 (1 + 2x) 3 (1 + 2x) 3<br />

31. y = 2x<br />

x +1<br />

⇒<br />

y 0 =<br />

(x +1)(2)− (2x)(1) 2<br />

=<br />

(x +1) 2 (x +1) . 2<br />

At (1, 1), y 0 = 1 2 , and an equation of the tangent line is y − 1= 1 2 (x − 1),ory = 1 2 x + 1 2 .<br />

33. y =2xe x ⇒ y 0 =2(x · e x + e x · 1) = 2e x (x +1).<br />

At (0, 0), y 0 =2e 0 (0 + 1) = 2 · 1 · 1=2, and an equation of the tangent line is y − 0=2(x − 0),ory =2x. Theslopeof<br />

the normal line is − 1 2 , so an equation of the normal line is y − 0=− 1 2 (x − 0),ory = − 1 2 x.

F.<br />

94 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

35. (a) y = f(x) = 1<br />

1+x 2 ⇒<br />

f 0 (x) = (1 + x2 )(0) − 1(2x)<br />

= −2x . So the slope of the<br />

(1 + x 2 ) 2 (1 + x 2 )<br />

2<br />

(b)<br />

tangent line at the point <br />

−1, 1 2 is f 0 (−1) = 2 2 = 1 2 2<br />

and its<br />

equation is y − 1 2 = 1 2 (x +1)or y = 1 2 x +1.<br />

37. (a) f(x) = ex<br />

⇒ f 0 (x) = x3 (e x ) − e x (3x 2 )<br />

= x2 e x (x − 3)<br />

= ex (x − 3)<br />

x 3 (x 3 ) 2 x 6 x 4<br />

(b)<br />

f 0 =0when f has a horizontal tangent line, f 0 is negative when<br />

f is decreasing, and f 0 is positive when f is increasing.<br />

39. (a) f(x) =(x − 1)e x ⇒ f 0 (x) =(x − 1)e x + e x (1) = e x (x − 1+1)=xe x .<br />

f 00 (x) =x(e x )+e x (1) = e x (x +1)<br />

(b)<br />

f 0 =0when f has a horizontal tangent and f 00 =0when f 0 has a<br />

horizontal tangent. f 0 is negative when f is decreasing and positive when f<br />

is increasing. f 00 is negative when f 0 is decreasing and positive when f 0 is<br />

increasing. f 00 is negative when f is concave down and positive when f is<br />

concave up.<br />

41. f(x) = x2<br />

1+x ⇒ f 0 (x) = (1 + x)(2x) − x2 (1)<br />

(1 + x) 2 = 2x +2x2 − x 2<br />

(1 + x) 2 = x2 +2x<br />

x 2 +2x +1<br />

⇒<br />

so f 00 (1) =<br />

f 00 (x) = (x2 +2x +1)(2x +2)− (x 2 +2x)(2x +2)<br />

(x 2 +2x +1) 2 = (2x +2)(x2 +2x +1− x 2 − 2x)<br />

[(x +1) 2 ] 2<br />

=<br />

2<br />

(1 + 1) 3 = 2 8 = 1 4 .<br />

2(x +1)(1) 2<br />

=<br />

(x +1) 4 (x +1) 3 ,<br />

43. We are given that f(5) = 1, f 0 (5) = 6, g(5) = −3,andg 0 (5) = 2.<br />

(a) (fg) 0 (5) = f(5)g 0 (5) + g(5)f 0 (5) = (1)(2) + (−3)(6) = 2 − 18 = −16<br />

0 f<br />

(b) (5) = g(5)f 0 (5) − f(5)g 0 (5) (−3)(6) − (1)(2)<br />

= = − 20 g<br />

[g(5)] 2 (−3) 2 9<br />

0 g<br />

(c) (5) = f(5)g0 (5) − g(5)f 0 (5)<br />

=<br />

f<br />

[f(5)] 2<br />

(1)(2) − (−3)(6)<br />

(1) 2 =20<br />

45. f(x) =e x g(x) ⇒ f 0 (x) =e x g 0 (x)+g(x)e x = e x [g 0 (x)+g(x)]. f 0 (0) = e 0 [g 0 (0) + g(0)] = 1(5 + 2) = 7

F.<br />

TX.10<br />

SECTION 3.2 THE PRODUCT AND QUOTIENT RULES ¤ 95<br />

47. (a) From the graphs of f and g, we obtain the following values: f(1) = 2 since the point (1, 2) is on the graph of f;<br />

g(1) = 1 since the point (1, 1) is on the graph of g; f 0 (1) = 2 since the slope of the line segment between (0, 0) and (2, 4)<br />

is 4 − 0<br />

2 − 0 =2; 0 − 4<br />

g0 (1) = −1 since the slope of the line segment between (−2, 4) and (2, 0) is<br />

2 − (−2) = −1.<br />

Now u(x) =f(x)g(x),sou 0 (1) = f(1)g 0 (1) + g(1) f 0 (1) = 2 · (−1) + 1 · 2=0.<br />

(b) v(x) =f(x)/g(x),sov 0 (5) = g(5)f 0 (5) − f(5)g 0 (5)<br />

= 2 <br />

− 1 3 − 3 · 2<br />

3<br />

= − 8 3<br />

[g(5)] 2 2 2 4 = − 2 3<br />

49. (a) y = xg(x) ⇒ y 0 = xg 0 (x)+g(x) · 1=xg 0 (x)+g(x)<br />

(b) y =<br />

x<br />

g(x)<br />

⇒<br />

y 0 = g(x) · 1 − xg0 (x)<br />

[g(x)] 2<br />

= g(x) − xg0 (x)<br />

[g(x)] 2<br />

(c) y = g(x)<br />

x<br />

⇒ y 0 = xg0 (x) − g(x) · 1<br />

(x) 2<br />

= xg0 (x) − g(x)<br />

x 2<br />

51. If y = f(x) = x<br />

x +1 ,thenf 0 (x +1)(1)− x(1) 1<br />

(x) = = .Whenx = a, the equation of the tangent line is<br />

(x +1) 2 (x +1)<br />

2<br />

y −<br />

a<br />

a +1 = 1<br />

a<br />

(x − a). This line passes through (1, 2) when 2 −<br />

(a +1)<br />

2<br />

a +1 = 1<br />

(1 − a) ⇔<br />

(a +1)<br />

2<br />

2(a +1) 2 − a(a +1)=1− a ⇔ 2a 2 +4a +2− a 2 − a − 1+a =0 ⇔ a 2 +4a +1=0.<br />

The quadratic formula gives the roots of this equation as a = −4 ± 4 2 − 4(1)(1)<br />

2(1)<br />

so there are two such tangent lines. Since<br />

f −2 ± √ 3 = −2 ± √ 3<br />

−2 ± √ 3+1 = −2 ± √ 3<br />

−1 ± √ 3 · −1 ∓ √ 3<br />

−1 ∓ √ 3<br />

<br />

the lines touch the curve at A<br />

= 2 ± 2 √ 3 ∓ √ 3 − 3<br />

1 − 3<br />

−2+ √ 3, 1 − √ 3<br />

2<br />

<br />

and B −2 − √ <br />

3, 1+√ 3<br />

≈ (−3.73, 1.37).<br />

2<br />

= −1 ± √ 3<br />

−2<br />

<br />

≈ (−0.27, −0.37)<br />

= 1 ∓ √ 3<br />

,<br />

2<br />

= −4 ± √ 12<br />

2<br />

= −2 ± √ 3,<br />

53. If P (t) denotes the population at time t and A(t) the average annual income, then T (t) =P (t)A(t) is the total personal<br />

income. The rate at which T (t) is rising is given by T 0 (t) =P (t)A 0 (t)+A(t)P 0 (t)<br />

⇒<br />

T 0 (1999) = P (1999)A 0 (1999) + A(1999)P 0 (1999) = (961,400)(\$1400/yr)+(\$30,593)(9200/yr)<br />

=\$1,345,960,000/yr +\$281,455,600/yr =\$1,627,415,600/yr<br />

So the total personal income was rising by about \$1.627 billion per year in 1999.<br />

The term P (t)A 0 (t) ≈ \$1.346 billion represents the portion of the rate of change of total income due to the existing<br />

population’s increasing income. The term A(t)P 0 (t) ≈ \$281 million represents the portion of the rate of change of total<br />

income due to increasing population.

F.<br />

96 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

We will sometimes use the form f 0 g + fg 0 rather than the form fg 0 + gf 0 for the Product Rule.<br />

55. (a) (fgh) 0 =[(fg)h] 0 =(fg) 0 h +(fg)h 0 =(f 0 g + fg 0 )h +(fg)h 0 = f 0 gh + fg 0 h + fgh 0<br />

(b) Putting f = g = h in part (a), we have d<br />

dx [f(x)]3 =(fff) 0 = f 0 ff + ff 0 f + fff 0 =3fff 0 =3[f(x)] 2 f 0 (x).<br />

(c)<br />

d<br />

dx (e3x )= d<br />

dx (ex ) 3 =3(e x ) 2 e x =3e 2x e x =3e 3x<br />

57. For f(x) =x 2 e x , f 0 (x) =x 2 e x + e x (2x) =e x (x 2 +2x). Similarly, we have<br />

f 00 (x) =e x (x 2 +4x +2)<br />

f 000 (x) =e x (x 2 +6x +6)<br />

f (4) (x) =e x (x 2 +8x + 12)<br />

f (5) (x) =e x (x 2 +10x +20)<br />

It appears that the coefficient of x in the quadratic term increases by 2 with each differentiation. The pattern for the<br />

constant terms seems to be 0=1· 0, 2=2· 1, 6=3· 2, 12 = 4 · 3, 20 = 5 · 4. So a reasonable guess is that<br />

f (n) (x) =e x [x 2 +2nx + n(n − 1)].<br />

Proof: Let S n be the statement that f (n) (x) =e x [x 2 +2nx + n(n − 1)].<br />

1. S 1 is true because f 0 (x) =e x (x 2 +2x).<br />

2. Assume that S k is true; that is, f (k) (x) =e x [x 2 +2kx + k(k − 1)]. Then<br />

f (k+1) (x) = d <br />

f (k) (x) = e x (2x +2k)+[x 2 +2kx + k(k − 1)]e x<br />

dx<br />

= e x [x 2 +(2k +2)x +(k 2 + k)] = e x [x 2 +2(k +1)x +(k +1)k]<br />

This shows that S k+1 is true.<br />

3. Therefore, by mathematical induction, S n is true for all n;thatis,f (n) (x) =e x [x 2 +2nx + n(n − 1)] for every<br />

positive integer n.<br />

3.3 Derivatives of Trigonometric Functions<br />

1. f(x) =3x 2 − 2cosx ⇒ f 0 (x) =6x − 2(− sin x) =6x +2sinx<br />

3. f(x) =sinx + 1 2 cot x ⇒ f 0 (x) =cosx − 1 2 csc2 x<br />

5. g(t) =t 3 cos t ⇒ g 0 (t) =t 3 (− sin t)+(cost) · 3t 2 =3t 2 cos t − t 3 sin t or t 2 (3 cos t − t sin t)<br />

7. h(θ) =cscθ + e θ cot θ ⇒ h 0 (θ) =− csc θ cot θ + e θ (− csc 2 θ)+(cotθ)e θ = − csc θ cot θ + e θ (cot θ − csc 2 θ)<br />

9. y =<br />

x<br />

2 − tan x ⇒ y0 = (2 − tan x)(1) − x(− sec2 x)<br />

= 2 − tan x + x sec2 x<br />

(2 − tan x) 2 (2 − tan x) 2<br />

11. f(θ) = sec θ<br />

1+secθ<br />

⇒<br />

f 0 (θ) =<br />

(1 + sec θ)(sec θ tan θ) − (sec θ)(sec θ tan θ) (sec θ tan θ) [(1+secθ) − sec θ] sec θ tan θ<br />

= =<br />

(1 + sec θ) 2 (1 + sec θ) 2 (1 + sec θ) 2

F.<br />

TX.10<br />

SECTION 3.3 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS ¤ 97<br />

13. y = sin x ⇒ y 0 = x2 cos x − (sin x)(2x) x(x cos x − 2sinx) x cos x − 2sinx<br />

x 2 (x 2 ) 2 = =<br />

x 4<br />

x 3<br />

15. Using Exercise 3.2.55(a), f(x) =xe x csc x ⇒<br />

17.<br />

19.<br />

f 0 (x)=(x) 0 e x csc x + x(e x ) 0 csc x + xe x (csc x) 0 =1e x csc x + xe x csc x + xe x (− cot x csc x)<br />

= e x csc x (1 + x − x cot x)<br />

<br />

d<br />

d 1 (sin x)(0) − 1(cos x)<br />

(csc x) = =<br />

dx dx sin x<br />

sin 2 = − cos x<br />

x<br />

sin 2 x = − 1<br />

sin x · cos x = − csc x cot x<br />

sin x<br />

d<br />

d<br />

(cot x) =<br />

dx dx<br />

cos x<br />

<br />

(sin x)(− sin x) − (cos x)(cos x)<br />

=<br />

sin x<br />

sin 2 = − sin2 x +cos 2 x<br />

x<br />

sin 2 = − 1<br />

x sin 2 x = − csc2 x<br />

21. y =secx ⇒ y 0 =secx tan x,soy 0 ( π 3 )=secπ 3 tan π 3 =2√ 3. An equation of the tangent line to the curve y =secx<br />

at the point π<br />

, 2 is y − 2=2 √ 3 √ √<br />

x − π 3 3 or y =2 3 x +2−<br />

2<br />

3 3 π.<br />

23. y = x +cosx ⇒ y 0 =1− sin x. At(0, 1), y 0 =1, and an equation of the tangent line is y − 1=1(x − 0),ory = x +1.<br />

25. (a) y =2x sin x ⇒ y 0 =2(x cos x +sinx · 1). At π<br />

2 ,π ,<br />

y 0 =2 π<br />

cos π<br />

2 2 +sinπ 2 =2(0+1)=2, and an equation of the<br />

tangent line is y − π =2 <br />

x − π 2 ,ory =2x.<br />

(b)<br />

27. (a) f(x) =secx − x ⇒ f 0 (x) =secx tan x − 1<br />

(b)<br />

Note that f 0 =0where f has a minimum. Also note that f 0 is negative<br />

when f is decreasing and f 0 is positive when f is increasing.<br />

29. H(θ) =θ sin θ ⇒ H 0 (θ) =θ (cos θ) +(sinθ) · 1=θ cos θ +sinθ ⇒<br />

H 00 (θ) =θ (− sin θ)+(cosθ) · 1+cosθ = −θ sin θ +2cosθ<br />

31. (a) f(x) = tan x − 1<br />

sec x<br />

⇒<br />

f 0 (x) = sec x(sec2 x) − (tan x − 1)(sec x tan x)<br />

(sec x) 2<br />

(b) f(x) = tan x − 1<br />

sec x<br />

=<br />

sin x<br />

cos x − 1<br />

=<br />

1<br />

cos x<br />

(c) From part (a), f 0 (x) = 1+tanx<br />

sec x<br />

sin x − cos x<br />

cos x<br />

1<br />

cos x<br />

= sec x(sec2 x − tan 2 x +tanx)<br />

sec 2 x<br />

= 1+tanx<br />

sec x<br />

=sinx − cos x ⇒ f 0 (x) =cosx − (− sin x) =cosx +sinx<br />

= 1<br />

sec x + tan x<br />

sec x =cosx +sinx, which is the expression for f 0 (x) in part (b).

F.<br />

98 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

33. f(x) =x +2sinx has a horizontal tangent when f 0 (x) =0 ⇔ 1+2cosx =0 ⇔ cos x = − 1 2<br />

⇔<br />

x = 2π +2πn or 4π +2πn,wheren is an integer. Note that 4π and 2π are ± π units from π.Thisallowsustowritethe<br />

3 3 3 3 3<br />

solutions in the more compact equivalent form (2n +1)π ± π 3<br />

, n an integer.<br />

35. (a) x(t) =8sint ⇒ v(t) =x 0 (t) =8cost ⇒ a(t) =x 00 (t) =−8sint<br />

(b) The mass at time t = 2π 3<br />

has position x √3 <br />

2π<br />

3 =8sin<br />

2π<br />

=8 3 2<br />

<br />

and acceleration a √3<br />

2π<br />

3 = −8sin<br />

2π<br />

= −8 3 2<br />

=4 √ 3,velocityv <br />

2π<br />

3 =8cos<br />

2π<br />

=8 <br />

3<br />

− 1 2 = −4,<br />

= −4 √ 3.Sincev <br />

2π<br />

3 < 0, the particle is moving to the left.<br />

37. From the diagram we can see that sin θ = x/10 ⇔ x =10sinθ. Wewanttofind the rate<br />

of change of x with respect to θ,thatis,dx/dθ. Taking the derivative of x =10sinθ,weget<br />

dx/dθ = 10(cos θ). Sowhenθ = π 3 , dx<br />

dθ =10cosπ 3 =10 1<br />

2<br />

<br />

39. lim<br />

x→0<br />

sin 3x<br />

x<br />

3sin3x<br />

=lim<br />

x→0 3x<br />

=3 lim<br />

3x→0<br />

sin 3x<br />

3x<br />

[multiply numerator and denominator by 3]<br />

[as x → 0, 3x → 0]<br />

sin θ<br />

=3lim<br />

θ→0 θ<br />

[let θ =3x]<br />

=3(1) [Equation 2]<br />

=3<br />

<br />

tan 6t sin 6t<br />

41. lim<br />

t→0 sin 2t =lim ·<br />

t→0 t<br />

sin 6t<br />

=6lim<br />

t→0 6t<br />

1<br />

cos 6t ·<br />

<br />

t<br />

6sin6t 1<br />

=lim · lim<br />

sin 2t t→0 6t t→0 cos 6t · lim<br />

t→0<br />

1<br />

· lim<br />

t→0 cos 6t · 1<br />

2 lim<br />

t→0<br />

<br />

sin(cos θ)<br />

sin lim cos θ<br />

θ→0<br />

43. lim =<br />

θ→0 sec θ lim sec θ = sin 1 =sin1<br />

1<br />

θ→0<br />

45. Divide numerator and denominator by θ. (sin θ also works.)<br />

lim<br />

θ→0<br />

47. lim<br />

x→π/4<br />

49. (a)<br />

(b)<br />

sin θ<br />

θ +tanθ =lim<br />

θ→0<br />

1 − tan x<br />

sin x − cos x =<br />

d<br />

dx tan x = d<br />

dx<br />

d<br />

dx sec x = d<br />

dx<br />

sin θ<br />

θ<br />

1+ sin θ<br />

θ<br />

lim<br />

x→π/4<br />

·<br />

1<br />

cos θ<br />

sin x<br />

cos x ⇒ sec2 x =<br />

1<br />

cos x<br />

=<br />

2t<br />

sin 2t =6(1)· 1<br />

1 · 1<br />

2 (1) = 3<br />

sin θ<br />

lim<br />

θ→0 θ<br />

sin θ<br />

1+lim lim<br />

θ→0 θ θ→0<br />

<br />

1 − sin x <br />

· cos x<br />

cos x<br />

(sin x − cos x) · cos x =<br />

lim<br />

x→π/4<br />

1<br />

cos θ<br />

cos x cos x − sin x (− sin x)<br />

cos 2 x<br />

=<br />

2t<br />

2sin2t<br />

1<br />

1+1· 1 = 1 2<br />

cos x − sin x<br />

(sin x − cos x)cosx =<br />

lim<br />

x→π/4<br />

−1<br />

cos x = −1<br />

1/ √ 2 = −√ 2<br />

= cos2 x +sin 2 x<br />

. Sosec 2 x = 1<br />

cos 2 x<br />

cos 2 x .<br />

(cos x)(0) − 1(− sin x)<br />

⇒ sec x tan x = . Sosec x tan x = sin x<br />

cos 2 x<br />

cos 2 x .

F.<br />

TX.10<br />

SECTION 3.4 THE CHAIN RULE ¤ 99<br />

(c)<br />

d<br />

d 1+cotx<br />

(sin x +cosx) =<br />

dx dx csc x<br />

⇒<br />

cos x − sin x = csc x (− csc2 x) − (1 + cot x)(− csc x cot x)<br />

csc 2 x<br />

= − csc2 x +cot 2 x +cotx<br />

csc x<br />

So cos x − sin x = cot x − 1<br />

csc x .<br />

= −1+cotx<br />

csc x<br />

= csc x [− csc2 x +(1+cotx) cotx]<br />

csc 2 x<br />

51. By the definition of radian measure, s = rθ,wherer is the radius of the circle. By drawing the bisector of the angle θ,wecan<br />

see that sin θ 2 = d/2<br />

r<br />

⇒ d =2r sin θ s<br />

.So lim<br />

2 θ→0 + d = lim<br />

θ→0 +<br />

rθ<br />

2r sin(θ/2) = lim<br />

θ→0 +<br />

2 · (θ/2)<br />

2sin(θ/2) =lim<br />

θ→0<br />

θ/2<br />

sin(θ/2) =1.<br />

sin x<br />

[This is just the reciprocal of the limit lim =1combined with the fact that as θ → 0, θ → 0 also.]<br />

x→0 x<br />

2<br />

3.4 The Chain Rule<br />

1. Let u = g(x) =4x and y = f(u) =sinu. Then dy<br />

dx = dy du<br />

=(cosu)(4) = 4 cos 4x.<br />

du dx<br />

3. Let u = g(x) =1− x 2 and y = f(u) =u 10 . Then dy<br />

dx = dy du<br />

du dx =(10u9 )(−2x) =−20x(1 − x 2 ) 9 .<br />

5. Let u = g(x) = √ x and y = f(u) =e u .Then dy<br />

dx = dy du<br />

du dx =(eu )<br />

7. F (x) =(x 4 +3x 2 − 2) 5 ⇒ F 0 (x) =5(x 4 +3x 2 − 2) 4 ·<br />

or 10x(x 4 +3x 2 − 2) 4 (2x 2 +3) <br />

<br />

1<br />

2 x−1/2 <br />

= e √x ·<br />

1<br />

2 √ x = e√ x<br />

2 √ x .<br />

d <br />

x 4 +3x 2 − 2 =5(x 4 +3x 2 − 2) 4 (4x 3 +6x)<br />

dx<br />

9. F (x) = 4√ 1+2x + x 3 =(1+2x + x 3 ) 1/4 ⇒<br />

F 0 (x) = 1 4 (1 + 2x + x3 ) −3/4 ·<br />

11. g(t) =<br />

2+3x 2<br />

=<br />

4 4 (1 + 2x + x 3 ) 3<br />

d<br />

dx (1 + 2x + x3 )=<br />

1<br />

4(1 + 2x + x 3 ) 3/4 · (2 + 3x2 )=<br />

2+3x 2<br />

4(1 + 2x + x 3 ) 3/4<br />

1<br />

(t 4 +1) 3 =(t4 +1) −3 ⇒ g 0 (t) =−3(t 4 +1) −4 (4t 3 )=−12t 3 (t 4 +1) −4 = −12t3<br />

(t 4 +1) 4<br />

13. y =cos(a 3 + x 3 ) ⇒ y 0 = − sin(a 3 + x 3 ) · 3x 2 [a 3 is just a constant] = −3x 2 sin(a 3 + x 3 )<br />

15. y = xe −kx ⇒ y 0 = x e −kx (−k) + e −kx · 1=e −kx (−kx +1)<br />

<br />

or (1 − kx)e<br />

−kx <br />

17. g(x) =(1+4x) 5 (3 + x − x 2 ) 8 ⇒<br />

g 0 (x) =(1+4x) 5 · 8(3 + x − x 2 ) 7 (1 − 2x)+(3+x − x 2 ) 8 · 5(1 + 4x) 4 · 4<br />

=4(1+4x) 4 (3 + x − x 2 ) 7 2(1 + 4x)(1 − 2x)+5(3+x − x 2 ) <br />

=4(1+4x) 4 (3 + x − x 2 ) 7 (2 + 4x − 16x 2 )+(15+5x − 5x 2 ) =4(1+4x) 4 (3 + x − x 2 ) 7 (17 + 9x − 21x 2 )

F.<br />

100 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

19. y =(2x − 5) 4 (8x 2 − 5) −3 ⇒<br />

y 0 =4(2x − 5) 3 (2)(8x 2 − 5) −3 +(2x − 5) 4 (−3)(8x 2 − 5) −4 (16x)<br />

=8(2x − 5) 3 (8x 2 − 5) −3 − 48x(2x − 5) 4 (8x 2 − 5) −4<br />

[This simplifies to 8(2x − 5) 3 (8x 2 − 5) −4 (−4x 2 +30x − 5).]<br />

21. y =<br />

x 2 3<br />

+1<br />

x 2 − 1<br />

⇒<br />

x<br />

y 0 2 2<br />

+1<br />

=3<br />

·<br />

x 2 − 1<br />

x 2 +1<br />

=3<br />

x 2 − 1<br />

d<br />

dx<br />

x 2 +1<br />

=3<br />

x 2 − 1<br />

23. y = e x cos x ⇒ y 0 = e x cos x ·<br />

25. F (z) =<br />

2<br />

· 2x[x2 − 1 − (x 2 +1)]<br />

(x 2 − 1) 2 =3<br />

1/2<br />

z − 1 z − 1<br />

z +1 = z +1<br />

F 0 (z) = 1 −1/2 z − 1<br />

·<br />

2 z +1<br />

d<br />

dz<br />

x 2 2<br />

+1<br />

· (x2 − 1)(2x) − (x 2 +1)(2x)<br />

x 2 − 1<br />

(x 2 − 1) 2<br />

x 2 2<br />

+1<br />

·<br />

x 2 − 1<br />

2x(−2)<br />

(x 2 − 1) = −12x(x2 +1) 2<br />

2 (x 2 − 1) 4<br />

d<br />

dx (x cos x) =ex cos x [x(− sin x)+(cosx) · 1] = e x cos x (cos x − x sin x)<br />

⇒<br />

z − 1<br />

= 1 z +1 2<br />

1/2 z +1<br />

·<br />

z − 1<br />

(z +1)(1)− (z − 1)(1)<br />

(z +1) 2<br />

= 1 (z +1) 1/2 z +1− z +1<br />

· = 1 (z +1) 1/2<br />

2 (z − 1)<br />

1/2<br />

(z +1) 2 2 (z − 1) · 2<br />

1/2 (z +1) = 1<br />

2 (z − 1) 1/2 (z +1) 3/2<br />

27. y =<br />

r<br />

√<br />

r2 +1<br />

⇒<br />

y 0 =<br />

√<br />

√<br />

r 2<br />

r2 +1(1)− r · 1<br />

2 (r2 +1) −1/2 r2 +1− √<br />

(2r)<br />

√<br />

r2 +1 r2 +1<br />

2<br />

= √<br />

r2 +1 2<br />

=<br />

√<br />

r2 +1 √ r 2 +1− r 2<br />

√<br />

r2 +1<br />

√<br />

r2 +1 2<br />

=<br />

r 2 +1 − r 2<br />

√<br />

r2 +1 1<br />

3<br />

=<br />

(r 2 +1) or 3/2 (r2 +1) −3/2<br />

Another solution: Write y as a product and make use of the Product Rule. y = r(r 2 +1) −1/2<br />

⇒<br />

y 0 = r · − 1 2 (r2 +1) −3/2 (2r)+(r 2 +1) −1/2 · 1=(r 2 +1) −3/2 [−r 2 +(r 2 +1) 1 ]=(r 2 +1) −3/2 (1) = (r 2 +1) −3/2 .<br />

The step that students usually have trouble with is factoring out (r 2 +1) −3/2 . But this is no different than factoring out x 2<br />

from x 2 + x 5 ; that is, we are just factoring out a factor with the smallest exponent that appears on it. In this case, − 3 2 is<br />

smaller than − 1 2 .<br />

29. y = sin(tan 2x) ⇒ y 0 =cos(tan2x) ·<br />

31. Using Formula 5 and the Chain Rule, y =2 sin πx ⇒<br />

y 0 =2 sin πx (ln 2) ·<br />

d<br />

dx (tan 2x) = cos(tan 2x) · d<br />

sec2 (2x) ·<br />

dx (2x) = 2 cos(tan 2x)sec2 (2x)<br />

d<br />

dx (sin πx) =2sin πx (ln 2) · cos πx · π =2 sin πx (π ln 2) cos πx

F.<br />

TX.10<br />

SECTION 3.4 THE CHAIN RULE ¤ 101<br />

33. y =sec 2 x +tan 2 x =(secx) 2 +(tanx) 2 ⇒<br />

y 0 =2(secx)(sec x tan x) + 2(tan x)(sec 2 x)=2sec 2 x tan x +2sec 2 x tan x =4sec 2 x tan x<br />

1 − e<br />

2x<br />

<br />

35. y =cos<br />

1+e 2x<br />

⇒<br />

1 − e<br />

y 0 2x<br />

<br />

= − sin<br />

·<br />

1+e 2x<br />

<br />

d 1 − e<br />

2x<br />

1 − e<br />

2x<br />

<br />

= − sin<br />

· (1 + e2x )(−2e 2x ) − (1 − e 2x )(2e 2x )<br />

dx 1+e 2x 1+e 2x (1 + e 2x ) 2<br />

1 − e<br />

2x<br />

1 − e<br />

2x<br />

<br />

= − sin<br />

· −2e2x (1 + e 2x )+(1− e 2x ) <br />

= − sin<br />

1+e 2x (1 + e 2x ) 2<br />

1+e 2x <br />

<br />

· −2e2x (2)<br />

(1 + e 2x ) = 4e 2x 1 − e<br />

2x<br />

<br />

2 (1 + e 2x ) · sin 2 1+e 2x<br />

37. y =cot 2 (sin θ) =[cot(sinθ)] 2 ⇒<br />

y 0 = 2[cot(sin θ)] ·<br />

d<br />

dθ [cot(sin θ)] = 2 cot(sin θ) · [− csc2 (sin θ) · cos θ] =−2cosθ cot(sin θ) csc 2 (sin θ)<br />

39. f(t) =tan(e t )+e tan t ⇒ f 0 (t) =sec 2 (e t ) · d<br />

dt (et )+e tan t · d<br />

dt (tan t) =sec2 (e t ) · e t + e tan t · sec 2 t<br />

<br />

41. f(t) =sin 2 e sin2 t<br />

= sin e<br />

t 2 sin2 ⇒<br />

<br />

f 0 (t)=2 sin e sin2 t<br />

· d<br />

<br />

=2sin e sin2 t<br />

cos<br />

<br />

dt sin<br />

<br />

e sin2 t<br />

<br />

e sin2 t<br />

<br />

=4sin e sin2 t<br />

cos e sin2 t<br />

e sin2t sin t cos t<br />

<br />

=2sin e sin2 t<br />

· cos e sin2 t<br />

· d<br />

· e sin2t · d<br />

<br />

dt sin2 t =2sin e sin2 t<br />

cos<br />

dt esin2 t<br />

<br />

e sin2 t<br />

<br />

e sin2t · 2sint cos t<br />

43. g(x) =(2ra rx + n) p ⇒<br />

g 0 (x) =p(2ra rx + n) p−1 ·<br />

45. y =cos sin(tan πx) = cos(sin(tan πx)) 1/2 ⇒<br />

y 0 = − sin(sin(tan πx)) 1/2 ·<br />

= − sin sin(tan πx)<br />

2 sin(tan πx)<br />

d<br />

dx (2rarx + n) =p(2ra rx + n) p−1 · 2ra rx (ln a) · r =2r 2 p(ln a)(2ra rx + n) p−1 a rx<br />

d<br />

dx (sin(tan πx))1/2 = − sin(sin(tan πx)) 1/2 · 1 (sin(tan d<br />

2 πx))−1/2 · (sin(tan πx))<br />

dx<br />

· cos(tan πx) ·<br />

= −π cos(tan πx)sec2 (πx)sin sin(tan πx)<br />

2 sin(tan πx)<br />

d<br />

dx tan πx = − sin sin(tan πx)<br />

2 · cos(tan πx) · sec 2 (πx) · π<br />

sin(tan πx)<br />

47. h(x) = √ x 2 +1 ⇒ h 0 (x) = 1 2 (x2 +1) −1/2 (2x) =<br />

x<br />

√<br />

x2 +1<br />

⇒<br />

h 00 (x) =<br />

√ <br />

<br />

1<br />

x2 +1· 1 − x<br />

2 (x2 +1) −1/2 (2x)<br />

√<br />

x2 +1 2<br />

=<br />

x 2 +1 −1/2 (x 2 +1)− x 2<br />

(x 2 +1) 1 =<br />

1<br />

(x 2 +1) 3/2

F.<br />

102 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

49. y = e αx sin βx ⇒ y 0 = e αx · β cos βx +sinβx · αe αx = e αx (β cos βx + α sin βx) ⇒<br />

y 00 = e αx (−β 2 sin βx + αβ cos βx)+(β cos βx + α sin βx) · αe αx<br />

= e αx (−β 2 sin βx + αβ cos βx + αβ cos βx + α 2 sin βx) =e αx (α 2 sin βx − β 2 sin βx +2αβ cos βx)<br />

= e αx (α 2 − β 2 )sinβx +2αβ cos βx <br />

51. y =(1+2x) 10 ⇒ y 0 =10(1+2x) 9 · 2=20(1+2x) 9 .<br />

At (0, 1), y 0 = 20(1 + 0) 9 =20, and an equation of the tangent line is y − 1=20(x − 0),ory =20x +1.<br />

53. y =sin(sinx) ⇒ y 0 =cos(sinx) · cos x. At (π, 0), y 0 =cos(sinπ) · cos π =cos(0)· (−1) = 1(−1) = −1,andan<br />

equation of the tangent line is y − 0=−1(x − π),ory = −x + π.<br />

55. (a) y =<br />

2<br />

⇒ y 0 = (1 + e−x )(0) − 2(−e −x ) 2e −x<br />

=<br />

1+e −x (1 + e −x ) 2 (1 + e −x ) . 2<br />

(b)<br />

At (0, 1), y 0 =<br />

2e 0<br />

(1 + e 0 ) 2 = 2(1)<br />

(1 + 1) 2 = 2 2 2 = 1 2 .Soanequationofthe<br />

tangent line is y − 1= 1 2 (x − 0) or y = 1 2 x +1.<br />

57. (a) f(x) =x √ 2 − x 2 = x(2 − x 2 ) 1/2 ⇒<br />

f 0 (x) =x · 1<br />

2 (2 − x2 ) −1/2 (−2x)+(2− x 2 ) 1/2 · 1=(2− x 2 ) −1/2 −x 2 +(2− x 2 ) = 2 − 2x2 √<br />

2 − x<br />

2<br />

(b)<br />

f 0 =0when f has a horizontal tangent line, f 0 is negative when f is<br />

decreasing, and f 0 is positive when f is increasing.<br />

59. For the tangent line to be horizontal, f 0 (x) =0. f(x) =2sinx +sin 2 x ⇒ f 0 (x) =2cosx +2sinx cos x =0 ⇔<br />

2cosx(1 + sin x) =0 ⇔ cos x =0or sin x = −1, sox = π 2 +2nπ or 3π 2<br />

+2nπ, wheren is any integer. Now<br />

f π<br />

2<br />

<br />

=3and f<br />

3π<br />

2<br />

where n is any integer.<br />

<br />

= −1, so the points on the curve with a horizontal tangent are<br />

π<br />

2 +2nπ, 3 and 3π<br />

2 +2nπ, −1 ,<br />

61. F (x) =f(g(x)) ⇒ F 0 (x) =f 0 (g(x)) · g 0 (x),soF 0 (5) = f 0 (g(5)) · g 0 (5) = f 0 (−2) · 6=4· 6=24<br />

63. (a) h(x) =f(g(x)) ⇒ h 0 (x) =f 0 (g(x)) · g 0 (x),soh 0 (1) = f 0 (g(1)) · g 0 (1) = f 0 (2) · 6=5· 6=30.<br />

(b) H(x) =g(f(x)) ⇒ H 0 (x) =g 0 (f(x)) · f 0 (x),soH 0 (1) = g 0 (f(1)) · f 0 (1) = g 0 (3) · 4=9· 4=36.<br />

65. (a) u(x) =f(g(x)) ⇒ u 0 (x) =f 0 (g(x))g 0 (x). Sou 0 (1) = f 0 (g(1))g 0 (1) = f 0 (3)g 0 (1). Tofind f 0 (3), note that f is<br />

linear from (2, 4) to (6, 3), so its slope is 3 − 4<br />

6 − 2 = − 1 4 .Tofind g0 (1), note that g is linear from (0, 6) to (2, 0), so its slope<br />

is 0 − 6<br />

2 − 0 = −3. Thus, f 0 (3)g 0 (1) = <br />

− 1 4 (−3) =<br />

3<br />

. 4<br />

(b) v(x) =g(f(x)) ⇒ v 0 (x) =g 0 (f(x))f 0 (x). Sov 0 (1) = g 0 (f(1))f 0 (1) = g 0 (2)f 0 (1), which does not exist since<br />

g 0 (2) does not exist.

F.<br />

TX.10<br />

SECTION 3.4 THE CHAIN RULE ¤ 103<br />

(c) w(x) =g(g(x)) ⇒ w 0 (x) =g 0 (g(x))g 0 (x). Sow 0 (1) = g 0 (g(1))g 0 (1) = g 0 (3)g 0 (1). Tofind g 0 (3), note that g is<br />

linear from (2, 0) to (5, 2),soitsslopeis 2 − 0<br />

5 − 2 = 2 3 . Thus, g0 (3)g 0 (1) = 2<br />

3<br />

<br />

(−3) = −2.<br />

67. (a) F (x) =f(e x ) ⇒ F 0 (x) =f 0 (e x ) d<br />

dx (ex )=f 0 (e x )e x<br />

(b) G(x) =e f(x) ⇒ G 0 (x) =e f(x) d<br />

dx f(x) =ef(x) f 0 (x)<br />

69. r(x) =f(g(h(x))) ⇒ r 0 (x) =f 0 (g(h(x))) · g 0 (h(x)) · h 0 (x),so<br />

r 0 (1) = f 0 (g(h(1))) · g 0 (h(1)) · h 0 (1) = f 0 (g(2)) · g 0 (2) · 4=f 0 (3) · 5 · 4=6· 5 · 4=120<br />

71. F (x) =f(3f(4f(x))) ⇒<br />

F 0 (x)=f 0 (3f(4f(x))) ·<br />

d<br />

dx (3f(4f(x))) = f 0 (3f(4f(x))) · 3f 0 d<br />

(4f(x)) ·<br />

dx (4f(x))<br />

= f 0 (3f(4f(x))) · 3f 0 (4f(x)) · 4f 0 (x), so<br />

F 0 (0) = f 0 (3f(4f(0))) · 3f 0 (4f(0)) · 4f 0 (0) = f 0 (3f(4 · 0)) · 3f 0 (4 · 0) · 4 · 2=f 0 (3 · 0) · 3 · 2 · 4 · 2=2· 3 · 2 · 4 · 2=96.<br />

73. y = Ae −x + Bxe −x ⇒<br />

y 0 = A(−e −x )+B[x(−e −x )+e −x · 1] = −Ae −x + Be −x − Bxe −x =(B − A)e −x − Bxe −x<br />

⇒<br />

y 00 =(B − A)(−e −x ) − B[x(−e −x )+e −x · 1] = (A − B)e −x − Be −x + Bxe −x =(A − 2B)e −x + Bxe −x ,<br />

so<br />

y 00 +2y 0 + y =(A − 2B)e −x + Bxe −x +2[(B − A)e −x − Bxe −x ]+Ae −x + Bxe −x<br />

=[(A − 2B)+2(B − A)+A]e −x +[B − 2B + B]xe −x =0.<br />

75. The use of D, D 2 , ..., D n is just a derivative notation (see text page 157). In general, Df(2x) =2f 0 (2x),<br />

D 2 f(2x) =4f 00 (2x), ..., D n f(2x) =2 n f (n) (2x). Sincef(x) =cosx and 50 = 4(12) + 2, wehave<br />

f (50) (x) =f (2) (x) =− cos x,soD 50 cos 2x = −2 50 cos 2x.<br />

77. s(t) =10+ 1 sin(10πt) ⇒ the velocity after t seconds is v(t) 4 =s0 (t) = 1 cos(10πt)(10π) = 5π cos(10πt) cm/s.<br />

4 2<br />

79. (a) B(t) =4.0+0.35 sin 2πt ⇒<br />

dB <br />

5.4 dt = 0.35 cos 2πt 2π<br />

= 0.7π 2πt<br />

cos<br />

5.4 5.4 5.4 5.4 = 7π 2πt<br />

cos<br />

54 5.4<br />

(b) At t =1, dB<br />

dt = 7π 2π<br />

cos<br />

54 5.4 ≈ 0.16.<br />

81. s(t) =2e −1.5t sin 2πt ⇒<br />

v(t) =s 0 (t) =2[e −1.5t (cos 2πt)(2π)+(sin2πt)e −1.5t (−1.5)] = 2e −1.5t (2π cos 2πt − 1.5sin2πt)

F.<br />

104 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

83. By the Chain Rule, a(t) = dv<br />

dt = dv ds<br />

ds dt = dv<br />

dv<br />

v(t) =v(t)<br />

ds ds .<br />

Thederivativedv/dt is the rate of change of the velocity<br />

with respect to time (in other words, the acceleration) whereas the derivative dv/ds is the rate of change of the velocity with<br />

respect to the displacement.<br />

85. (a) Using a calculator or CAS, we obtain the model Q = ab t with a ≈ 100.0124369 and b ≈ 0.000045145933.<br />

(b) Use Q 0 (t) =ab t ln b (from Formula 5) with the values of a and b from part (a) to get Q 0 (0.04) ≈−670.63 μA.<br />

The result of Example 2 in Section 2.1 was −670 μA.<br />

87. (a) Derive gives g 0 (t) =<br />

45(t − 2)8<br />

without simplifying. With either Maple or Mathematica, we first get<br />

(2t +1)<br />

10<br />

g 0 (t − 2)8 (t − 2)9<br />

(t) =9 − 18 , and the simplification command results in the expression given by Derive.<br />

(2t +1)<br />

9<br />

(2t +1)<br />

10<br />

(b) Derive gives y 0 =2(x 3 − x +1) 3 (2x +1) 4 (17x 3 +6x 2 − 9x +3)without simplifying. With either Maple or<br />

Mathematica, we first get y 0 =10(2x +1) 4 (x 3 − x +1) 4 +4(2x +1) 5 (x 3 − x +1) 3 (3x 2 − 1). Ifweuse<br />

Mathematica’s Factor or Simplify,orMaple’sfactor, we get the above expression, but Maple’s simplify gives<br />

the polynomial expansion instead. For locating horizontal tangents, the factored form is the most helpful.<br />

89. (a) If f is even, then f(x) =f(−x). Using the Chain Rule to differentiate this equation, we get<br />

f 0 (x) =f 0 (−x) d<br />

dx (−x) =−f 0 (−x). Thus,f 0 (−x) =−f 0 (x),sof 0 is odd.<br />

(b) If f is odd, then f(x) =−f(−x).<br />

even.<br />

91. (a)<br />

(b)<br />

d<br />

dx (sinn x cos nx) =n sin n−1 x cos x cos nx +sin n x (−n sin nx)<br />

Differentiating this equation, we get f 0 (x) =−f 0 (−x)(−1) = f 0 (−x),sof 0 is<br />

[Product Rule]<br />

= n sin n−1 x (cos nx cos x − sin nx sin x) [factor out n sin n−1 x]<br />

= n sin n−1 x cos(nx + x) [Addition Formula for cosine]<br />

= n sin n−1 x cos[(n +1)x] [factor out x]<br />

d<br />

dx (cosn x cos nx) =n cos n−1 x (− sin x)cosnx +cos n x (−n sin nx)<br />

[Product Rule]<br />

= −n cos n−1 x (cos nx sin x +sinnx cos x) [factor out −n cos n−1 x]<br />

= −n cos n−1 x sin(nx + x) [Addition Formula for sine]<br />

= −n cos n−1 x sin[(n +1)x] [factor out x]<br />

93. Since θ ◦ = <br />

π<br />

d<br />

180 θ rad, we have<br />

dθ (sin θ◦ )= d <br />

sin<br />

π<br />

180<br />

dθ<br />

θ = π cos π θ = π cos 180 180 180 θ◦ .<br />

95. The Chain Rule says that dy<br />

dx = dy<br />

du<br />

dy<br />

d 2 y<br />

dx = d<br />

2 dx<br />

dx<br />

du<br />

dx ,so<br />

<br />

= d<br />

dx<br />

dy du<br />

du dx<br />

d<br />

=<br />

dx<br />

dy du<br />

du<br />

dx + dy d<br />

du dx<br />

du<br />

dx<br />

[Product Rule]<br />

d<br />

=<br />

du<br />

dy<br />

du<br />

du du<br />

dx dx + dy<br />

<br />

d 2 u<br />

du dx = d2 y du<br />

2 du 2 dx<br />

2<br />

+ dy d 2 u<br />

du dx 2

F.<br />

TX.10<br />

SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 105<br />

3.5 Implicit Differentiation<br />

d<br />

1. (a)<br />

dx (xy +2x +3x2 )= d<br />

dx (4) ⇒ (x · y0 + y · 1)+2+6x =0 ⇒ xy 0 = −y − 2 − 6x ⇒<br />

y 0 −y − 2 − 6x<br />

= or y 0 = −6 − y +2<br />

x<br />

x .<br />

(b) xy +2x +3x 2 =4 ⇒ xy =4− 2x − 3x 2 ⇒ y =<br />

3. (a)<br />

5.<br />

(c) From part (a), y 0 =<br />

−y − 2 − 6x<br />

x<br />

=<br />

−(4/x − 2 − 3x) − 2 − 6x<br />

x<br />

4 − 2x − 3x2<br />

x<br />

=<br />

= 4 x − 2 − 3x,soy0 = − 4 x 2 − 3.<br />

−4/x − 3x<br />

x<br />

= − 4 x 2 − 3.<br />

<br />

d 1<br />

dx x + 1 <br />

= d<br />

y dx (1) ⇒ − 1 x − 1<br />

2 y 2 y0 =0 ⇒ − 1 y 2 y0 = 1 ⇒ y 0 = − y2<br />

x 2 x 2<br />

(b) 1 x + 1 y =1 ⇒ 1 y =1− 1 x = x − 1<br />

x<br />

(c) y 0 = − y2<br />

2<br />

− 1)]<br />

= −[x/(x<br />

x2 x 2<br />

x 2<br />

⇒ y = x (x − 1)(1) − (x)(1)<br />

x − 1 ,soy0 = = −1<br />

(x − 1) 2 (x − 1) . 2<br />

= −<br />

x 2 (x − 1) = − 1<br />

2 (x − 1) 2<br />

d <br />

x 3 + y 3 = d<br />

dx<br />

dx (1) ⇒ 3x2 +3y 2 · y 0 =0 ⇒ 3y 2 y 0 = −3x 2 ⇒ y 0 = − x2<br />

y 2<br />

7.<br />

d<br />

dx (x2 + xy − y 2 )= d<br />

dx (4) ⇒ 2x + x · y0 + y · 1 − 2yy 0 =0 ⇒<br />

xy 0 − 2yy 0 = −2x − y ⇒ (x − 2y) y 0 = −2x − y ⇒ y 0 = −2x − y<br />

x − 2y<br />

= 2x + y<br />

2y − x<br />

9.<br />

11.<br />

13.<br />

15.<br />

d <br />

x 4 (x + y) = d <br />

y 2 (3x − y) ⇒ x 4 (1 + y 0 )+(x + y) · 4x 3 = y 2 (3 − y 0 )+(3x − y) · 2yy 0 ⇒<br />

dx<br />

dx<br />

x 4 + x 4 y 0 +4x 4 +4x 3 y =3y 2 − y 2 y 0 +6xy y 0 − 2y 2 y 0 ⇒ x 4 y 0 +3y 2 y 0 − 6xy y 0 =3y 2 − 5x 4 − 4x 3 y ⇒<br />

(x 4 +3y 2 − 6xy) y 0 =3y 2 − 5x 4 − 4x 3 y ⇒ y 0 = 3y2 − 5x 4 − 4x 3 y<br />

x 4 +3y 2 − 6xy<br />

d<br />

dx (x2 y 2 + x sin y) = d<br />

dx (4) ⇒ x2 · 2yy 0 + y 2 · 2x + x cos y · y 0 +siny · 1=0 ⇒<br />

2x 2 yy 0 + x cos y · y 0 = −2xy 2 − sin y ⇒ (2x 2 y + x cos y)y 0 = −2xy 2 − sin y ⇒ y 0 = −2xy2 − sin y<br />

2x 2 y + x cos y<br />

d<br />

(4 cos x sin y) =<br />

d<br />

dx<br />

dx (1) ⇒ 4[cosx · cos y · y0 +siny · (− sin x)] = 0 ⇒<br />

y 0 (4 cos x cos y) =4sinx sin y ⇒ y 0 =<br />

d<br />

dx (ex/y )=<br />

d<br />

dx (x − y) ⇒ ex/y ·<br />

4sinx sin y<br />

=tanx tan y<br />

4cosx cos y<br />

<br />

d x<br />

=1− y 0 ⇒<br />

dx y<br />

e x/y · y · 1 − x · y0<br />

=1− y 0 ⇒ e x/y · 1<br />

y 2 y − xex/y<br />

· y 0 =1− y 0 ⇒ y 0 − xex/y<br />

· y 0 =1− ex/y<br />

y 2 y 2<br />

y<br />

⇒<br />

<br />

y 0 1 − xex/y<br />

= y − ex/y<br />

y 2 y<br />

⇒ y 0 =<br />

y − e x/y<br />

y<br />

= y(y − ex/y )<br />

y 2 − xe x/y y 2 − xe x/y<br />

y 2

F.<br />

106 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

17.<br />

<br />

xy =1+x 2 y ⇒ 1 2 (xy)−1/2 (xy 0 + y · 1) = 0 + x 2 y 0 + y · 2x ⇒<br />

19.<br />

21.<br />

23.<br />

<br />

<br />

y 0 x<br />

2 xy − x2 =2xy −<br />

y<br />

2 xy<br />

⇒<br />

<br />

y 0 x − 2x 2 <br />

xy<br />

2 xy<br />

= 4xy xy − y<br />

2 xy<br />

x<br />

2 xy y0 + y<br />

2 xy = x2 y 0 +2xy ⇒<br />

⇒<br />

y 0 = 4xy xy − y<br />

x − 2x 2 xy<br />

d<br />

dx (ey cos x) = d<br />

dx [1 + sin(xy)] ⇒ ey (− sin x)+cosx · e y · y 0 =cos(xy) · (xy 0 + y · 1) ⇒<br />

−e y sin x + e y cos x · y 0 = x cos(xy) · y 0 + y cos(xy) ⇒ e y cos x · y 0 − x cos(xy) · y 0 = e y sin x + y cos(xy) ⇒<br />

[e y cos x − x cos(xy)] y 0 = e y sin x + y cos(xy) ⇒ y 0 = ey sin x + y cos(xy)<br />

e y cos x − x cos(xy)<br />

d <br />

f(x)+x 2 [f(x)] 3 = d<br />

dx<br />

dx (10) ⇒ f 0 (x)+x 2 · 3[f(x)] 2 · f 0 (x)+[f(x)] 3 · 2x =0. Ifx =1,wehave<br />

f 0 (1) + 1 2 · 3[f(1)] 2 · f 0 (1) + [f(1)] 3 · 2(1) = 0 ⇒ f 0 (1) + 1 · 3 · 2 2 · f 0 (1) + 2 3 · 2=0 ⇒<br />

f 0 (1) + 12f 0 (1) = −16 ⇒ 13f 0 (1) = −16 ⇒ f 0 (1) = − 16<br />

13 .<br />

d<br />

dy (x4 y 2 − x 3 y +2xy 3 )= d<br />

dy (0) ⇒ x4 · 2y + y 2 · 4x 3 x 0 − (x 3 · 1+y · 3x 2 x 0 )+2(x · 3y 2 + y 3 · x 0 )=0 ⇒<br />

4x 3 y 2 x 0 − 3x 2 yx 0 +2y 3 x 0 = −2x 4 y + x 3 − 6xy 2 ⇒ (4x 3 y 2 − 3x 2 y +2y 3 ) x 0 = −2x 4 y + x 3 − 6xy 2 ⇒<br />

x 0 = dx<br />

dy = −2x4 y + x 3 − 6xy 2<br />

4x 3 y 2 − 3x 2 y +2y 3<br />

25. x 2 + xy + y 2 =3 ⇒ 2x + xy 0 + y · 1+2yy 0 =0 ⇒ xy 0 +2yy 0 = −2x − y ⇒ y 0 (x +2y) =−2x − y ⇒<br />

y 0 = −2x − y<br />

x +2y . Whenx =1and y =1,wehavey0 = −2 − 1<br />

1+2· 1 = −3 = −1, so an equation of the tangent line is<br />

3<br />

y − 1=−1(x − 1) or y = −x +2.<br />

27. x 2 + y 2 =(2x 2 +2y 2 − x) 2 ⇒ 2x +2yy 0 =2(2x 2 +2y 2 − x)(4x +4yy 0 − 1). Whenx =0and y = 1 ,wehave<br />

2<br />

0+y 0 =2( 1 2 )(2y0 − 1) ⇒ y 0 =2y 0 − 1 ⇒ y 0 =1, so an equation of the tangent line is y − 1 =1(x − 0)<br />

2<br />

or y = x + 1 . 2<br />

29. 2(x 2 + y 2 ) 2 =25(x 2 − y 2 ) ⇒ 4(x 2 + y 2 )(2x +2yy 0 ) = 25(2x − 2yy 0 ) ⇒<br />

4(x + yy 0 )(x 2 + y 2 )=25(x − yy 0 ) ⇒ 4yy 0 (x 2 + y 2 )+25yy 0 =25x − 4x(x 2 + y 2 ) ⇒<br />

y 0 = 25x − 4x(x2 + y 2 )<br />

25y +4y(x 2 + y 2 ) . Whenx =3and y =1,wehavey0 75 − 120<br />

= = − 45 = − 9 ,<br />

25 + 40 65 13<br />

so an equation of the tangent line is y − 1=− 9<br />

13 (x − 3) or y = − 9 13 x + 40<br />

13 .<br />

31. (a) y 2 =5x 4 − x 2 ⇒ 2yy 0 =5(4x 3 ) − 2x ⇒ y 0 = 10x3 − x<br />

.<br />

y<br />

(b)<br />

So at the point (1, 2) we have y 0 = 10(1)3 − 1<br />

2<br />

of the tangent line is y − 2= 9 2 (x − 1) or y = 9 2 x − 5 2 .<br />

= 9 , and an equation<br />

2

F.<br />

TX.10<br />

SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 107<br />

33. 9x 2 + y 2 =9 ⇒ 18x +2yy 0 =0 ⇒ 2yy 0 = −18x ⇒ y 0 = −9x/y ⇒<br />

<br />

y · 1 − x · y<br />

y 00 0 y − x(−9x/y)<br />

= −9<br />

= −9<br />

= −9 · y2 +9x 2 9<br />

= −9 · [since x and y must satisfy the original<br />

y 2<br />

y 2<br />

y 3 y 3<br />

equation, 9x 2 + y 2 =9].Thus,y 00 = −81/y 3 .<br />

35. x 3 + y 3 =1 ⇒ 3x 2 +3y 2 y 0 =0 ⇒ y 0 = − x2<br />

y 2<br />

y 00 = − y2 (2x) − x 2 · 2yy 0<br />

= − 2xy2 − 2x 2 y(−x 2 /y 2 )<br />

= − 2xy4 +2x 4 y<br />

= − 2xy(y3 + x 3 )<br />

= − 2x<br />

(y 2 ) 2 y 4<br />

y 6 y 6 y , 5<br />

since x and y must satisfy the original equation, x 3 + y 3 =1.<br />

37. (a) There are eight points with horizontal tangents: four at x ≈ 1.57735 and<br />

four at x ≈ 0.42265.<br />

(b) y 0 =<br />

3x 2 − 6x +2<br />

2(2y 3 − 3y 2 − y +1)<br />

⇒<br />

⇒ y 0 = −1 at (0, 1) and y 0 = 1 at (0, 2).<br />

3<br />

Equations of the tangent lines are y = −x +1and y = 1 3 x +2.<br />

(c) y 0 =0 ⇒ 3x 2 − 6x +2=0 ⇒ x =1± 1 3<br />

√<br />

3<br />

(d) By multiplying the right side of the equation by x − 3, we obtain the first<br />

graph. By modifying the equation in other ways, we can generate the other<br />

graphs.<br />

y(y 2 − 1)(y − 2)<br />

= x(x − 1)(x − 2)(x − 3)<br />

y(y 2 − 4)(y − 2)<br />

= x(x − 1)(x − 2)<br />

y(y +1)(y 2 − 1)(y − 2)<br />

= x(x − 1)(x − 2)<br />

(y +1)(y 2 − 1)(y − 2)<br />

=(x − 1)(x − 2)<br />

x(y +1)(y 2 − 1)(y − 2)<br />

= y(x − 1)(x − 2)<br />

y(y 2 +1)(y − 2)<br />

= x(x 2 − 1)(x − 2)<br />

y(y +1)(y 2 − 2)<br />

= x(x − 1)(x 2 − 2)

F.<br />

108 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

39. From Exercise 29, a tangent to the lemniscate will be horizontal if y 0 =0 ⇒ 25x − 4x(x 2 + y 2 )=0 ⇒<br />

x[25 − 4(x 2 + y 2 )] = 0 ⇒ x 2 + y 2 = 25 4<br />

(1). (Note that when x is 0, y is also 0, and there is no horizontal tangent<br />

at the origin.) Substituting 25 for 4<br />

x2 + y 2 in the equation of the lemniscate, 2(x 2 + y 2 ) 2 =25(x 2 − y 2 ),weget<br />

<br />

x 2 − y 2 = 25 (2). Solving (1)and(2), we have x 2 = 75 and 8 16 y2 = 25 , so the four points are ± 5√ 3<br />

, ± 5 .<br />

16 4 4<br />

41.<br />

x 2<br />

a − y2<br />

2x<br />

=1 ⇒ 2 b2 a − 2yy0 =0 ⇒ y 0 = b2 x<br />

2 b 2 a 2 y<br />

⇒<br />

an equation of the tangent line at (x 0 ,y 0 ) is<br />

y − y 0 = b2 x 0<br />

(x − x<br />

a 2 0 ). Multiplying both sides by y0 y0y<br />

gives<br />

y 0 b2 b − y2 0<br />

2 b 2<br />

= x0x<br />

a 2 − x2 0<br />

a 2 .Since(x 0,y 0 ) lies on the hyperbola,<br />

we have x0x<br />

a 2<br />

− y0y<br />

b 2 = x2 0<br />

a 2 − y2 0<br />

b 2 =1.<br />

43. If the circle has radius r, its equation is x 2 + y 2 = r 2 ⇒ 2x +2yy 0 =0 ⇒ y 0 = − x , so the slope of the tangent line<br />

y<br />

at P (x 0 ,y 0 ) is − x 0<br />

−1<br />

. The negative reciprocal of that slope is = y 0<br />

, which is the slope of OP, so the tangent line at<br />

y 0 −x 0 /y 0 x 0<br />

P is perpendicular to the radius OP.<br />

45. y =tan √ −1 x ⇒ y 0 1<br />

= √ 2 ·<br />

1+ x<br />

d<br />

√ <br />

x = 1<br />

dx 1+x<br />

<br />

1<br />

2 x−1/2 <br />

=<br />

1<br />

2 √ x (1 + x)<br />

47. y =sin −1 (2x +1) ⇒<br />

y 0 =<br />

1<br />

· d<br />

1 − (2x +1)<br />

2 dx (2x +1)= 1<br />

<br />

1 − (4x2 +4x +1) · 2= 2<br />

√<br />

−4x2 − 4x = 1<br />

√<br />

−x2 − x<br />

49. G(x) = √ 1 − x 2 arccos x ⇒ G 0 (x) = √ 1 − x 2 ·<br />

51. h(t) =cot −1 (t) +cot −1 (1/t) ⇒<br />

h 0 (t) =− 1<br />

1+t − 1<br />

2 1+(1/t) · d 1<br />

2 dt t = − 1<br />

1+t − t2<br />

2 t 2 +1 ·<br />

−1<br />

√ +arccosx · 1<br />

1 − x<br />

2 2 (1 − x2 ) −1/2 (−2x) =−1 − x √ arccos x<br />

1 − x<br />

2<br />

− 1 t 2 <br />

= − 1<br />

1+t 2 + 1<br />

t 2 +1 =0.<br />

Note that this makes sense because h(t) = π 2 for t>0 and h(t) =3π 2<br />

for t

F.<br />

TX.10<br />

SECTION 3.5 IMPLICIT DIFFERENTIATION ¤ 109<br />

57. Let y =cos −1 x.Thencos y = x and 0 ≤ y ≤ π ⇒ −sin y dy<br />

dx =1<br />

dy<br />

dx = − 1<br />

sin y = − 1<br />

<br />

1 − cos2 y = − 1<br />

√<br />

1 − x<br />

2 .<br />

⇒<br />

[Notethatsin y ≥ 0 for 0 ≤ y ≤ π.]<br />

59. x 2 + y 2 = r 2 is a circle with center O and ax + by =0is a line through O [assume a<br />

and b are not both zero]. x 2 + y 2 = r 2 ⇒ 2x +2yy 0 =0 ⇒ y 0 = −x/y,sothe<br />

slope of the tangent line at P 0 (x 0,y 0) is −x 0/y 0. The slope of the line OP 0 is y 0/x 0,<br />

which is the negative reciprocal of −x 0/y 0. Hence, the curves are orthogonal, and the<br />

families of curves are orthogonal trajectories of each other.<br />

61. y = cx 2 ⇒ y 0 =2cx and x 2 +2y 2 = k [assume k>0] ⇒ 2x +4yy 0 =0 ⇒<br />

2yy 0 = −x ⇒ y 0 = − x<br />

2(y) = − x<br />

2(cx 2 ) = − 1 , so the curves are orthogonal if<br />

2cx<br />

c 6= 0.Ifc =0, then the horizontal line y = cx 2 =0intersects x 2 +2y 2 = k orthogonally<br />

<br />

at ± √ <br />

k, 0 , since the ellipse x 2 +2y 2 = k has vertical tangents at those two points.<br />

63. To find the points at which the ellipse x 2 − xy + y 2 =3crosses the x-axis, let y =0and solve for x.<br />

y =0 ⇒ x 2 − x(0) + 0 2 =3 ⇔ x = ± √ 3. So the graph of the ellipse crosses the x-axis at the points ± √ 3, 0 .<br />

Using implicit differentiation to find y 0 ,weget2x − xy 0 − y +2yy 0 =0 ⇒ y 0 (2y − x) =y − 2x ⇔ y 0 = y − 2x<br />

2y − x .<br />

So y 0 at √ 3, 0 is 0 − 2 √ 3<br />

2(0) − √ 3 =2and y0 at − √ 3, 0 is 0+2√ 3<br />

2(0) + √ =2. Thus, the tangent lines at these points are parallel.<br />

3<br />

65. x 2 y 2 + xy =2 ⇒ x 2 · 2yy 0 + y 2 · 2x + x · y 0 + y · 1=0 ⇔ y 0 (2x 2 y + x) =−2xy 2 − y ⇔<br />

y 0 = − 2xy2 + y<br />

2x 2 y + x .So− 2xy2 + y<br />

2x 2 y + x = −1 ⇔ 2xy2 + y =2x 2 y + x ⇔ y(2xy +1)=x(2xy +1) ⇔<br />

y(2xy +1)− x(2xy +1)=0 ⇔ (2xy +1)(y − x) =0 ⇔ xy = − 1 2 or y = x.Butxy = − 1 2<br />

⇒<br />

x 2 y 2 + xy = 1 4 − 1 2 6=2,sowemusthavex = y. Then x2 y 2 + xy =2 ⇒ x 4 + x 2 =2 ⇔ x 4 + x 2 − 2=0 ⇔<br />

(x 2 +2)(x 2 − 1) = 0. Sox 2 = −2, which is impossible, or x 2 =1 ⇔ x = ±1. Sincex = y, the points on the curve<br />

where the tangent line has a slope of −1 are (−1, −1) and (1, 1).<br />

67. (a) If y = f −1 (x),thenf(y) =x. Differentiating implicitly with respect to x and remembering that y is a function of x,<br />

we get f 0 (y) dy dy<br />

=1,so<br />

dx dx = 1<br />

f 0 (y)<br />

⇒ f −10 (x) =<br />

1<br />

f 0 (f −1 (x)) .<br />

(b) f(4) = 5 ⇒ f −1 (5) = 4. Bypart(a), f −10 (5) =<br />

1<br />

f 0 (f −1 (5)) = 1<br />

f 0 (4) =1 <br />

2<br />

3 =<br />

3<br />

. 2

F.<br />

110 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

69. x 2 +4y 2 =5 ⇒ 2x +4(2yy 0 )=0 ⇒ y 0 = − x .Nowleth be the height of the lamp, and let (a, b) be the point of<br />

4y<br />

tangency of the line passing through the points (3,h) and (−5, 0). Thislinehasslope(h − 0)/[3 − (−5)] = 1 8 h.Butthe<br />

slope of the tangent line through the point (a, b) can be expressed as y 0 = − a 4b ,oras b − 0<br />

a − (−5) = b [since the line<br />

a +5<br />

passes through (−5, 0) and (a, b)], so − a 4b =<br />

b<br />

a +5<br />

⇔ 4b 2 = −a 2 − 5a ⇔ a 2 +4b 2 = −5a. Buta 2 +4b 2 =5<br />

[since (a, b) is on the ellipse], so 5=−5a ⇔ a = −1. Then4b 2 = −a 2 − 5a = −1 − 5(−1) = 4 ⇒ b =1, since the<br />

point is on the top half of the ellipse. So h 8 =<br />

x-axis.<br />

b<br />

a +5 = 1<br />

−1+5 = 1 4<br />

⇒<br />

h =2. So the lamp is located 2 units above the<br />

3.6 Derivatives of Logarithmic Functions<br />

1. The differentiation formula for logarithmic functions,<br />

3. f(x) =sin(lnx) ⇒ f 0 (x) =cos(lnx) ·<br />

5. f(x) =log 2 (1 − 3x) ⇒ f 0 (x) =<br />

d<br />

dx (log a x) = 1 ,issimplestwhena = e because ln e =1.<br />

x ln a<br />

d<br />

1 cos(ln x)<br />

ln x =cos(lnx) · =<br />

dx x x<br />

1<br />

(1 − 3x)ln2<br />

d<br />

−3<br />

(1 − 3x) =<br />

dx<br />

(1 − 3x) ln2 or 3<br />

(3x − 1) ln 2<br />

7. f(x) = 5√ ln x =(lnx) 1/5 ⇒ f 0 (x) = 1 5 (ln x)−4/5 d<br />

dx (ln x) = 1<br />

5(ln x) 4/5 · 1<br />

x = 1<br />

5x 5 (ln x) 4<br />

9. f(x) =sinx ln(5x) ⇒ f 0 (x) =sinx ·<br />

11. F (t) =ln<br />

F 0 (t) =3·<br />

1<br />

5x · d<br />

sin x · 5<br />

(5x)+ln(5x) · cos x = +cosx ln(5x) = sin x +cosx ln(5x)<br />

dx 5x<br />

x<br />

(2t +1)3<br />

(3t − 1) 4 =ln(2t +1)3 − ln(3t − 1) 4 =3ln(2t +1)− 4ln(3t − 1) ⇒<br />

1<br />

2t +1 · 2 − 4 · 1<br />

3t − 1 · 3= 6<br />

2t +1 − 12<br />

3t − 1 , or combined, −6(t +3)<br />

(2t +1)(3t − 1) .<br />

13. g(x) =ln x √ x 2 − 1 =lnx +ln(x 2 − 1) 1/2 =lnx + 1 2 ln(x2 − 1) ⇒<br />

g 0 (x) = 1 x + 1 2 ·<br />

1<br />

x 2 − 1 · 2x = 1 x + x<br />

x 2 − 1 = x2 − 1+x · x<br />

= 2x2 − 1<br />

x(x 2 − 1) x(x 2 − 1)<br />

15. f(u) =<br />

ln u<br />

1+ln(2u)<br />

⇒<br />

f 0 (u) =<br />

[1 + ln(2u)] · 1<br />

u − ln u · 1<br />

2u · 2<br />

[1 + ln(2u)] 2 =<br />

17. y =ln 2 − x − 5x 2 ⇒ y 0 =<br />

1<br />

u<br />

[1 + ln(2u) − ln u] 1+(ln2+lnu) − ln u<br />

= =<br />

[1 + ln(2u)] 2 u[1 + ln(2u)] 2<br />

1<br />

−10x − 1<br />

· (−1 − 10x) =<br />

2 − x − 5x2 2 − x − 5x or 10x +1<br />

2 5x 2 + x − 2<br />

19. y =ln(e −x + xe −x )=ln(e −x (1 + x)) = ln(e −x )+ln(1+x) =−x +ln(1+x) ⇒<br />

y 0 = −1+ 1 −1 − x +1<br />

= = − x<br />

1+x 1+x 1+x<br />

1+ln2<br />

u[1 + ln(2u)] 2

F.<br />

TX.10<br />

SECTION 3.6 DERIVATIVES OF LOGARITHMIC FUNCTIONS ¤ 111<br />

√<br />

21. y =2x log 10 x =2x log10 x 1/2 =2x · 1 log 2 10 x = x log 10 x ⇒ y 0 1<br />

= x ·<br />

x ln 10 +log 10 x · 1= 1<br />

ln 10 +log 10 x<br />

Note:<br />

1<br />

ln 10 = ln e<br />

ln 10 =log 10 e, so the answer could be written as 1<br />

ln 10 +log 10 x =log 10 e +log 10 x =log 10 ex.<br />

23. y = x 2 ln(2x) ⇒ y 0 = x 2 ·<br />

y 00 =1+2x ·<br />

1 · 2+ln(2x) · (2x) =x +2x ln(2x)<br />

2x ⇒<br />

1 · 2+ln(2x) · 2=1+2+2ln(2x) =3+2ln(2x)<br />

2x<br />

25. y =ln x + √ 1+x 2 ⇒<br />

y 0 1<br />

=<br />

x + √ 1+x 2<br />

=<br />

<br />

1<br />

x + √ 1+<br />

1+x 2<br />

d √ <br />

x + 1+x<br />

2<br />

=<br />

dx<br />

<br />

x<br />

√ = 1+x<br />

2<br />

1<br />

<br />

<br />

x + √ 1+ 1<br />

1+x (1 + 2 2 x2 ) −1/2 (2x)<br />

1<br />

x + √ 1+x 2 ·<br />

√<br />

1+x2 + x<br />

√<br />

1+x<br />

2<br />

=<br />

1<br />

√<br />

1+x<br />

2<br />

⇒<br />

y 00 = − 1 2 (1 + x2 ) −3/2 (2x) =<br />

−x<br />

(1 + x 2 ) 3/2<br />

27. f(x) =<br />

x<br />

1 − ln(x − 1)<br />

⇒<br />

−1<br />

[1 − ln(x − 1)] · 1 − x ·<br />

f 0 (x) =<br />

x − 1<br />

[1 − ln(x − 1)] 2 =<br />

=<br />

2x − 1 − (x − 1) ln(x − 1)<br />

(x − 1)[1 − ln(x − 1)] 2<br />

(x − 1)[1 − ln(x − 1)] + x<br />

x − 1<br />

x − 1 − (x − 1) ln(x − 1) + x<br />

=<br />

[1 − ln(x − 1)] 2 (x − 1)[1 − ln(x − 1)] 2<br />

Dom(f) ={x | x − 1 > 0 and 1 − ln(x − 1) 6= 0} = {x | x>1 and ln(x − 1) 6= 1}<br />

= x | x>1 and x − 1 6=e 1 = {x | x>1 and x 6= 1+e} =(1, 1+e) ∪ (1 + e, ∞)<br />

29. f(x) =ln(x 2 − 2x) ⇒ f 0 (x) =<br />

1<br />

2(x − 1)<br />

(2x − 2) =<br />

x 2 − 2x x(x − 2) .<br />

Dom(f) ={x | x(x − 2) > 0} =(−∞, 0) ∪ (2, ∞).<br />

31. f(x) = ln x ⇒ f 0 (x) = x2 (1/x) − (ln x)(2x) x − 2x ln x x(1 − 2lnx)<br />

= = = 1 − 2lnx ,<br />

x 2 (x 2 ) 2 x 4<br />

x 4<br />

x 3<br />

so f 0 (1) = 1 − 2ln1 = 1 − 2 · 0 =1.<br />

1 3 1<br />

<br />

33. y =ln xe x2 =lnx +lne x2 =lnx + x 2 ⇒ y 0 = 1 +2x. At(1, 1), the slope of the tangent line is<br />

x<br />

y 0 (1) = 1 + 2 = 3, and an equation of the tangent line is y − 1=3(x − 1),ory =3x − 2.<br />

35. f(x) =sinx +lnx ⇒ f 0 (x) =cosx +1/x.<br />

This is reasonable, because the graph shows that f increases when f 0 is<br />

positive, and f 0 (x) =0when f has a horizontal tangent.

F.<br />

112 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

37. y =(2x +1) 5 (x 4 − 3) 6 ⇒ ln y =ln (2x +1) 5 (x 4 − 3) 6 ⇒ ln y =5ln(2x +1)+6ln(x 4 − 3) ⇒<br />

1<br />

1<br />

y y0 =5·<br />

2x +1 · 2+6· 1<br />

x 4 − 3 · 4x3<br />

10<br />

y 0 = y<br />

2x +1 + 24x3<br />

x 4 − 3<br />

<br />

⇒<br />

<br />

10<br />

=(2x +1) 5 (x 4 − 3) 6 2x +1 + 24x3 .<br />

x 4 − 3<br />

[The answer could be simplified to y 0 =2(2x +1) 4 (x 4 − 3) 5 (29x 4 +12x 3 − 15), but this is unnecessary.]<br />

39. y = sin2 x tan 4 x<br />

(x 2 +1) 2 ⇒ ln y =ln(sin 2 x tan 4 x) − ln(x 2 +1) 2 ⇒<br />

ln y =ln(sinx) 2 +ln(tanx) 4 − ln(x 2 +1) 2 ⇒ ln y =2ln|sin x| +4ln|tan x| − 2ln(x 2 +1) ⇒<br />

<br />

1 1<br />

y y0 =2·<br />

sin x · cos x +4· 1<br />

tan x · 1<br />

sec2 x − 2 ·<br />

x 2 +1 · 2x ⇒ y0 = sin2 x tan 4 x<br />

2cotx + 4sec2 x<br />

(x 2 +1) 2 tan x − 4x <br />

x 2 +1<br />

41. y = x x ⇒ ln y =lnx x ⇒ ln y = x ln x ⇒ y 0 /y = x(1/x)+(lnx) · 1 ⇒ y 0 = y(1 + ln x) ⇒<br />

y 0 = x x (1 + ln x)<br />

43. y = x sin x ⇒ ln y =lnx sin x ⇒ ln y =sinx ln x ⇒ y0<br />

y =(sinx) · 1 +(lnx)(cos x)<br />

x ⇒<br />

<br />

<br />

sin x<br />

sin x<br />

y 0 = y<br />

x<br />

+lnx cos x ⇒ y 0 = x sin x x<br />

+lnx cos x<br />

45. y =(cosx) x ⇒ ln y =ln(cosx) x ⇒ ln y = x ln cos x ⇒ 1 y y0 = x ·<br />

<br />

y 0 = y ln cos x − x sin x <br />

cos x<br />

⇒ y 0 =(cosx) x (ln cos x − x tan x)<br />

1 · (− sin x)+lncosx · 1<br />

cos x ⇒<br />

47. y =(tanx) 1/x ⇒ ln y =ln(tanx) 1/x ⇒ ln y = 1 ln tan x ⇒<br />

x<br />

1<br />

y y0 = 1 x · 1<br />

tan x · sec2 x +lntanx ·<br />

− 1 <br />

<br />

sec<br />

⇒ y 0 2 x ln tan x<br />

= y −<br />

x 2 x tan x x 2<br />

y 0 =(tanx) 1/x sec 2 x<br />

x tan x<br />

49. y =ln(x 2 + y 2 ) ⇒ y 0 =<br />

<br />

ln tan x<br />

−<br />

x 2<br />

1<br />

x 2 + y 2<br />

or y 0 = (tan x) 1/x · 1<br />

x<br />

x 2 y 0 + y 2 y 0 − 2yy 0 =2x ⇒ (x 2 + y 2 − 2y)y 0 =2x ⇒ y 0 =<br />

51. f(x) =ln(x − 1) ⇒ f 0 (x) =<br />

<br />

csc x sec x −<br />

⇒<br />

<br />

ln tan x<br />

x<br />

d<br />

dx (x2 + y 2 ) ⇒ y 0 2x +2yy0<br />

= ⇒ x 2 y 0 + y 2 y 0 =2x +2yy 0 ⇒<br />

x 2 + y 2<br />

2x<br />

x 2 + y 2 − 2y<br />

1<br />

(x − 1) =(x − 1)−1 ⇒ f 00 (x) =−(x − 1) −2 ⇒ f 000 (x) =2(x − 1) −3 ⇒<br />

f (4) (x) =−2 · 3(x − 1) −4 ⇒ ··· ⇒ f (n) (x) =(−1) n−1 · 2 · 3 · 4 ·····(n − 1)(x − 1) −n =(−1) n−1 (n − 1)!<br />

(x − 1) n<br />

53. If f(x) =ln(1+x), thenf 0 (x) = 1<br />

1+x ,sof 0 (0) = 1.<br />

ln(1 + x) f(x)<br />

Thus, lim =lim<br />

x→0 x x→0 x<br />

= lim f(x) − f(0)<br />

= f 0 (0) = 1.<br />

x→0 x − 0

F.<br />

SECTION TX.10 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 113<br />

3.7 Rates of Change in the Natural and Social Sciences<br />

1. (a) s = f(t) =t 3 − 12t 2 +36t ⇒ v(t) =f 0 (t) =3t 2 − 24t +36<br />

(b) v(3) = 27 − 72 + 36 = −9 ft/s<br />

(c) The particle is at rest when v(t) =0. 3t 2 − 24t +36=0 ⇔ 3(t − 2)(t − 6) = 0 ⇔ t =2sor6 s.<br />

(d) The particle is moving in the positive direction when v(t) > 0. 3(t − 2)(t − 6) > 0 ⇔ 0 ≤ t6.<br />

(e) Since the particle is moving in the positive direction and in the (f )<br />

negative direction, we need to calculate the distance traveled in the<br />

intervals [0, 2], [2, 6],and[6, 8] separately.<br />

|f(2) − f(0)| = |32 − 0| =32.<br />

|f(6) − f(2)| = |0 − 32| =32.<br />

|f(8) − f(6)| = |32 − 0| =32.<br />

The total distance is 32 + 32 + 32 = 96 ft.<br />

(g) v(t) =3t 2 − 24t +36<br />

a(t) =v 0 (t) =6t − 24.<br />

⇒<br />

(h )<br />

a(3) = 6(3) − 24 = −6(ft/s)/s orft/s 2 .<br />

(i) The particle is speeding up when v and a havethesamesign.Thisoccurswhen2

F.<br />

114 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

(i) The particle is speeding up when v and a have the same sign. This occurs when 0

F.<br />

SECTION TX.10 3.7 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES ¤ 115<br />

(c) The circumference is C(r) =2πr = A 0 (r). Thefigure suggests that if ∆r is small,<br />

then the change in the area of the circle (a ring around the outside) is approximately equal<br />

to its circumference times ∆r. Straightening out this ring gives us a shape that is approximately<br />

rectangular with length 2πr and width ∆r, so∆A ≈ 2πr(∆r). Algebraically,<br />

∆A = A(r + ∆r) − A(r) =π(r + ∆r) 2 − πr 2 =2πr(∆r)+π(∆r) 2 .<br />

So we see that if ∆r is small, then ∆A ≈ 2πr(∆r) and therefore, ∆A/∆r ≈ 2πr.<br />

15. S(r) =4πr 2 ⇒ S 0 (r) =8πr ⇒<br />

(a) S 0 (1) = 8π ft 2 /ft (b) S 0 (2) = 16π ft 2 /ft (c) S 0 (3) = 24π ft 2 /ft<br />

As the radius increases, the surface area grows at an increasing rate. In fact, the rate of change is linear with respect to the<br />

17. The mass is f(x) =3x 2 , so the linear density at x is ρ(x) =f 0 (x) =6x.<br />

(a) ρ(1) = 6 kg/m (b) ρ(2) = 12kg/m (c) ρ(3) = 18 kg/m<br />

Since ρ is an increasing function, the density will be the highest at the right end of the rod and lowest at the left end.<br />

19. The quantity of charge is Q(t) =t 3 − 2t 2 +6t +2,sothecurrentisQ 0 (t) =3t 2 − 4t +6.<br />

(a) Q 0 (0.5) = 3(0.5) 2 − 4(0.5) + 6 = 4.75 A<br />

(b) Q 0 (1) = 3(1) 2 − 4(1) + 6 = 5 A<br />

The current is lowest when Q 0 has a minimum. Q 00 (t) =6t − 4 < 0 when t< 2 3 . So the current decreases when t< 2 3 and<br />

increases when t> 2 3 . Thus, the current is lowest at t = 2 3 s.<br />

21. (a) To find the rate of change of volume with respect to pressure, we first solve for V in terms of P .<br />

PV = C ⇒ V = C P ⇒ dV<br />

dP = − C P 2 .<br />

(b) From the formula for dV/dP in part (a), we see that as P increases, the absolute value of dV/dP decreases.<br />

Thus, the volume is decreasing more rapidly at the beginning.<br />

(c) β = − 1 V<br />

dV<br />

dP = − 1 − C <br />

=<br />

V P 2<br />

C<br />

(PV)P =<br />

C<br />

CP = 1 P<br />

23. In Example 6, the population function was n =2 t n 0. Since we are tripling instead of doubling and the initial population is<br />

400, the population function is n(t) =400· 3 t . The rate of growth is n 0 (t) =400· 3 t · ln 3, so the rate of growth after<br />

2.5 hours is n 0 (2.5) = 400 · 3 2.5 · ln 3 ≈ 6850 bacteria/hour.<br />

1860 − 1750<br />

25. (a) 1920: m 1 =<br />

1920 − 1910 = 110<br />

2070 − 1860<br />

=11, m2 =<br />

10 1930 − 1920 = 210<br />

10 =21,<br />

(m 1 + m 2 )/ 2 = (11 + 21)/2 =16million/year<br />

4450 − 3710<br />

1980: m 1 =<br />

1980 − 1970 = 740<br />

10 =74, m 5280 − 4450<br />

2 =<br />

1990 − 1980 = 830<br />

10 =83,<br />

(m 1 + m 2)/ 2 = (74 + 83)/2 =78.5 million/year<br />

(b) P (t) =at 3 + bt 2 + ct + d (in millions of people), where a ≈ 0.0012937063, b ≈−7.061421911, c ≈ 12,822.97902,<br />

and d ≈−7,743,770.396.<br />

(c) P (t) =at 3 + bt 2 + ct + d ⇒ P 0 (t) =3at 2 +2bt + c (in millions of people per year)

F.<br />

116 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

(d) P 0 (1920) = 3(0.0012937063)(1920) 2 +2(−7.061421911)(1920) + 12,822.97902<br />

≈ 14.48 million/year [smaller than the answer in part (a), but close to it]<br />

P 0 (1980) ≈ 75.29 million/year (smaller, but close)<br />

(e) P 0 (1985) ≈ 81.62 million/year, so the rate of growth in 1985 was about 81.62 million/year.<br />

27. (a) Using v = P<br />

4ηl (R2 − r 2 ) with R =0.01, l =3, P =3000,andη =0.027,wehavev as a function of r:<br />

v(r) = 3000<br />

4(0.027)3 (0.012 − r 2 ). v(0) = 0.925 cm/s, v(0.005) = 0.694 cm/s, v(0.01) = 0.<br />

(b) v(r) = P<br />

4ηl (R2 − r 2 ) ⇒ v 0 (r) = P<br />

Pr<br />

(−2r) =−<br />

4ηl 2ηl .<br />

Whenl =3, P =3000,andη =0.027,wehave<br />

v 0 (r) =−<br />

3000r<br />

2(0.027)3 . v0 (0) = 0, v 0 (0.005) = −92.592 (cm/s)/cm, and v 0 (0.01) = −185.185 (cm/s)/cm.<br />

(c) The velocity is greatest where r =0(at the center) and the velocity is changing most where r = R =0.01 cm<br />

(at the edge).<br />

29. (a) C(x) = 1200 + 12x − 0.1x 2 +0.0005x 3 ⇒ C 0 (x) =12− 0.2x +0.0015x 2 \$/yard, which is the marginal cost<br />

function.<br />

(b) C 0 (200) = 12 − 0.2(200) + 0.0015(200) 2 =\$32/yard, and this is the rate at which costs are increasing with respect to<br />

the production level when x =200.<br />

C 0 (200) predicts the cost of producing the 201st yard.<br />

(c) The cost of manufacturing the 201st yard of fabric is C(201) − C(200) = 3632.2005 − 3600 ≈ \$32.20,whichis<br />

approximately C 0 (200).<br />

31. (a) A(x) = p(x)<br />

x<br />

⇒ A 0 (x) = xp0 (x) − p(x) · 1<br />

x 2<br />

= xp0 (x) − p(x)<br />

x 2 .<br />

A 0 (x) > 0 ⇒ A(x) is increasing; that is, the average productivity increases as the size of the workforce increases.<br />

(b) p 0 (x) is greater than the average productivity ⇒ p 0 (x) >A(x) ⇒ p 0 (x) > p(x)<br />

x<br />

xp 0 (x) − p(x) > 0 ⇒ xp0 (x) − p(x)<br />

x 2 > 0 ⇒ A 0 (x) > 0.<br />

⇒ xp 0 (x) >p(x) ⇒<br />

33. PV = nRT ⇒ T = PV<br />

nR = PV<br />

(10)(0.0821) = 1 (PV). Using the Product Rule, we have<br />

0.821<br />

dT<br />

dt = 1<br />

0.821 [P (t)V 0 (t)+V (t)P 0 (t)] = 1 [(8)(−0.15) + (10)(0.10)] ≈−0.2436 K/min.<br />

0.821<br />

35. (a) If the populations are stable, then the growth rates are neither positive nor negative; that is, dC<br />

dt =0and dW dt<br />

(b) “The caribou go extinct” means that the population is zero, or mathematically, C =0.<br />

=0.<br />

(c)Wehavetheequations dC<br />

dt = aC − bCW and dW dt<br />

= −cW + dCW .LetdC/dt = dW/dt =0, a =0.05, b =0.001,<br />

c =0.05,andd =0.0001 to obtain 0.05C − 0.001CW =0 (1) and −0.05W +0.0001CW =0 (2). Adding10 times<br />

(2) to (1) eliminates the CW-terms and gives us 0.05C − 0.5W =0 ⇒ C =10W . Substituting C =10W into (1)

F.<br />

TX.10<br />

SECTION 3.8 EXPONENTIAL GROWTH AND DECAY ¤ 117<br />

results in 0.05(10W ) − 0.001(10W )W =0 ⇔ 0.5W − 0.01W 2 =0 ⇔ 50W − W 2 =0 ⇔<br />

W (50 − W )=0 ⇔ W =0or 50. SinceC =10W , C =0or 500. Thus, the population pairs (C, W ) that lead to<br />

stable populations are (0, 0) and (500, 50).Soitispossibleforthetwospeciestoliveinharmony.<br />

3.8 Exponential Growth and Decay<br />

1. The relative growth rate is 1 P<br />

dP<br />

dt<br />

Thus, P (6) = 2e 0.7944(6) ≈ 234.99 or about 235 members.<br />

=0.7944,so<br />

dP<br />

dt =0.7944P and, by Theorem 2, P (t) =P (0)e0.7944t =2e 0.7944t .<br />

3. (a) By Theorem 2, P (t) =P (0)e kt = 100e kt . NowP (1) = 100e k(1) = 420 ⇒ e k = 420<br />

100<br />

⇒ k =ln4.2.<br />

So P (t) =100e (ln 4.2)t =100(4.2) t .<br />

(b) P (3) = 100(4.2) 3 =7408.8 ≈ 7409 bacteria<br />

(c) dP/dt = kP ⇒ P 0 (3) = k · P (3) = (ln 4.2) 100(4.2) 3 [from part (a)] ≈ 10,632 bacteria/hour<br />

(d) P (t) = 100(4.2) t =10,000 ⇒ (4.2) t =100 ⇒ t =(ln100)/(ln 4.2) ≈ 3.2 hours<br />

5. (a) Let the population (in millions) in the year t be P (t). Since the initial time is the year 1750, we substitute t − 1750 for t in<br />

Theorem 2, so the exponential model gives P (t) =P (1750)e k(t−1750) .ThenP (1800) = 980 = 790e k(1800−1750)<br />

⇒<br />

980<br />

= 790 ek(50) ⇒ ln 980 =50k ⇒ k = 1 980<br />

ln ≈ 0.0043104. Sowiththismodel,wehave<br />

790 50 790<br />

P (1900) = 790e k(1900−1750) ≈ 1508 million, and P (1950) = 790e k(1950−1750) ≈ 1871 million. Both of these<br />

estimates are much too low.<br />

(b) In this case, the exponential model gives P (t) =P (1850)e k(t−1850) ⇒ P (1900) = 1650 = 1260e k(1900−1850) ⇒<br />

ln 1650 = k(50) ⇒ k = 1 1650<br />

ln ≈ 0.005393. Sowiththismodel,weestimate<br />

1260 50 1260<br />

P (1950) = 1260e k(1950−1850) ≈ 2161 million. This is still too low, but closer than the estimate of P (1950) in part (a).<br />

(c) The exponential model gives P (t) =P (1900)e k(t−1900) ⇒ P (1950) = 2560 = 1650e k(1950−1900) ⇒<br />

ln 2560 = k(50) ⇒ k = 1 2560<br />

1650 50<br />

ln<br />

1650<br />

≈ 0.008785. With this model, we estimate<br />

P (2000) = 1650e k(2000−1900) ≈ 3972 million. This is much too low. The discrepancy is explained by the fact that the<br />

world birth rate (average yearly number of births per person) is about the same as always, whereas the mortality rate<br />

(especially the infant mortality rate) is much lower, owing mostly to advances in medical science and to the wars in the first<br />

part of the twentieth century. The exponential model assumes, among other things, that the birth and mortality rates will<br />

remain constant.<br />

7. (a) If y =[N 2O 5] then by Theorem 2, dy<br />

dt = −0.0005y ⇒ y(t) =y(0)e−0.0005t = Ce −0.0005t .<br />

(b) y(t) =Ce −0.0005t =0.9C ⇒ e −0.0005t =0.9 ⇒ −0.0005t =ln0.9 ⇒ t = −2000 ln 0.9 ≈ 211 s

F.<br />

118 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

9. (a) If y(t) is the mass (in mg) remaining after t years, then y(t) =y(0)e kt =100e kt .<br />

y(30) = 100e 30k = 1 (100) ⇒ 2 e30k = 1 2<br />

⇒ k = −(ln 2)/30 ⇒ y(t) = 100e −(ln 2)t/30 = 100 · 2 −t/30<br />

(b) y(100) = 100 · 2 −100/30 ≈ 9.92 mg<br />

(c) 100e −(ln 2)t/30 =1 ⇒ −(ln 2)t/30 = ln 1<br />

ln 0.01<br />

100<br />

⇒ t = −30<br />

ln 2<br />

≈ 199.3 years<br />

11. Let y(t) be the level of radioactivity. Thus, y(t) =y(0)e −kt and k is determined by using the half-life:<br />

y(5730) = 1 y(0) ⇒ 2 y(0)e−k(5730) = 1 y(0) ⇒ 2 e−5730k = 1 ⇒ −5730k =ln 1 ⇒ k = − ln 1 2<br />

2 2<br />

5730 = ln 2<br />

5730 .<br />

If 74% of the 14 C remains, then we know that y(t) =0.74y(0) ⇒ 0.74 = e −t(ln 2)/5730 ⇒ ln 0.74 = − t ln 2 ⇒<br />

5730<br />

5730(ln 0.74)<br />

t = − ≈ 2489 ≈ 2500 years.<br />

ln 2<br />

13. (a) Using Newton’s Law of Cooling, dT<br />

dt = k(T − T s), wehave dT = k(T − 75). Nowlety = T − 75, so<br />

dt<br />

y(0) = T (0) − 75 = 185 − 75 = 110,soy is a solution of the initial-value problem dy/dt = ky with y(0) = 110 and by<br />

Theorem 2 we have y(t) =y(0)e kt =110e kt .<br />

y(30) = 110e 30k =150− 75 ⇒ e 30k = 75 = 15<br />

110 22<br />

⇒ k = 1 15<br />

30<br />

ln ,soy(t) = 110e 30 1 t ln( 15<br />

22 )<br />

22 and<br />

y(45) = 110e 45<br />

30<br />

ln( 15<br />

22 ) ≈ 62 ◦ F.Thus,T (45) ≈ 62 + 75 = 137 ◦ F.<br />

(b) T (t) =100 ⇒ y(t) =25. y(t) = 110e 30 1 t ln( 15<br />

1<br />

22 ) =25 ⇒ e 30<br />

t ln( 15<br />

22 ) =<br />

25<br />

110<br />

⇒ 1 15 25<br />

30<br />

t ln<br />

22<br />

=ln110 ⇒<br />

25<br />

30 ln<br />

110<br />

t =<br />

ln 15 ≈ 116 min.<br />

22<br />

dT<br />

15. = k(T − 20). Lettingy = T − 20, wegetdy<br />

dt dt = ky,soy(t) =y(0)ekt . y(0) = T (0) − 20 = 5 − 20 = −15, so<br />

y(25) = y(0)e 25k = −15e 25k ,andy(25) = T (25) − 20 = 10 − 20 = −10,so−15e 25k = −10 ⇒ e 25k = 2 . Thus,<br />

3<br />

25k =ln <br />

2<br />

3 and k =<br />

1<br />

ln <br />

2<br />

25 3 ,soy(t) =y(0)e kt = −15e (1/25) ln(2/3)t .Moresimply,e 25k = 2 ⇒ e k = <br />

2 1/25<br />

⇒<br />

3 3<br />

e kt = <br />

2 t/25<br />

⇒ y(t) =−15 · <br />

2 t/25<br />

.<br />

3<br />

3<br />

(a) T (50) = 20 + y(50) = 20 − 15 · <br />

2 50/25<br />

=20− 15 · <br />

2 2<br />

=20− 20 =13.¯3 ◦ C<br />

3<br />

3<br />

3<br />

(b) 15 = T (t) =20+y(t) =20− 15 · <br />

2 t/25<br />

3<br />

⇒ 15 · <br />

2 t/25<br />

3<br />

=5 ⇒ <br />

2 t/25<br />

3<br />

= 1 3<br />

⇒<br />

(t/25) ln <br />

2<br />

3 =ln 1<br />

<br />

⇒ t =25ln <br />

1<br />

3<br />

3 ln 2<br />

<br />

3 ≈ 67.74 min.<br />

17. (a) Let P (h) be the pressure at altitude h. ThendP/dh = kP ⇒ P (h) =P (0)e kh =101.3e kh .<br />

P (1000) = 101.3e 1000k =87.14 ⇒ 1000k =ln <br />

87.14<br />

⇒ k = 1 ln <br />

87.14<br />

⇒<br />

101.3<br />

1000 101.3<br />

P (h) =101.3 e 1000 1 h ln( 87.14<br />

101.3) ,soP (3000) = 101.3e<br />

3ln( 87.14<br />

101.3 ) ≈ 64.5 kPa.<br />

(b) P (6187) = 101.3 e 6187 ln( 87.14<br />

1000 101.3 ) ≈ 39.9 kPa

F.<br />

TX.10<br />

19. (a) Using A = A 0<br />

<br />

1+ r n<br />

nt<br />

with A0 = 3000, r =0.05,andt =5,wehave:<br />

SECTION 3.9 RELATED RATES ¤ 119<br />

(i) Annually: n =1; A =3000 <br />

1+ 0.05 1·5<br />

1<br />

= \$3828.84<br />

(ii) Semiannually: n =2; A =3000 <br />

1+ 0.05 2·5<br />

2<br />

= \$3840.25<br />

(iii) Monthly: n =12; A =3000 1+ 0.05<br />

12<br />

(iv) Weekly: n =52; A =3000 1+ 0.05<br />

52<br />

(v) Daily: n =365; A =3000 1+ 0.05<br />

365<br />

12·5<br />

= \$3850.08<br />

52·5<br />

= \$3851.61<br />

365·5<br />

= \$3852.01<br />

(vi) Continuously: A =3000e (0.05)5 = \$3852.08<br />

(b) dA/dt =0.05A and A(0) = 3000.<br />

3.9 Related Rates<br />

1. V = x 3 ⇒ dV<br />

dt = dV dx dx<br />

dx dt =3x2 dt<br />

3. Let s denote the side of a square. The square’s area A is given by A = s 2 . Differentiating with respect to t gives us<br />

dA<br />

dt<br />

ds<br />

ds dA<br />

=2s .WhenA =16, s =4. Substitution 4 for s and 6 for gives us<br />

dt dt dt = 2(4)(6) = 48 cm2 /s.<br />

5. V = πr 2 h = π(5) 2 h =25πh ⇒ dV<br />

dt<br />

=25π<br />

dh<br />

dt<br />

⇒<br />

3=25π dh<br />

dt<br />

⇒<br />

dh<br />

dt = 3<br />

25π m/min.<br />

7. y = x 3 +2x ⇒ dy<br />

dt = dy dx<br />

dx dt =(3x2 + 2)(5) = 5(3x 2 +2).Whenx =2, dy<br />

dt =5(14)=70.<br />

9. z 2 = x 2 + y 2 ⇒ 2z dz dx dy<br />

=2x +2y<br />

dt dt dt<br />

z 2 =5 2 +12 2 ⇒ z 2 =169 ⇒ z = ±13. For dx<br />

dt<br />

⇒<br />

dz<br />

dt = 1 <br />

x dx<br />

z dt + y dy <br />

.Whenx =5and y =12,<br />

dt<br />

dy dz<br />

=2and =3,<br />

dt dt = 1 (5 · 2+12· 3) = ±46<br />

±13 13 .<br />

11. (a) Given: a plane flying horizontally at an altitude of 1 mi and a speed of 500 mi/h passes directly over a radar station.<br />

If we let t be time (in hours) and x be the horizontal distance traveled by the plane (in mi), then we are given<br />

that dx/dt = 500 mi/h.<br />

(b) Unknown: the rate at which the distance from the plane to the station is increasing<br />

(c)<br />

when it is 2 mi from the station. If we let y be the distance from the plane to the station,<br />

then we want to find dy/dt when y =2mi.<br />

(d) By the Pythagorean Theorem, y 2 = x 2 +1 ⇒ 2y (dy/dt) =2x (dx/dt).<br />

(e) dy<br />

dt = x dx<br />

y dt = x y (500). Sincey2 = x 2 +1,wheny =2, x = √ 3,so dy<br />

√<br />

3<br />

dt = 2 (500) = 250 √ 3 ≈ 433 mi/h.<br />

13. (a) Given: a man 6 ft tall walks away from a street light mounted on a 15-ft-tall pole at a rate of 5 ft/s. If we let t be time (in s)<br />

and x be the distance from the pole to the man (in ft), then we are given that dx/dt =5ft/s.

F.<br />

120 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

(b) Unknown: the rate at which the tip of his shadow is moving when he is 40 ft<br />

from the pole. If we let y be the distance from the man to the tip of his<br />

(c)<br />

shadow(inft),thenwewanttofind d (x + y) when x =40ft.<br />

dt<br />

(d) By similar triangles, 15<br />

6 = x + y<br />

y<br />

⇒ 15y =6x +6y ⇒ 9y =6x ⇒ y = 2 3 x.<br />

(e) The tip of the shadow moves at a rate of d dt (x + y) = d dt<br />

15. We are given that dx<br />

dt<br />

2z dz dx dy<br />

=2x +2y<br />

dt dt dt<br />

17. We are given that dx<br />

dt<br />

x + 2 <br />

3 x = 5 dx<br />

3 dt = 5 (5) = 25 3 3<br />

ft/s.<br />

=60mi/hand<br />

dy<br />

dt =25mi/h. z2 = x 2 + y 2<br />

⇒<br />

z dz<br />

dt = x dx<br />

dt + y dy<br />

dt<br />

⇒<br />

⇒<br />

dz<br />

dt = 1 <br />

x dx<br />

z dt + y dy <br />

.<br />

dt<br />

After 2 hours, x = 2 (60) = 120 and y =2(25)=50 ⇒ z = √ 120 2 +50 2 = 130,<br />

so dz<br />

dt = 1 <br />

x dx<br />

z dt + y dy <br />

120(60) + 50(25)<br />

= =65mi/h.<br />

dt<br />

130<br />

2z dz<br />

dx<br />

dt =2(x + y) dt + dy<br />

dt<br />

dy<br />

=4ft/sand<br />

dt =5ft/s. z2 =(x + y) 2 + 500 2 ⇒<br />

<br />

. 15 minutes after the woman starts, we have<br />

x =(4ft/s)(20 min)(60 s/min) = 4800 ft and y =5· 15 · 60 = 4500<br />

⇒<br />

z = (4800 + 4500) 2 +500 2 = √ 86,740,000,so<br />

dz<br />

dt = x + y dx<br />

z dt + dy <br />

4800 + 4500<br />

= √ (4 + 5) = √ 837 ≈ 8.99 ft/s.<br />

dt 86,740,000 8674<br />

19. A = 1 dh<br />

bh,whereb is the base and h is the altitude. We are given that<br />

2<br />

dt<br />

Product Rule, we have dA<br />

dt = 1 2<br />

b =20,so2= 1 <br />

20 · 1+10 db<br />

2<br />

dt<br />

<br />

b dh<br />

dt + h db<br />

dt<br />

<br />

⇒<br />

=1cm/min and<br />

dA<br />

dt =2cm2 /min. Using the<br />

<br />

.Whenh =10and A = 100,wehave100 = 1 b(10) 2 ⇒ 1 b =10 2 ⇒<br />

4=20+10 db<br />

dt<br />

⇒<br />

db<br />

dt = 4 − 20 = −1.6 cm/min.<br />

10<br />

21. We are given that dx<br />

dt<br />

2z dz<br />

dx<br />

dt =2(x + y) dt + dy<br />

dt<br />

dy<br />

=35km/hand<br />

dt =25km/h. z2 =(x + y) 2 + 100 2 ⇒<br />

<br />

.At4:00PM, x =4(35)=140and y = 4(25) = 100<br />

⇒<br />

z = (140 + 100) 2 +100 2 = √ 67,600 = 260,so<br />

dz<br />

dt = x + y dx<br />

z dt + dy <br />

140 + 100<br />

= (35 + 25) = 720 ≈ 55.4 km/h.<br />

dt 260<br />

13

F.<br />

TX.10<br />

SECTION 3.9 RELATED RATES ¤ 121<br />

23. If C = the rate at which water is pumped in, then dV<br />

dt = C − 10,000,where<br />

V = 1 3 πr2 h is the volume at time t. By similar triangles, r 2 = h 6<br />

⇒ r = 1 3 h ⇒<br />

V = 1 π 1<br />

3 3 h2 h = π 27 h3 ⇒ dV<br />

dt = π dh<br />

9 h2 .Whenh =200cm,<br />

dt<br />

dh<br />

dt =20cm/min,soC − 10,000 = π 9 (200)2 (20) ⇒ C =10,000 + 800,000 π ≈ 289,253 cm 3 /min.<br />

9<br />

25. The figure is labeled in meters. The area A of a trapezoid is<br />

1<br />

(base1 + base2)(height), and the volume V of the 10-meter-long trough is 10A.<br />

2<br />

Thus, the volume of the trapezoid with height h is V =(10) 1 2 [0.3+(0.3+2a)]h.<br />

By similar triangles, a h = 0.25<br />

0.5 = 1 2 ,so2a = h ⇒ V =5(0.6+h)h =3h +5h2 .<br />

Now dV<br />

dt = dV dh<br />

dh dt<br />

⇒<br />

0.2 =(3+10h) dh<br />

dt<br />

dh<br />

dt = 0.2<br />

3 + 10(0.3) = 0.2<br />

6 m/min = 1 10<br />

m/min or<br />

30 3 cm/min.<br />

⇒<br />

dh<br />

dt = 0.2 .Whenh =0.3,<br />

3+10h<br />

27. We are given that dV<br />

dt =30ft3 /min. V = 1 3 πr2 h = 1 2 h<br />

3 π h = πh3<br />

2 12<br />

⇒<br />

dV<br />

dt = dV dh<br />

dh dt<br />

⇒<br />

30 = πh2<br />

4<br />

dh<br />

dt<br />

⇒<br />

dh<br />

dt = 120<br />

πh . 2<br />

When h =10ft, dh<br />

dt = 120<br />

10 2 π = 6 ≈ 0.38 ft/min.<br />

5π<br />

29. A = 1 2 bh,butb =5mandsin θ = h 4<br />

⇒<br />

h =4sinθ,soA = 1 (5)(4 sin θ) =10sinθ.<br />

2<br />

We are given dθ<br />

dA<br />

dt<br />

When θ = π 3 , dA<br />

dt =0.6 cos π 3<br />

dθ<br />

=(10cosθ)(0.06) = 0.6cosθ.<br />

dt<br />

dt = dA<br />

dθ<br />

<br />

=(0.6) <br />

1<br />

2 =0.3 m 2 /s.<br />

31. Differentiating both sides of PV = C with respect to t and using the Product Rule gives us P dV<br />

dt + V dP dt =0<br />

dV<br />

dt = −V P<br />

dP<br />

dP<br />

.WhenV = 600, P = 150 and<br />

dt dt<br />

decreasing at a rate of 80 cm 3 /min.<br />

33. With R 1 =80and R 2 = 100,<br />

dV<br />

=20,sowehave<br />

dt = − 600 (20) = −80. Thus, the volume is<br />

150<br />

1<br />

R = 1 + 1 = 1 R 1 R 2 80 + 1<br />

100 = 180<br />

8000 = 9 400<br />

,soR =<br />

400<br />

with respect to t,wehave− 1 dR<br />

R 2 dt = − 1 dR 1<br />

− 1 dR 2<br />

R1<br />

2 dt R2<br />

2 dt<br />

R 2 = 100, dR <br />

dt = 4002 1<br />

9 2 80 (0.3) + 1 <br />

2 100 (0.2) = 107 ≈ 0.132 Ω/s.<br />

2 810<br />

⇒<br />

dR 1<br />

dt = dR 1<br />

R2 + 1 R1<br />

2 dt R2<br />

2<br />

dR 2<br />

dt<br />

⇒<br />

9 . Differentiating 1 R = 1 + 1 R 1 R 2<br />

<br />

.WhenR 1 =80and

F.<br />

122 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

35. We are given dθ/dt =2 ◦ /min = π 90<br />

rad/min. By the Law of Cosines,<br />

x 2 =12 2 +15 2 − 2(12)(15) cos θ =369− 360 cos θ<br />

2x dx<br />

dt<br />

=360sinθ<br />

dθ<br />

dt<br />

⇒<br />

dx<br />

dt<br />

=<br />

180 sin θ<br />

x<br />

x = √ 369 − 360 cos 60 ◦ = √ 189 = 3 √ 21,so dx<br />

dt<br />

⇒<br />

dθ<br />

dt .Whenθ =60◦ ,<br />

=<br />

180 sin 60◦<br />

3 √ 21<br />

37. (a) By the Pythagorean Theorem, 4000 2 + y 2 = 2 . Differentiating with respect to t,<br />

we obtain 2y dy<br />

dt<br />

d<br />

dy<br />

=2 . We know that =600ft/s, so when y =3000ft,<br />

dt dt<br />

= √ 4000 2 + 3000 2 = √ 25,000,000 = 5000 ft<br />

and d<br />

dt = y dy<br />

<br />

(b) Here tan θ =<br />

dt = 3000<br />

5000<br />

y<br />

4000<br />

⇒<br />

(600) =<br />

1800<br />

5<br />

=360ft/s.<br />

d<br />

dt (tan θ) = d y<br />

<br />

dt 4000<br />

⇒<br />

π<br />

90 = π √ √<br />

3 7 π<br />

3 √ 21 = ≈ 0.396 m/min.<br />

21<br />

sec 2 θ dθ<br />

dt = 1 dy<br />

4000 dt<br />

⇒<br />

dθ<br />

dt = cos2 θ dy<br />

4000 dt .When<br />

y =3000ft, dy<br />

4000<br />

=600ft/s, =5000and cos θ = = 4000<br />

dt 5000 = 4 dθ<br />

,so<br />

5 dt = (4/5)2 (600) = 0.096 rad/s.<br />

4000<br />

39. cot θ = x 5<br />

dx<br />

dt = 5π 6<br />

⇒<br />

−csc 2 θ dθ<br />

dt = 1 dx<br />

5 dt<br />

⇒<br />

2 2√3<br />

= 10 π km/min [≈ 130 mi/h]<br />

9<br />

<br />

− csc π 2 <br />

− π <br />

= 1 dx<br />

3 6 5 dt<br />

⇒<br />

41. We are given that dx<br />

dt<br />

=300km/h. By the Law of Cosines,<br />

y 2 = x 2 +1 2 − 2(1)(x) cos 120 ◦ = x 2 +1− 2x − 1 2<br />

= x 2 + x +1,so<br />

2y dy dx<br />

=2x<br />

dt dt + dx<br />

dt<br />

⇒<br />

dy<br />

dt<br />

=<br />

2x +1<br />

2y<br />

dx<br />

300<br />

.After1 minute, x =<br />

60<br />

dt =5km ⇒<br />

y = √ 5 2 +5+1= √ 31 km ⇒ dy<br />

dt = 2(5) + 1<br />

2 √ 1650<br />

(300) = √ ≈ 296 km/h.<br />

31 31<br />

43. Let the distance between the runner and the friend be .ThenbytheLawofCosines,<br />

2 =200 2 +100 2 − 2 · 200 · 100 · cos θ =50,000 − 40,000 cos θ (). Differentiating<br />

implicitly with respect to t,weobtain2 d<br />

dt<br />

dθ<br />

= −40,000(− sin θ) .NowifD is the<br />

dt<br />

distance run when the angle is θ radians, then by the formula for the length of an arc<br />

on a circle, s = rθ,wehaveD = 100θ,soθ = 1<br />

100 D ⇒ dθ<br />

dt = 1 dD<br />

100 dt = 7 . To substitute into the expression for<br />

100<br />

d<br />

dt , we must know sin θ atthetimewhen =200,whichwefind from (): 2002 =50,000 − 40,000 cos θ<br />

<br />

cos θ = 1 ⇒ sin θ = 1 − <br />

1 2<br />

= √ 15<br />

d<br />

. Substituting, we get 2(200)<br />

4 4 4<br />

dt =40,000 √ <br />

15 7<br />

<br />

⇒<br />

4 100<br />

d/dt = 7 √ 15<br />

4<br />

≈ 6.78 m/s. Whether the distance between them is increasing or decreasing depends on the direction in which<br />

the runner is running.<br />

F.<br />

TX.10SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 123<br />

3.10 Linear Approximations and Differentials<br />

1. f(x) =x 4 +3x 2 ⇒ f 0 (x) =4x 3 +6x,sof(−1) = 4 and f 0 (−1) = −10.<br />

Thus, L(x) =f(−1) + f 0 (−1)(x − (−1)) = 4 + (−10)(x +1)=−10x − 6.<br />

3. f(x) =cosx ⇒ f 0 (x) =− sin x, sof π<br />

2 =0and f<br />

0 π<br />

Thus, L(x) =f π<br />

2<br />

+ f<br />

0 π<br />

2<br />

5. f(x) = √ 1 − x ⇒ f 0 (x) =<br />

x −<br />

π<br />

2<br />

=0− 1<br />

x −<br />

π<br />

2<br />

2<br />

= −1.<br />

= −x +<br />

π<br />

2 .<br />

−1<br />

2 √ 1 − x ,sof(0) = 1 and f 0 (0) = − 1 2 .<br />

Therefore,<br />

√ 1 − x = f(x) ≈ f(0) + f 0 (0)(x − 0) = 1 + − 1 2<br />

<br />

(x − 0) = 1 −<br />

1<br />

2 x.<br />

So √ 0.9 = √ 1 − 0.1 ≈ 1 − 1 (0.1) = 0.95<br />

2<br />

and √ 0.99 = √ 1 − 0.01 ≈ 1 − 1 (0.01) = 0.995.<br />

2<br />

7. f(x) = 3√ 1 − x =(1− x) 1/3 ⇒ f 0 (x) =− 1 3 (1 − x)−2/3 ,sof(0) = 1<br />

and f 0 (0) = − 1 .Thus,f(x) ≈ f(0) + f 0 3<br />

(0)(x − 0) = 1 − 1 x.Weneed<br />

3<br />

3√ 1 − x − 0.1 < 1 −<br />

1<br />

x< 3√ 3<br />

1 − x +0.1, which is true when<br />

−1.204

F.<br />

124 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

17. (a) y =tanx ⇒ dy =sec 2 xdx<br />

(b) When x = π/4 and dx = −0.1, dy =[sec(π/4)] 2 (−0.1) = √ 2 2<br />

(−0.1) = −0.2.<br />

19. y = f(x) =2x − x 2 , x =2, ∆x = −0.4 ⇒<br />

∆y = f(1.6) − f(2) = 0.64 − 0=0.64<br />

dy =(2− 2x) dx =(2− 4)(−0.4) = 0.8<br />

21. y = f(x) =2/x, x =4, ∆x =1 ⇒<br />

∆y = f(5) − f(4) = 2 5 − 2 4 = −0.1<br />

dy = − 2 x dx = − 2 (1) = −0.125<br />

2 42 23. To estimate (2.001) 5 ,we’llfind the linearization of f(x) =x 5 at a =2.Sincef 0 (x) =5x 4 , f(2) = 32,andf 0 (2) = 80,<br />

we have L(x) = 32 + 80(x − 2) = 80x − 128. Thus, x 5 ≈ 80x − 128 when x is near 2 ,so<br />

(2.001) 5 ≈ 80(2.001) − 128 = 160.08 − 128 = 32.08.<br />

<br />

25. To estimate (8.06) 2/3 ,we’llfind the linearization of f(x) =x 2/3 at a =8.Sincef 0 (x) = 2 3 x−1/3 =2/ 3 3√ <br />

x ,<br />

f(8) = 4, andf 0 (8) = 1 ,wehaveL(x) 3 =4+1 (x − 8) = 1 x + 4 . Thus, 3 3 3 x2/3 ≈ 1 x + 4 when x is near 8, so<br />

3 3<br />

(8.06) 2/3 ≈ 1 3 (8.06) + 4 3 = 12.06<br />

3<br />

=4.02.<br />

27. y = f(x) =tanx ⇒ dy =sec 2 xdx.Whenx =45 ◦ and dx = −1 ◦ ,<br />

dy =sec 2 45 ◦ (−π/180) = √ 2 2<br />

(−π/180) = −π/90,sotan 44 ◦ = f(44 ◦ ) ≈ f(45 ◦ )+dy =1− π/90 ≈ 0.965.<br />

29. y = f(x) =secx ⇒ f 0 (x) =secx tan x, sof(0) = 1 and f 0 (0) = 1 · 0=0. The linear approximation of f at 0 is<br />

f(0) + f 0 (0)(x − 0) = 1 + 0(x) =1.Since0.08 is close to 0, approximating sec 0.08 with 1 is reasonable.<br />

31. y = f(x) =lnx ⇒ f 0 (x) =1/x, sof(1) = 0 and f 0 (1) = 1. The linear approximation of f at 1 is<br />

f(1) + f 0 (1)(x − 1) = 0 + 1(x − 1) = x − 1. Nowf(1.05) = ln 1.05 ≈ 1.05 − 1=0.05, so the approximation<br />

is reasonable.<br />

33. (a) If x is the edge length, then V = x 3 ⇒ dV =3x 2 dx. Whenx =30and dx =0.1, dV =3(30) 2 (0.1) = 270,sothe<br />

maximum possible error in computing the volume of the cube is about 270 cm 3 . The relative error is calculated by dividing<br />

the change in V , ∆V ,byV . We approximate ∆V with dV .<br />

Relative error = ∆V<br />

V ≈ dV V = 3x2 dx<br />

=3 dx 0.1<br />

x 3 x =3 =0.01.<br />

30<br />

Percentage error = relative error × 100% = 0.01 × 100% = 1%.

F.<br />

TX.10SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 125<br />

(b) S =6x 2 ⇒ dS =12xdx.Whenx =30and dx =0.1, dS = 12(30)(0.1) = 36, so the maximum possible error in<br />

computing the surface area of the cube is about 36 cm 2 .<br />

Relative error = ∆S<br />

S<br />

≈ dS S = 12xdx =2 dx 0.1<br />

6x 2 x =2 30<br />

=0.006.<br />

Percentage error = relative error × 100% = 0.006 × 100% = 0.6%.<br />

35. (a)Forasphereofradiusr, the circumference is C =2πr and the surface area is S =4πr 2 ,so<br />

r = C 2 C<br />

2π ⇒ S =4π = C2<br />

⇒ dS = 2 2π π<br />

π CdC.WhenC =84and dC =0.5, dS = 2 84<br />

(84)(0.5) =<br />

π π ,<br />

so the maximum error is about 84<br />

π ≈ 27 cm2 . Relative error ≈ dS S = 84/π<br />

84 2 /π = 1<br />

84 ≈ 0.012<br />

(b) V = 4 3 πr3 = 4 3 π C<br />

2π<br />

3<br />

= C3<br />

6π 2 ⇒ dV = 1<br />

2π 2 C2 dC. WhenC =84and dC =0.5,<br />

dV = 1<br />

2π 2 (84)2 (0.5) = 1764<br />

1764<br />

, so the maximum error is about ≈ 179 cm 3 .<br />

π2 π 2<br />

The relative error is approximately dV V = 1764/π2<br />

(84) 3 /(6π 2 ) = 1 56 ≈ 0.018.<br />

37. (a) V = πr 2 h ⇒ ∆V ≈ dV =2πrh dr =2πrh ∆r<br />

(b) The error is<br />

∆V − dV =[π(r + ∆r) 2 h − πr 2 h] − 2πrh ∆r = πr 2 h +2πrh ∆r + π(∆r) 2 h − πr 2 h − 2πrh ∆r = π(∆r) 2 h.<br />

39. V = RI ⇒ I = V R ⇒ dI = − V ∆I<br />

dR. The relative error in calculating I is ≈ dI<br />

R2 I I = −(V/R2 ) dR<br />

= − dR V/R R .<br />

Hence, the relative error in calculating I is approximately the same (in magnitude) as the relative error in R.<br />

41. (a) dc = dc dx =0dx =0<br />

dx<br />

(c) d(u + v) = d<br />

du<br />

dx (u + v) dx = dx + dv<br />

dx<br />

(d) d(uv) = d <br />

dx (uv) dx = u dv<br />

dx + v du<br />

dx<br />

u<br />

<br />

(e) d = d u<br />

v du<br />

dx = dx − u dv<br />

dx dx =<br />

v dx v<br />

v 2<br />

(f ) d (x n )= d<br />

dx (xn ) dx = nx n−1 dx<br />

<br />

dx = du dv<br />

dx + dx = du + dv<br />

dx dx<br />

<br />

dx = u dv du<br />

dx + v dx = udv+ vdu<br />

dx dx<br />

v du dv<br />

dx − u<br />

dx dx dx vdu− udv<br />

=<br />

v 2<br />

v 2<br />

43. (a) The graph shows that f 0 (1) = 2,soL(x) =f(1) + f 0 (1)(x − 1) = 5 + 2(x − 1) = 2x +3.<br />

f(0.9) ≈ L(0.9) = 4.8 and f(1.1) ≈ L(1.1) = 5.2.<br />

d<br />

du<br />

(b) d(cu) = (cu) dx = c dx = cdu<br />

dx dx<br />

(b) From the graph, we see that f 0 (x) is positive and decreasing. This means that the slopes of the tangent lines are positive,<br />

but the tangents are becoming less steep. So the tangent lines lie above the curve. Thus, the estimates in part (a) are too<br />

large.

F.<br />

126 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

3.11 Hyperbolic Functions<br />

1. (a) sinh 0 = 1 2 (e0 − e 0 )=0 (b) cosh 0 = 1 2 (e0 + e 0 )= 1 2<br />

(1 + 1) = 1<br />

3. (a) sinh(ln 2) = eln 2 − e −ln 2<br />

2<br />

= eln 2 − (e ln 2 ) −1<br />

2<br />

= 2 − 2−1<br />

2<br />

= 2 − 1 2<br />

2<br />

= 3 4<br />

(b) sinh 2 = 1 2 (e2 − e −2 ) ≈ 3.62686<br />

5. (a) sech 0 = 1<br />

cosh 0 = 1 1 =1 (b) cosh−1 1=0because cosh 0 = 1.<br />

7. sinh(−x) = 1 2 [e−x − e −(−x) ]= 1 2 (e−x − e x )=− 1 2 (e−x − e x )=− sinh x<br />

9. cosh x +sinhx = 1 2 (ex + e −x )+ 1 2 (ex − e −x )= 1 2 (2ex )=e x<br />

11. sinh x cosh y +coshx sinh y = 1<br />

2 (ex − e −x ) 1<br />

2 (ey + e −y ) + 1<br />

2 (ex + e −x ) 1<br />

2 (ey − e −y ) <br />

= 1 4 [(ex+y + e x−y − e −x+y − e −x−y )+(e x+y − e x−y + e −x+y − e −x−y )]<br />

= 1 4 (2ex+y − 2e −x−y )= 1 2 [ex+y − e −(x+y) ]=sinh(x + y)<br />

13. Divide both sides of the identity cosh 2 x − sinh 2 x =1by sinh 2 x:<br />

cosh 2 x<br />

sinh 2 x − sinh2 x<br />

sinh 2 x = 1<br />

sinh 2 x ⇔ coth2 x − 1=csch 2 x.<br />

15. Putting y = x in the result from Exercise 11, we have<br />

sinh 2x =sinh(x + x) =sinhx cosh x +coshx sinh x =2sinhx cosh x.<br />

17. tanh(ln x) =<br />

sinh(ln x)<br />

cosh(ln x) = (eln x − e − ln x )/2<br />

(e ln x + e − ln x )/2 = x − (eln x ) −1 x − x−1 x − 1/x<br />

= =<br />

x +(e ln x )<br />

−1<br />

x + x−1 x +1/x = (x2 − 1)/x<br />

(x 2 +1)/x = x2 − 1<br />

x 2 +1<br />

19. By Exercise 9, (cosh x +sinhx) n =(e x ) n = e nx =coshnx +sinhnx.<br />

21. sech x = 1<br />

cosh x ⇒ sech x = 1<br />

5/3 = 3 5 .<br />

cosh 2 x − sinh 2 x =1 ⇒ sinh 2 x =cosh 2 x − 1= <br />

5 2<br />

− 1= 16 ⇒ sinh x = 4 [because x>0].<br />

3<br />

9 3<br />

csch x = 1<br />

sinh x ⇒ csch x = 1<br />

4/3 = 3 4 .<br />

tanh x = sinh x<br />

cosh x<br />

⇒ tanh x =<br />

4/3<br />

5/3 = 4 5 .<br />

coth x = 1<br />

tanh x ⇒ coth x = 1<br />

4/5 = 5 4 .<br />

e x − e −x<br />

23. (a) lim tanh x = lim<br />

x→∞ x→∞ e x + e<br />

(b) lim tanh x =<br />

x→−∞<br />

lim<br />

x→−∞<br />

−x ·<br />

e−x<br />

= lim<br />

e−x e x − e −x ex<br />

·<br />

e x + e−x e = x<br />

x→∞<br />

lim<br />

x→−∞<br />

1 − e −2x<br />

1+e −2x = 1 − 0<br />

1+0 =1<br />

e 2x − 1<br />

e 2x +1 = 0 − 1<br />

0+1 = −1

F.<br />

TX.10<br />

SECTION 3.11 HYPERBOLIC FUNCTIONS ¤ 127<br />

e x − e −x<br />

(c) lim sinh x = lim = ∞<br />

x→∞ x→∞ 2<br />

(d) lim sinh x =<br />

x→−∞<br />

lim<br />

x→−∞<br />

e x − e −x<br />

2<br />

= −∞<br />

2<br />

(e) lim sech x = lim =0<br />

x→∞ x→∞ e x + e−x e x + e −x<br />

(f ) lim coth x = lim<br />

x→∞ x→∞ e x − e<br />

(g)<br />

(h)<br />

−x ·<br />

e−x<br />

= lim<br />

e−x x→∞<br />

1+e −2x 1+0<br />

= =1 [Or: Use part (a)]<br />

1 − e−2x 1 − 0<br />

cosh x<br />

lim coth x = lim = ∞,sincesinh x → 0 through positive values and cosh x → 1.<br />

x→0 + x→0 + sinh x<br />

cosh x<br />

lim coth x = lim = −∞,sincesinh x → 0 through negative values and cosh x → 1.<br />

x→0− x→0 − sinh x<br />

(i) lim csch x =<br />

x→−∞<br />

lim<br />

x→−∞<br />

2<br />

=0<br />

e x − e−x 25. Let y =sinh −1 x.Thensinh y = x and, by Example 1(a), cosh 2 y − sinh 2 y =1 ⇒ [with cosh y>0]<br />

cosh y = 1+sinh 2 y = √ 1+x 2 .SobyExercise9,e y =sinhy +coshy = x + √ 1+x 2 ⇒ y =ln x + √ 1+x 2 .<br />

27. (a) Let y =tanh −1 x.Thenx =tanhy = sinh y<br />

cosh y = (ey − e −y )/2<br />

(e y + e −y )/2 · ey<br />

e = e2y − 1<br />

y e 2y +1<br />

1+x = e 2y − xe 2y ⇒ 1+x = e 2y (1 − x) ⇒ e 2y = 1+x<br />

1+x<br />

1 − x ⇒ 2y =ln 1 − x<br />

(b) Let y =tanh −1 x.Thenx =tanhy, so from Exercise 18 we have<br />

e 2y = 1+tanhy<br />

1 − tanh y = 1+x<br />

1+x<br />

1 − x ⇒ 2y =ln 1 − x<br />

⇒ y = 1 2 ln 1+x<br />

1 − x<br />

29. (a) Let y =cosh −1 x.Thencosh y = x and y ≥ 0 ⇒ sinh y dy<br />

dx =1<br />

dy<br />

dx = 1<br />

sinh y = 1<br />

<br />

cosh 2 y − 1 = 1<br />

√<br />

x2 − 1<br />

⇒<br />

<br />

.<br />

[since sinh y ≥ 0 for y ≥ 0]. Or: Use Formula 4.<br />

⇒ xe 2y + x = e 2y − 1 ⇒<br />

<br />

1+x<br />

⇒ y = 1 ln .<br />

2<br />

1 − x<br />

(b) Let y =tanh −1 x.Thentanh y = x ⇒ sech 2 y dy<br />

dy<br />

=1 ⇒<br />

dx dx = 1<br />

sech 2 y = 1<br />

1 − tanh 2 y = 1<br />

1 − x . 2<br />

Or: Use Formula 5.<br />

(c) Let y =csch −1 x.Thencsch y = x ⇒ −csch y coth y dy<br />

dy<br />

=1 ⇒<br />

dx dx = − 1<br />

. By Exercise 13,<br />

csch y coth y<br />

coth y = ± csch 2 y +1=± √ x 2 +1.Ifx>0,thencoth y>0,socoth y = √ x 2 +1.Ifx0.]<br />

1 − x2 ⇒

F.<br />

128 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

(e) Let y =coth −1 x.Thencoth y = x ⇒ −csch 2 y dy<br />

dy<br />

=1 ⇒<br />

dx dx = − 1<br />

csch 2 y = 1<br />

1 − coth 2 y = 1<br />

1 − x 2<br />

by Exercise 13.<br />

31. f(x) =x sinh x − cosh x ⇒ f 0 (x) =x (sinh x) 0 +sinhx · 1 − sinh x = x cosh x<br />

33. h(x) =ln(coshx) ⇒ h 0 (x) = 1<br />

cosh x (cosh x)0 = sinh x<br />

cosh x =tanhx<br />

35. y = e cosh 3x ⇒ y 0 = e cosh 3x · sinh 3x · 3=3e cosh 3x sinh 3x<br />

37. f(t) =sech 2 (e t )=[sech(e t )] 2 ⇒<br />

f 0 (t) =2[sech(e t )] [sech(e t )] 0 =2sech(e t ) − sech(e t )tanh(e t ) · e t = −2e t sech 2 (e t )tanh(e t )<br />

39. y = arctan(tanh x) ⇒ y 0 =<br />

1<br />

1+(tanhx) (tanh 2 x)0 =<br />

sech2 x<br />

1+tanh 2 x<br />

41. G(x) = 1 − cosh x<br />

1+coshx<br />

⇒<br />

G 0 (x) =<br />

(1 + cosh x)(− sinh x) − (1 − cosh x)(sinh x) − sinh x − sinh x cosh x − sinh x +sinhx cosh x<br />

=<br />

(1 + cosh x) 2 (1 + cosh x) 2<br />

= −2sinhx<br />

(1 + cosh x) 2<br />

43. y =tanh −1√ x ⇒ y 0 1<br />

= √ 2 · 1<br />

1<br />

2 x−1/2 =<br />

1 − x 2 √ x (1 − x)<br />

45. y = x sinh −1 (x/3) − √ 9+x 2 ⇒<br />

x<br />

<br />

y 0 =sinh −1 1/3<br />

+ x <br />

3<br />

−<br />

1+(x/3)<br />

2<br />

47. y =coth −1√ x 2 +1 ⇒ y 0 1<br />

=<br />

1 − (x 2 +1)<br />

2x<br />

x<br />

<br />

2 √ 9+x 2 =sinh−1 +<br />

3<br />

49. As the depth d of the water gets large, the fraction 2πd<br />

L<br />

approaches 1. Thus, v =<br />

x<br />

√<br />

9+x<br />

2 −<br />

2x<br />

2 √ x 2 +1 = − 1<br />

x √ x 2 +1<br />

<br />

gL 2πd gL gL<br />

2π tanh ≈<br />

L 2π (1) = 2π .<br />

x<br />

x<br />

<br />

√<br />

9+x<br />

2 =sinh−1 3<br />

2πd<br />

gets large, and from Figure 3 or Exercise 23(a), tanh L<br />

51. (a) y =20cosh(x/20) − 15 ⇒ y 0 = 20 sinh(x/20) · 1 = sinh(x/20). Since the right pole is positioned at x =7,<br />

20<br />

we have y 0 (7) = sinh 7<br />

20 ≈ 0.3572.<br />

(b) If α is the angle between the tangent line and the x-axis, then tan α = slope of the line =sinh 7 20 ,so<br />

α =tan −1 sinh 7<br />

20<br />

≈ 0.343 rad ≈ 19.66 ◦ .Thus,theanglebetweenthelineandthepoleisθ =90 ◦ − α ≈ 70.34 ◦ .<br />

53. (a) y = A sinh mx + B cosh mx ⇒ y 0 = mA cosh mx + mB sinh mx ⇒<br />

y 00 = m 2 A sinh mx + m 2 B cosh mx = m 2 (A sinh mx + B cosh mx) =m 2 y

F.<br />

TX.10<br />

CHAPTER 3 REVIEW ¤ 129<br />

(b) From part (a), a solution of y 00 =9y is y(x) =A sinh 3x + B cosh 3x. So−4 =y(0) = A sinh 0 + B cosh 0 = B,so<br />

B = −4. Nowy 0 (x) =3A cosh 3x − 12 sinh 3x ⇒ 6=y 0 (0) = 3A ⇒ A =2,soy =2sinh3x − 4cosh3x.<br />

55. The tangent to y =coshx has slope 1 when y 0 =sinhx =1 ⇒ x =sinh −1 1=ln 1+ √ 2 ,byEquation3.<br />

Since sinh x =1and y =coshx = 1+sinh 2 x,wehavecosh x = √ 2. The point is ln 1+ √ 2 , √ 2 .<br />

57. If ae x + be −x = α cosh(x + β) [or α sinh(x + β)], then<br />

<br />

ae x + be −x = α 2 e x+β ± e −x−β <br />

= α 2 e x e β ± e −x e −β = α<br />

eβ e x ± α<br />

e−β e −x . Comparing coefficients of e x<br />

2 2<br />

and e −x ,wehavea = α 2 eβ (1) and b = ± α 2 e−β (2). Weneedtofind α and β. Dividing equation (1) by equation (2)<br />

gives us a b = ±e2β ⇒ () 2β =ln ± a b<br />

<br />

⇒ β = 1 2 ln ± a b<br />

. Solving equations (1) and (2) for e β gives us<br />

e β = 2a α and eβ = ± α 2b ,so 2a α = ± α 2b<br />

⇒ α 2 = ±4ab ⇒ α =2 √ ±ab.<br />

() If a b > 0, weusethe+ sign and obtain a cosh function, whereas if a b<br />

< 0, weusethe− sign and obtain a sinh<br />

function.<br />

In summary, if a and b havethesamesign,wehaveae x + be −x =2 √ ab cosh x + 1 2 ln a b<br />

<br />

, whereas, if a and b have the<br />

opposite sign, then ae x + be −x =2 √ −ab sinh x + 1 2 ln − a b<br />

.<br />

3 Review<br />

1. (a) The Power Rule: If n is any real number, then d<br />

dx (xn )=nx n−1 . The derivative of a variable base raised to a constant<br />

power is the power times the base raised to the power minus one.<br />

(b) The Constant Multiple Rule: If c is a constant and f is a differentiable function, then d<br />

dx [cf(x)] = c d<br />

dx f(x).<br />

The derivative of a constant times a function is the constant times the derivative of the function.<br />

(c) The Sum Rule: If f and g are both differentiable, then d<br />

dx [f(x)+g(x)] = d<br />

dx f(x)+ d g(x). The derivative of a sum<br />

dx<br />

of functions is the sum of the derivatives.<br />

(d) The Difference Rule: If f and g are both differentiable, then d<br />

d<br />

[f(x) − g(x)] =<br />

dx dx f(x) − d g(x). The derivative of a<br />

dx<br />

difference of functions is the difference of the derivatives.<br />

(e) The Product Rule: If f and g are both differentiable, then d<br />

d<br />

[f(x) g(x)] = f(x)<br />

dx dx g(x)+g(x) d f(x). The<br />

dx<br />

derivative of a product of two functions is the first function times the derivative of the second function plus the second<br />

function times the derivative of the first function.

F.<br />

130 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

(f ) The Quotient Rule: If f and g are both differentiable, then d<br />

dx<br />

f(x)<br />

g(x) d<br />

d<br />

f(x) − f(x)<br />

= dx dx g(x)<br />

g(x)<br />

[ g(x)] 2 .<br />

The derivative of a quotient of functions is the denominator times the derivative of the numerator minus the numerator<br />

times the derivative of the denominator, all divided by the square of the denominator.<br />

(g) The Chain Rule: If f and g are both differentiable and F = f ◦ g is the composite function defined by F (x) =f(g(x)),<br />

then F is differentiable and F 0 is given by the product F 0 (x) =f 0 (g(x)) g 0 (x). The derivative of a composite function is<br />

the derivative of the outer function evaluated at the inner function times the derivative of the inner function.<br />

2. (a) y = x n ⇒ y 0 = nx n−1 (b) y = e x ⇒ y 0 = e x<br />

(c) y = a x ⇒ y 0 = a x ln a (d) y =lnx ⇒ y 0 =1/x<br />

(e) y =log a x ⇒ y 0 =1/(x ln a) (f ) y =sinx ⇒ y 0 =cosx<br />

(g) y =cosx ⇒ y 0 = − sin x (h) y =tanx ⇒ y 0 =sec 2 x<br />

(i) y =cscx ⇒ y 0 = − csc x cot x (j) y =secx ⇒ y 0 =secx tan x<br />

(k) y =cotx ⇒ y 0 = − csc 2 x (l) y =sin −1 x ⇒ y 0 =1/ √ 1 − x 2<br />

(m) y =cos −1 x ⇒ y 0 = −1/ √ 1 − x 2 (n) y =tan −1 x ⇒ y 0 =1/(1 + x 2 )<br />

(o) y =sinhx ⇒ y 0 =coshx (p) y =coshx ⇒ y 0 =sinhx<br />

(q) y =tanhx ⇒ y 0 =sech 2 x (r) y =sinh −1 x ⇒ y 0 =1/ √ 1+x 2<br />

(s) y =cosh −1 x ⇒ y 0 =1/ √ x 2 − 1 (t) y =tanh −1 x ⇒ y 0 =1/(1 − x 2 )<br />

e h − 1<br />

3. (a) e is the number such that lim =1.<br />

h→0 h<br />

(b) e =lim<br />

x→0<br />

(1 + x) 1/x<br />

(c) The differentiation formula for y = a x [y 0 = a x ln a] is simplest when a = e because ln e =1.<br />

(d) The differentiation formula for y =log a x [y 0 =1/(x ln a)] issimplestwhena = e because ln e =1.<br />

4. (a) Implicit differentiation consists of differentiating both sides of an equation involving x and y with respect to x,andthen<br />

solving the resulting equation for y 0 .<br />

(b) Logarithmic differentiation consists of taking natural logarithms of both sides of an equation y = f(x), simplifying,<br />

differentiating implicitly with respect to x, and then solving the resulting equation for y 0 .<br />

5. (a) The linearization L of f at x = a is L(x) =f(a)+f 0 (a)(x − a).<br />

(b) If y = f(x), then the differential dy is given by dy = f 0 (x) dx.<br />

(c)SeeFigure5inSection3.10.

F.<br />

TX.10<br />

CHAPTER 3 REVIEW ¤ 131<br />

1. True. This is the Sum Rule.<br />

3. True. This is the Chain Rule.<br />

5. False.<br />

√ <br />

d<br />

√ f 0 x<br />

dx f x =<br />

2 √ by the Chain Rule.<br />

x<br />

7. False.<br />

9. True.<br />

d<br />

dx 10x =10 x ln 10<br />

d<br />

dx (tan2 x)=2tanx sec 2 x,and d<br />

dx (sec2 x)=2secx (sec x tan x) =2tanx sec 2 x.<br />

Or:<br />

d<br />

dx (sec2 x)= d<br />

dx (1 + tan2 x)= d<br />

dx (tan2 x).<br />

11. True. g(x) =x 5 ⇒ g 0 (x) =5x 4 ⇒ g 0 (2) = 5(2) 4 =80,andbythedefinition of the derivative,<br />

g(x) − g(2)<br />

lim<br />

= g 0 (2) = 80.<br />

x→2 x − 2<br />

1. y =(x 4 − 3x 2 +5) 3 ⇒<br />

y 0 =3(x 4 − 3x 2 +5) 2<br />

d<br />

dx (x4 − 3x 2 +5)=3(x 4 − 3x 2 +5) 2 (4x 3 − 6x) =6x(x 4 − 3x 2 +5) 2 (2x 2 − 3)<br />

3. y = √ x + 1<br />

3√<br />

x<br />

4 = x1/2 + x −4/3 ⇒ y 0 = 1 2 x−1/2 − 4 3 x−7/3 = 1<br />

2 √ x − 4<br />

3 3√ x 7<br />

5. y =2x √ x 2 +1 ⇒<br />

y 0 =2x · 1<br />

2 (x2 +1) −1/2 (2x)+ √ x 2 +1(2)= √ 2x2<br />

x2 +1 +2√ x 2 +1= 2x2 +2(x 2 +1)<br />

√ = 2(2x2 +1)<br />

√<br />

x2 +1 x2 +1<br />

7. y = e sin 2θ ⇒ y 0 = e sin 2θ d<br />

dθ (sin 2θ) =esin 2θ sin 2θ<br />

(cos 2θ)(2) = 2 cos 2θe<br />

9. y = t<br />

1 − t 2 ⇒ y 0 = (1 − t2 )(1) − t(−2t)<br />

(1 − t 2 ) 2 = 1 − t2 +2t 2<br />

(1 − t 2 ) 2 = t2 +1<br />

(1 − t 2 ) 2<br />

11. y = √ x cos √ x ⇒<br />

y 0 = √ <br />

x cos √ x<br />

0<br />

+cos<br />

√<br />

x<br />

√<br />

x<br />

0<br />

=<br />

√<br />

x<br />

<br />

− sin √ x<br />

=<br />

− √ 1 2 x−1/2 x sin √ x +cos √ <br />

x = cos √ x − √ x sin √ x<br />

2 √ x<br />

<br />

1<br />

2 x−1/2 <br />

+cos √ x<br />

<br />

1<br />

2 x−1/2 <br />

13. y = e1/x<br />

⇒ y 0 = x2 (e 1/x ) 0 − e 1/x x 2 0<br />

= x2 (e 1/x )(−1/x 2 ) − e 1/x (2x)<br />

= −e1/x (1 + 2x)<br />

x 2 (x 2 ) 2 x 4 x 4

F.<br />

132 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

15.<br />

d<br />

dx (xy4 + x 2 y)= d<br />

dx (x +3y) ⇒ x · 4y3 y 0 + y 4 · 1+x 2 · y 0 + y · 2x =1+3y 0 ⇒<br />

y 0 (4xy 3 + x 2 − 3) = 1 − y 4 − 2xy ⇒ y 0 = 1 − y4 − 2xy<br />

4xy 3 + x 2 − 3<br />

17. y =<br />

sec 2θ<br />

1+tan2θ<br />

⇒<br />

y 0 = (1 + tan 2θ)(sec 2θ tan 2θ · 2) − (sec 2θ)(sec2 2θ · 2)<br />

= 2sec2θ [(1 + tan 2θ)tan2θ − sec2 2θ]<br />

(1 + tan 2θ) 2 (1 + tan 2θ) 2<br />

= 2sec2θ (tan 2θ +tan2 2θ − sec 2 2θ) 2sec2θ (tan 2θ − 1) <br />

= 1+tan 2 x =sec 2 x <br />

(1 + tan 2θ) 2 (1 + tan 2θ) 2<br />

19. y = e cx (c sin x − cos x) ⇒<br />

y 0 = e cx (c cos x +sinx)+ce cx (c sin x − cos x) =e cx (c 2 sin x − c cos x + c cos x +sinx)<br />

= e cx (c 2 sin x +sinx) =e cx sin x (c 2 +1)<br />

21. y =3 x ln x ⇒ y 0 =3 x ln x · ln 3 ·<br />

<br />

d<br />

dx (x ln x) =3x ln x · ln 3 x · 1 <br />

x +lnx · 1 =3 x ln x · ln 3(1 + ln x)<br />

23. y =(1− x −1 ) −1 ⇒<br />

y 0 = −1(1 − x −1 ) −2 [−(−1x −2 )] = −(1 − 1/x) −2 x −2 = −((x − 1)/x) −2 x −2 = −(x − 1) −2<br />

25. sin(xy) =x 2 − y ⇒ cos(xy)(xy 0 + y · 1) = 2x − y 0 ⇒ x cos(xy)y 0 + y 0 =2x − y cos(xy) ⇒<br />

y 0 [x cos(xy)+1]=2x − y cos(xy) ⇒ y 0 =<br />

2x − y cos(xy)<br />

x cos(xy)+1<br />

27. y =log 5 (1 + 2x) ⇒ y 0 =<br />

1 d<br />

(1 + 2x) ln5dx (1 + 2x) = 2<br />

(1 + 2x) ln5<br />

29. y =lnsinx − 1 2 sin2 x ⇒ y 0 = 1<br />

sin x · cos x − 1 · 2sinx · cos x =cotx − sin x cos x<br />

2<br />

31. y = x tan −1 (4x) ⇒ y 0 = x ·<br />

1<br />

1+(4x) · 4x<br />

2 4+tan−1 (4x) · 1=<br />

1+16x 2 +tan−1 (4x)<br />

33. y =ln|sec 5x +tan5x| ⇒<br />

y 0 =<br />

1<br />

sec 5x +tan5x (sec 5x tan 5x · 5sec5x (tan 5x +sec5x)<br />

5+sec2 5x · 5) = =5sec5x<br />

sec 5x +tan5x<br />

35. y =cot(3x 2 +5) ⇒ y 0 = − csc 2 (3x 2 + 5)(6x) =−6x csc 2 (3x 2 +5)<br />

37. y =sin tan √ 1+x 3 ⇒ y 0 =cos tan √ 1+x 3 sec 2 √ 1+x 3 3x 2 2 √ 1+x 3 <br />

39. y =tan 2 (sin θ) =[tan(sinθ)] 2 ⇒ y 0 =2[tan(sinθ)] · sec 2 (sin θ) · cos θ

F.<br />

TX.10<br />

CHAPTER 3 REVIEW ¤ 133<br />

√ x +1(2− x)<br />

5<br />

41. y =<br />

⇒ ln y = 1<br />

(x +3) 7 2<br />

√ x +1(2− x)<br />

y 0 5<br />

<br />

1<br />

=<br />

(x +3) 7 2(x +1) − 5<br />

2 − x − 7 <br />

x +3<br />

ln(x +1)+5ln(2− x) − 7ln(x +3) ⇒<br />

y0<br />

y = 1<br />

2(x +1) + −5<br />

2 − x − 7<br />

or y 0 = (2 − x)4 (3x 2 − 55x − 52)<br />

2 √ x +1(x +3) 8 .<br />

x +3<br />

⇒<br />

43. y = x sinh(x 2 ) ⇒ y 0 = x cosh(x 2 ) · 2x +sinh(x 2 ) · 1=2x 2 cosh(x 2 )+sinh(x 2 )<br />

45. y =ln(cosh3x) ⇒ y 0 =(1/ cosh 3x)(sinh 3x)(3) = 3 tanh 3x<br />

47. y =cosh −1 (sinh x) ⇒ y 0 =<br />

<br />

49. y =cos e √ <br />

tan 3x<br />

⇒<br />

1<br />

<br />

(sinh x)2 − 1 · cosh x = cosh x<br />

<br />

sinh 2 x − 1<br />

<br />

y 0 = − sin e √ <br />

tan 3x<br />

· e √ 0 <br />

tan 3x = − sin e √ <br />

tan 3x<br />

e √ tan 3x · 1 (tan 2 3x)−1/2 · sec 2 (3x) · 3<br />

<br />

−3sin e √ <br />

tan 3x<br />

e √ tan 3x sec 2 (3x)<br />

=<br />

2 √ tan 3x<br />

51. f(t) = √ 4t +1 ⇒ f 0 (t) = 1 2 (4t +1)−1/2 · 4=2(4t +1) −1/2 ⇒<br />

f 00 (t) =2(− 1 2 )(4t +1)−3/2 · 4=−4/(4t +1) 3/2 ,sof 00 (2) = −4/9 3/2 = − 4<br />

27 .<br />

53. x 6 + y 6 =1 ⇒ 6x 5 +6y 5 y 0 =0 ⇒ y 0 = −x 5 /y 5 ⇒<br />

y 00 = − y5 (5x 4 ) − x 5 (5y 4 y 0 )<br />

= − 5x4 y 4 y − x(−x 5 /y 5 ) = − 5x4 (y 6 + x 6 )/y 5<br />

= − 5x4<br />

(y 5 ) 2 y 10<br />

y 6<br />

y 11<br />

55. We firstshowitistrueforn =1: f(x) =xe x ⇒ f 0 (x) =xe x + e x =(x +1)e x . We now assume it is true<br />

for n = k: f (k) (x) =(x + k)e x . With this assumption, we must show it is true for n = k +1:<br />

f (k+1) (x) = d <br />

f (k) (x) = d<br />

dx<br />

dx [(x + k)ex ]=(x + k)e x + e x =[(x + k)+1]e x =[x +(k +1)]e x .<br />

Therefore, f (n) (x) =(x + n)e x by mathematical induction.<br />

57. y =4sin 2 x ⇒ y 0 =4· 2sinx cos x. At π<br />

6 , 1 , y 0 =8· 1<br />

2 · √3<br />

2 =2√ 3, so an equation of the tangent line<br />

is y − 1=2 √ 3 x − π 6<br />

<br />

,ory =2<br />

√<br />

3 x +1− π<br />

√<br />

3/3.<br />

59. y = √ 1+4sinx ⇒ y 0 = 1 2 (1 + 4 sin x)−1/2 · 4cosx =<br />

2cosx<br />

√ 1+4sinx<br />

.<br />

At (0, 1), y 0 = 2 √<br />

1<br />

=2, so an equation of the tangent line is y − 1=2(x − 0),ory =2x +1.<br />

61. y =(2+x)e −x ⇒ y 0 =(2+x)(−e −x )+e −x · 1=e −x [−(2 + x)+1]=e −x (−x − 1).<br />

At (0, 2), y 0 =1(−1) = −1, so an equation of the tangent line is y − 2=−1(x − 0),ory = −x +2.<br />

The slope of the normal line is 1, so an equation of the normal line is y − 2=1(x − 0),ory = x +2.

F.<br />

134 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

63. (a) f(x) =x √ 5 − x ⇒<br />

<br />

1<br />

f 0 (x) =x<br />

2 (5 − x)−1/2 (−1) + √ −x<br />

5 − x =<br />

2 √ 5 − x + √ 5 − x · 2 √ 5 − x<br />

2 √ 5 − x = −x<br />

2 √ 2(5 − x)<br />

+<br />

5 − x 2 √ 5 − x<br />

=<br />

−x +10− 2x<br />

2 √ 5 − x<br />

=<br />

10 − 3x<br />

2 √ 5 − x<br />

(b) At (1, 2): f 0 (1) = 7 4 .<br />

(c)<br />

So an equation of the tangent line is y − 2= 7 4 (x − 1) or y = 7 4 x + 1 4 .<br />

At (4, 4): f 0 (4) = − 2 2 = −1.<br />

So an equation of the tangent line is y − 4=−1(x − 4) or y = −x +8.<br />

(d)<br />

The graphs look reasonable, since f 0 is positive where f has tangents with<br />

positive slope, and f 0 is negative where f has tangents with negative slope.<br />

65. y =sinx +cosx ⇒ y 0 =cosx − sin x =0 ⇔ cos x =sinx and 0 ≤ x ≤ 2π ⇔ x = π or 5π , so the points<br />

4 4<br />

are π<br />

4 , √ 2 and 5π<br />

4 , −√ 2 .<br />

67. f(x) =(x − a)(x − b)(x − c) ⇒ f 0 (x) =(x − b)(x − c)+(x − a)(x − c)+(x − a)(x − b).<br />

So f 0 (x)<br />

f(x)<br />

=<br />

(x − b)(x − c)+(x − a)(x − c)+(x − a)(x − b)<br />

(x − a)(x − b)(x − c)<br />

= 1<br />

x − a + 1<br />

x − b + 1<br />

x − c .<br />

Or: f(x) =(x − a)(x − b)(x − c) ⇒ ln |f(x)| =ln|x − a| +ln|x − b| +ln|x − c| ⇒<br />

f 0 (x)<br />

f(x) = 1<br />

x − a + 1<br />

x − b + 1<br />

x − c<br />

69. (a) h(x) =f(x) g(x) ⇒ h 0 (x) =f(x) g 0 (x)+g(x) f 0 (x) ⇒<br />

h 0 (2) = f(2) g 0 (2) + g(2) f 0 (2) = (3)(4) + (5)(−2) = 12 − 10 = 2<br />

(b) F (x) =f(g(x)) ⇒ F 0 (x) =f 0 (g(x)) g 0 (x) ⇒ F 0 (2) = f 0 (g(2)) g 0 (2) = f 0 (5)(4) = 11 · 4=44<br />

71. f(x) =x 2 g(x) ⇒ f 0 (x) =x 2 g 0 (x)+g(x)(2x) =x[xg 0 (x)+2g(x)]<br />

73. f(x) =[g(x)] 2 ⇒ f 0 (x) =2[g(x)] · g 0 (x) =2g(x) g 0 (x)<br />

75. f(x) =g(e x ) ⇒ f 0 (x) =g 0 (e x ) e x<br />

77. f(x) =ln|g(x)| ⇒ f 0 (x) = 1<br />

g(x) g0 (x) = g0 (x)<br />

g(x)

F.<br />

TX.10<br />

CHAPTER 3 REVIEW ¤ 135<br />

79. h(x) =<br />

f(x) g(x)<br />

f(x)+g(x)<br />

⇒<br />

h 0 (x) = [f(x)+g(x)] [f(x) g0 (x)+g(x) f 0 (x)] − f(x) g(x)[f 0 (x)+g 0 (x)]<br />

[f(x)+g(x)] 2<br />

= [f(x)]2 g 0 (x)+f(x) g(x) f 0 (x)+f(x) g(x) g 0 (x)+[g(x)] 2 f 0 (x) − f(x) g(x) f 0 (x) − f(x) g(x) g 0 (x)<br />

[f(x)+g(x)] 2<br />

= f 0 (x)[g(x)] 2 + g 0 (x)[f(x)] 2<br />

[f(x)+g(x)] 2<br />

81. Using the Chain Rule repeatedly, h(x) =f(g(sin 4x)) ⇒<br />

h 0 (x) =f 0 (g(sin 4x)) ·<br />

83. y =[ln(x +4)] 2 ⇒ y 0 =2[ln(x +4)] 1 ·<br />

d<br />

dx (g(sin 4x)) = f 0 (g(sin 4x)) · g 0 d<br />

(sin 4x) ·<br />

dx (sin 4x) =f 0 (g(sin 4x))g 0 (sin 4x)(cos 4x)(4).<br />

1<br />

+4)<br />

· 1=2ln(x and y 0 =0 ⇔ ln(x +4)=0 ⇔<br />

x +4 x +4<br />

x +4=e 0 ⇒ x +4=1 ⇔ x = −3, so the tangent is horizontal at the point (−3, 0).<br />

85. y = f(x) =ax 2 + bx + c ⇒ f 0 (x) =2ax + b. We know that f 0 (−1) = 6 and f 0 (5) = −2, so−2a + b =6and<br />

10a + b = −2. Subtracting the first equation from the second gives 12a = −8 ⇒ a = − 2 . Substituting − 2 for a in the<br />

3 3<br />

first equation gives b = 14 3 .Nowf(1) = 4 ⇒ 4=a + b + c,soc =4+2 3 − 14<br />

3 =0and hence, f(x) =− 2 3 x2 + 14<br />

3 x.<br />

87. s(t) =Ae −ct cos(ωt + δ) ⇒<br />

v(t) =s 0 (t) =A{e −ct [−ω sin(ωt + δ)] + cos(ωt + δ)(−ce −ct )} = −Ae −ct [ω sin(ωt + δ)+c cos(ωt + δ)]<br />

⇒<br />

a(t) =v 0 (t) =−A{e −ct [ω 2 cos(ωt + δ) − cω sin(ωt + δ)] + [ω sin(ωt + δ)+c cos(ωt + δ)](−ce −ct )}<br />

= −Ae −ct [ω 2 cos(ωt + δ) − cω sin(ωt + δ) − cω sin(ωt + δ) − c 2 cos(ωt + δ)]<br />

= −Ae −ct [(ω 2 − c 2 )cos(ωt + δ) − 2cω sin(ωt + δ)] = Ae −ct [(c 2 − ω 2 )cos(ωt + δ)+2cω sin(ωt + δ)]<br />

89. (a) y = t 3 − 12t +3 ⇒ v(t) =y 0 =3t 2 − 12 ⇒ a(t) =v 0 (t) =6t<br />

(b) v(t) =3(t 2 − 4) > 0 when t>2,soitmovesupwardwhent>2 anddownwardwhen0 ≤ t2. The particle is slowing down when v and a have opposite<br />

signs; that is, when 0

F.<br />

136 ¤ CHAPTER 3 DIFFERENTIATION RULES<br />

TX.10<br />

93. (a) y(t) =y(0)e kt =200e kt ⇒ y(0.5) = 200e 0.5k =360 ⇒ e 0.5k =1.8 ⇒ 0.5k =ln1.8 ⇒<br />

k =2ln1.8 =ln(1.8) 2 =ln3.24 ⇒ y(t) = 200e (ln 3.24)t = 200(3.24) t<br />

(b) y(4) = 200(3.24) 4 ≈ 22,040 bacteria<br />

(c) y 0 (t) = 200(3.24) t · ln 3.24,soy 0 (4) = 200(3.24) 4 · ln 3.24 ≈ 25,910 bacteria per hour<br />

(d) 200(3.24) t =10,000 ⇒ (3.24) t =50 ⇒ t ln 3.24 = ln 50 ⇒ t =ln50/ ln 3.24 ≈ 3.33 hours<br />

95. (a) C 0 (t) =−kC(t) ⇒ C(t) =C(0)e −kt by Theorem 9.4.2. But C(0) = C 0 ,soC(t) =C 0 e −kt .<br />

(b) C(30) = 1 2 C0 since the concentration is reduced by half. Thus, 1 2 C0 = C0e−30k ⇒ ln 1 2 = −30k ⇒<br />

k = − 1 ln 1 = 1 ln 2. Since10% of the original concentration remains if 90% is eliminated, we want the value of t<br />

30 2 30<br />

such that C(t) = 1 10 C 0. Therefore,<br />

1<br />

C 10 0 = C 0 e −t(ln 2)/30 ⇒ ln 0.1 =−t(ln 2)/30 ⇒ t = − 30 ln 0.1 ≈ 100 h.<br />

ln 2<br />

97. If x = edge length, then V = x 3 ⇒ dV/dt =3x 2 dx/dt =10 ⇒ dx/dt =10/(3x 2 ) and S =6x 2 ⇒<br />

dS/dt =(12x) dx/dt =12x[10/(3x 2 )] = 40/x. Whenx =30, dS/dt = 40<br />

30 = 4 3 cm2 /min.<br />

99. Given dh/dt =5and dx/dt =15, find dz/dt. z 2 = x 2 + h 2 ⇒<br />

2z dz dx dh<br />

=2x +2h<br />

dt dt dt<br />

⇒<br />

dz<br />

dt = 1 (15x +5h). Whent =3,<br />

z<br />

h =45+3(5)=60and x = 15(3) = 45 ⇒ z = √ 45 2 +60 2 =75,<br />

so dz<br />

dt = 1 [15(45) + 5(60)] = 13 ft/s.<br />

75<br />

101. We are given dθ/dt = −0.25 rad/h. tan θ =400/x ⇒<br />

x =400cotθ ⇒ dx<br />

dt = −400 csc2 θ dθ<br />

dt .Whenθ = π , 6<br />

dx<br />

dt = −400(2)2 (−0.25) = 400 ft/h.<br />

103. (a) f(x) = 3√ 1+3x =(1+3x) 1/3 ⇒ f 0 (x) =(1+3x) −2/3 , so the linearization of f at a =0is<br />

L(x) =f(0) + f 0 (0)(x − 0) = 1 1/3 +1 −2/3 x =1+x. Thus,<br />

3√<br />

1.03 =<br />

3 1+3(0.01) ≈ 1+(0.01) = 1.01.<br />

3√ 1+3x ≈ 1+x<br />

⇒<br />

(b) The linear approximation is 3√ 1+3x ≈ 1+x,sofortherequiredaccuracy<br />

we want 3√ 1+3x − 0.1 < 1+x< 3√ 1+3x +0.1. From the graph,<br />

itappearsthatthisistruewhen−0.23

F.<br />

TX.10<br />

CHAPTER 3 REVIEW ¤ 137<br />

105. A = x 2 + 1 π 1<br />

2 2 x2 = <br />

1+ π 8 x<br />

2<br />

⇒ dA = 2+ π 4<br />

<br />

xdx.Whenx =60<br />

and dx =0.1, dA = 2+ π 4<br />

approximately 12 + 3π 2 ≈ 16.7 cm2 .<br />

√ <br />

4<br />

16 + h − 2 d<br />

107. lim<br />

=<br />

h→0 h dx<br />

<br />

60(0.1) = 12 +<br />

3π<br />

2<br />

, so the maximum error is<br />

4√<br />

x<br />

<br />

x =16<br />

= 1 4 x−3/4 x<br />

=16<br />

=<br />

1<br />

4 √ 4<br />

16 3 = 1<br />

32<br />

√ √ √ √ √ √ <br />

1+tanx − 1+sinx 1+tanx − 1+sinx 1+tanx + 1+sinx<br />

109. lim<br />

=lim<br />

x→0 x 3<br />

x→0 x √ 3 1+tanx + √ 1+sinx <br />

(1 + tan x) − (1 + sin x)<br />

=lim<br />

x→0 x 3√ 1+tanx + √ 1+sinx =lim sin x (1/ cos x − 1)<br />

x→0 x 3√ 1+tanx + √ 1+sinx · cos x<br />

cos x<br />

sin x (1 − cos x)<br />

=lim<br />

x→0 x 3√ 1+tanx + √ 1+sinx cos x · 1+cosx<br />

1+cosx<br />

sin x · sin 2 x<br />

=lim<br />

x→0 x 3√ 1+tanx + √ 1+sinx cos x (1 + cos x)<br />

<br />

= lim<br />

x→0<br />

=1 3 ·<br />

3<br />

sin x<br />

1<br />

lim √ √ <br />

x x→0 1+tanx + 1+sinx cos x (1 + cos x)<br />

1<br />

√<br />

1+<br />

√<br />

1<br />

<br />

· 1 · (1 + 1)<br />

= 1 4<br />

111.<br />

d<br />

dx [f(2x)] = x2 ⇒ f 0 (2x) · 2=x 2 ⇒ f 0 (2x) = 1 2 x2 .Lett =2x. Thenf 0 (t) = 1 1<br />

2 2 t2 = 1 8 t2 ,sof 0 (x) = 1 8 x2 .

F.<br />

TX.10

F.<br />

TX.10<br />

PROBLEMS PLUS<br />

1. Let a be the x-coordinate of Q. Since the derivative of y =1− x 2 is y 0 = −2x,theslopeatQ is −2a. But since the triangle<br />

is equilateral, AO/OC = √ 3/1, so the slope at Q is − √ 3. Therefore, we must have that −2a = − √ 3 ⇒ a = √ 3<br />

Thus, the point Q has coordinates<br />

√3<br />

√3<br />

, 1 − 2 2<br />

2 √3 <br />

<br />

= , 1 andbysymmetry,P has coordinates − √ <br />

3<br />

, 1 .<br />

2 4<br />

2 4<br />

2 .<br />

3. We must show that r (in the figure) is halfway between p and q,thatis,<br />

r =(p + q)/2. For the parabola y = ax 2 + bx + c, the slope of the tangent line is<br />

given by y 0 =2ax + b. An equation of the tangent line at x = p is<br />

y − (ap 2 + bp + c) =(2ap + b)(x − p). Solving for y gives us<br />

y =(2ap + b)x − 2ap 2 − bp +(ap 2 + bp + c)<br />

or y =(2ap + b)x + c − ap 2 (1)<br />

Similarly, an equation of the tangent line at x = q is<br />

y =(2aq + b)x + c − aq 2 (2)<br />

We can eliminate y and solve for x by subtracting equation (1) from equation (2).<br />

[(2aq + b) − (2ap + b)]x − aq 2 + ap 2 =0<br />

(2aq − 2ap)x = aq 2 − ap 2<br />

2a(q − p)x = a(q 2 − p 2 )<br />

x =<br />

a(q + p)(q − p)<br />

2a(q − p)<br />

= p + q<br />

2<br />

Thus, the x-coordinate of the point of intersection of the two tangent lines, namely r,is(p + q)/2.<br />

5. Let y =tan −1 x.Thentan y = x, so from the triangle we see that<br />

sin(tan −1 x)=siny =<br />

sin(tan −1 (sinh x)) =<br />

x<br />

√ . Using this fact we have that<br />

1+x<br />

2<br />

sinh x<br />

<br />

1+sinh 2 x = sinh x<br />

cosh x =tanhx.<br />

Hence, sin −1 (tanh x) =sin −1 (sin(tan −1 (sinh x))) = tan −1 (sinh x).<br />

7. We use mathematical induction. Let S n be the statement that<br />

S 1 is true because<br />

d n<br />

dx n (sin4 x +cos 4 x)=4 n−1 cos(4x + nπ/2).<br />

d<br />

dx (sin4 x +cos 4 x)=4sin 3 x cos x − 4cos 3 x sin x =4sinx cos x sin 2 x − cos 2 x x<br />

= −4sinx cos x cos 2x = −2sin2x cos 2 = − sin 4x =sin(−4x)<br />

=cos π<br />

− (−4x) 2<br />

=cos π<br />

+4x 2<br />

=4 n−1 cos <br />

4x + n π 2 when n =1<br />

[continued]<br />

139

F.<br />

140 ¤ CHAPTER 3 PROBLEMS PLUS<br />

TX.10<br />

d k <br />

Now assume S k is true, that is, sin 4 x +cos 4 x =4 k−1 cos <br />

4x + k π<br />

dx k 2 .Then<br />

d k+1<br />

dx k+1 (sin4 x +cos 4 x)= d <br />

d<br />

k<br />

dx dx k (sin4 x +cos 4 x) = d 4 k−1 cos <br />

4x + k π 2<br />

dx<br />

which shows that S k+1 is true.<br />

Therefore,<br />

Another proof: First write<br />

= −4 k−1 sin <br />

4x + k π d<br />

2 ·<br />

dx<br />

=4 k sin −4x − k π 2<br />

<br />

4x + k<br />

π<br />

2 = −4 k sin <br />

4x + k π 2<br />

=4 k cos π<br />

2 − −4x − k π 2<br />

=4 k cos 4x +(k +1) π 2<br />

d n<br />

dx n (sin4 x +cos 4 x)=4 n−1 cos <br />

4x + n π 2 for every positive integer n, by mathematical induction.<br />

sin 4 x +cos 4 x =(sin 2 x +cos 2 x) 2 − 2sin 2 x cos 2 x =1− 1 2 sin2 2x =1− 1 (1 − cos 4x) = 3 + 1 cos 4x<br />

4 4 4<br />

<br />

Then we have dn<br />

dx n (sin4 x +cos 4 x)= dn 3<br />

dx n 4 + 1 <br />

4 cos 4x = 1 <br />

4 · 4n cos 4x + n π <br />

=4 n−1 cos 4x + n π <br />

.<br />

2<br />

2<br />

9. We must find a value x 0 such that the normal lines to the parabola y = x 2 at x = ±x 0 intersectatapointoneunitfromthe<br />

points <br />

±x 0 ,x 2 0 . The normals to y = x 2 at x = ±x 0 have slopes − 1 and pass through <br />

±x 0 ,x 2 0 respectively, so the<br />

±2x 0<br />

normals have the equations y − x 2 0 = − 1<br />

2x 0<br />

(x − x 0) and y − x 2 0 = 1<br />

2x 0<br />

(x + x 0). The common y-intercept is x 2 0 + 1 2 .<br />

We want to find the value of x 0 for which the distance from 0,x 2 0 + 1 2<br />

<br />

to<br />

<br />

x0,x 2 0<br />

equals 1. The square of the distance is<br />

(x 0 − 0) 2 + x 2 0 − x 2 0 + 1 2<br />

the center of the circle is at 0, 5 4<br />

.<br />

2<br />

= x 2 0 + 1 4 =1 ⇔ x 0 = ± √ 3<br />

2 .Forthesevaluesofx 0,they-intercept is x 2 0 + 1 2 = 5 4 ,so<br />

Another solution: Let the center of the circle be (0,a). Then the equation of the circle is x 2 +(y − a) 2 =1.<br />

Solving with the equation of the parabola, y = x 2 ,wegetx 2 +(x 2 − a) 2 =1 ⇔ x 2 + x 4 − 2ax 2 + a 2 =1 ⇔<br />

x 4 +(1− 2a)x 2 + a 2 − 1=0. The parabola and the circle will be tangent to each other when this quadratic equation in x 2<br />

<br />

has equal roots; that is, when the discriminant is 0. Thus,(1 − 2a) 2 − 4(a 2 − 1) = 0<br />

⇔<br />

1 − 4a +4a 2 − 4a 2 +4=0 ⇔ 4a =5,soa = 5 4 . The center of the circle is 0, 5 4<br />

<br />

.<br />

11. We can assume without loss of generality that θ =0at time t =0,sothatθ =12πt rad. [The angular velocity of the wheel<br />

is 360 rpm =360· (2π rad)/(60 s) =12π rad/s.] Then the position of A as a function of time is<br />

A =(40cosθ, 40 sin θ) = (40 cos 12πt, 40 sin 12πt),sosin α =<br />

y 40 sin θ<br />

= = sin θ = 1 sin 12πt.<br />

1.2 m 120 3 3<br />

(a) Differentiating the expression for sin α,wegetcos α · dα<br />

dt = 1 3 · 12π · cos 12πt =4π cos θ. Whenθ = π 3 ,wehave<br />

sin α = 1 √<br />

<br />

√ 2 <br />

3<br />

3 11<br />

3 sin θ = 6 ,socos α = dα<br />

1 − = and<br />

6 12 dt = 4π cos π 3<br />

cos α = 2π<br />

= 4π √ 3<br />

11/12 11

F.<br />

TX.10<br />

CHAPTER 3 PROBLEMS PLUS ¤ 141<br />

(b) By the Law of Cosines, |AP | 2 = |OA| 2 + |OP| 2 − 2 |OA||OP| cos θ<br />

⇒<br />

120 2 =40 2 + |OP| 2 − 2 · 40 |OP| cos θ ⇒ |OP| 2 − (80 cos θ) |OP| − 12,800 = 0 ⇒<br />

|OP| = 1 2<br />

<br />

80 cos θ ±<br />

√ 6400 cos2 θ +51,200 =40cosθ ± 40 √ cos 2 θ +8=40 cos θ + √ 8+cos 2 θ cm<br />

[since |OP| > 0]. As a check, note that |OP| =160cm when θ =0and |OP| =80 √ 2 cm when θ = π 2 .<br />

(c) By part (b), the x-coordinate of P is given by x =40 cos θ + √ 8+cos 2 θ ,so<br />

dx<br />

dt = dx<br />

dθ<br />

<br />

dθ<br />

dt =40 − sin θ −<br />

<br />

<br />

2cosθ sin θ<br />

2 √ · 12π = −480π sin θ 1+<br />

8+cos 2 θ<br />

In particular, dx/dt =0cm/swhenθ =0and dx/dt = −480π cm/swhenθ = π 2 .<br />

13. Consider the statement that<br />

d<br />

dx (eax sin bx) =ae ax sin bx + be ax cos bx,and<br />

d n<br />

dx n (eax sin bx) =r n e ax sin(bx + nθ). Forn =1,<br />

<br />

cos θ<br />

√ cm/s.<br />

8+cos2 θ<br />

re ax sin(bx + θ) =re ax [sin bx cos θ +cosbx sin θ] =re ax a<br />

r sin bx + b r cos bx <br />

= ae ax sin bx + be ax cos bx<br />

since tan θ = b a<br />

⇒ sin θ = b r and cos θ = a . So the statement is true for n =1.<br />

r<br />

But<br />

Assume it is true for n = k. Then<br />

d k+1<br />

dx k+1 (eax sin bx)= d <br />

r k e ax sin(bx + kθ) = r k ae ax sin(bx + kθ)+r k e ax b cos(bx + kθ)<br />

dx<br />

= r k e ax [a sin(bx + kθ)+b cos(bx + kθ)]<br />

sin[bx +(k +1)θ] =sin[(bx + kθ)+θ] =sin(bx + kθ)cosθ +sinθ cos(bx + kθ) = a sin(bx + kθ)+ b cos(bx + kθ).<br />

r r<br />

Hence, a sin(bx + kθ)+b cos(bx + kθ) =r sin[bx +(k +1)θ].So<br />

d k+1<br />

dx k+1 (eax sin bx) =r k e ax [a sin(bx+kθ)+b cos(bx+kθ)] = r k e ax [r sin(bx+(k +1)θ)] = r k+1 e ax [sin(bx+(k +1)θ)].<br />

Therefore, the statement is true for all n by mathematical induction.<br />

15. It seems from the figure that as P approaches the point (0, 2) from the right, x T →∞and y T → 2 + .AsP approaches the<br />

point (3, 0) from the left, it appears that x T → 3 + and y T →∞.Soweguessthatx T ∈ (3, ∞) and y T ∈ (2, ∞). Itis<br />

more difficult to estimate the range of values for x N and y N. We might perhaps guess that x N ∈ (0, 3),<br />

and y N ∈ (−∞, 0) or (−2, 0).<br />

In order to actually solve the problem, we implicitly differentiate the equation of the ellipse to find the equation of the<br />

tangent line: x2<br />

9 + y2<br />

4 =1 ⇒ 2x 9 + 2y 4 y0 =0,soy 0 = − 4 9<br />

x<br />

. So at the point (x0,y0) on the ellipse, an equation of the<br />

y

F.<br />

142 ¤ CHAPTER 3 PROBLEMS PLUS<br />

TX.10<br />

tangent line is y − y 0 = − 4 x 0<br />

(x − x 0) or 4x 0x +9y 0y =4x 2 0 +9y<br />

9 y<br />

0. 2 This can be written as x 0x<br />

0 9 + y 0y<br />

4 = x2 0<br />

9 + y2 0<br />

4 =1,<br />

because (x 0 ,y 0 ) lies on the ellipse. So an equation of the tangent line is x 0x<br />

9 + y 0y<br />

4 =1.<br />

Therefore, the x-intercept x T for the tangent line is given by x 0x T<br />

9<br />

=1 ⇔ x T = 9 x 0<br />

,andthey-intercept y T is given<br />

by y0yT<br />

4<br />

=1 ⇔ y T = 4 y 0<br />

.<br />

So as x 0 takes on all values in (0, 3), x T takes on all values in (3, ∞),andasy 0 takes on all values in (0, 2), y T takes on<br />

all values in (2, ∞).<br />

1<br />

At the point (x 0,y 0) on the ellipse, the slope of the normal line is −<br />

y 0 (x = 9 ,andits<br />

0,y 0) 4 x 0<br />

y 0<br />

y 0<br />

equation is y − y 0 = 9 (x − x 0). Sothex-intercept x N for the normal line is given by 0 − y 0 = 9 (x N − x 0)<br />

4 x 0 4 x 0<br />

y 0<br />

⇒<br />

x N = − 4x0<br />

9 + x 0 = 5x0<br />

9 ,andthey-intercept y N is given by y N − y 0 = 9 4<br />

y 0<br />

(0 − x 0 ) ⇒ y N = − 9y0<br />

x 0 4 + y 0 = − 5y0<br />

4 .<br />

So as x 0 takes on all values in (0, 3), x N takes on all values in 0, 5 3<br />

<br />

,andasy0 takes on all values in (0, 2), y N takes on<br />

all values in − 5 2 , 0 .<br />

17. (a) If the two lines L 1 and L 2 have slopes m 1 and m 2 and angles of<br />

inclination φ 1 and φ 2 ,thenm 1 =tanφ 1 and m 2 =tanφ 2 . The triangle<br />

in the figure shows that φ 1 + α +(180 ◦ − φ 2 ) = 180 ◦ and so<br />

α = φ 2 − φ 1 . Therefore, using the identity for tan(x − y),wehave<br />

tan α =tan(φ 2 − φ 1 )= tan φ 2 − tan φ 1<br />

m2 − m1<br />

and so tan α = .<br />

1+tanφ 2 tan φ 1 1+m 1m 2<br />

(b) (i) The parabolas intersect when x 2 =(x − 2) 2 ⇒ x =1.Ify = x 2 ,theny 0 =2x, so the slope of the tangent<br />

to y = x 2 at (1, 1) is m 1 =2(1)=2.Ify =(x − 2) 2 ,theny 0 =2(x − 2), so the slope of the tangent to<br />

y =(x − 2) 2 at (1, 1) is m 2 =2(1− 2) = −2. Therefore, tan α =<br />

so α =tan −1 4<br />

3<br />

≈ 53 ◦ [or 127 ◦ ].<br />

m2 − m1<br />

= −2 − 2<br />

1+m 1m 2 1+2(−2) = 4 3 and<br />

(ii) x 2 − y 2 =3and x 2 − 4x + y 2 +3=0intersect when x 2 − 4x +(x 2 − 3) + 3 = 0 ⇔ 2x(x − 2) = 0 ⇒<br />

x =0or 2,but0 is extraneous. If x =2,theny = ±1.Ifx 2 − y 2 =3then 2x − 2yy 0 =0 ⇒ y 0 = x/y and<br />

x 2 − 4x + y 2 +3=0 ⇒ 2x − 4+2yy 0 =0 ⇒ y 0 = 2 − x .At(2, 1) the slopes are m 1 =2and<br />

y<br />

m 2 =0,so tan α = 0 − 2<br />

1+2· 0 = −2 ⇒ α ≈ 117◦ .At(2, −1) the slopes are m 1 = −2 and m 2 =0,<br />

so tan α =<br />

0 − (−2)<br />

1+(−2) (0) =2 ⇒ α ≈ 63◦ [or 117 ◦ ].

F.<br />

TX.10<br />

CHAPTER 3 PROBLEMS PLUS ¤ 143<br />

19. Since ∠ROQ = ∠OQP = θ, the triangle QOR is isosceles, so<br />

|QR| = |RO| = x. By the Law of Cosines, x 2 = x 2 + r 2 − 2rx cos θ. Hence,<br />

2rx cos θ = r 2 ,sox =<br />

sin θ = y/r), and hence x →<br />

r2<br />

2r cos θ = r<br />

2cosθ .Notethatasy → 0+ , θ → 0 + (since<br />

r<br />

2cos0 = r .Thus,asP is taken closer and closer<br />

2<br />

to the x-axis, the point R approaches the midpoint of the radius AO.<br />

21. lim<br />

x→0<br />

sin(a +2x) − 2sin(a + x)+sina<br />

x 2<br />

= lim<br />

x→0<br />

sin a cos 2x +cosa sin 2x − 2sina cos x − 2cosa sin x +sina<br />

x 2<br />

= lim<br />

x→0<br />

sin a (cos 2x − 2cosx +1)+cosa (sin 2x − 2sinx)<br />

x 2<br />

= lim<br />

x→0<br />

sin a (2 cos 2 x − 1 − 2cosx +1)+cosa (2 sin x cos x − 2sinx)<br />

x 2<br />

= lim<br />

x→0<br />

sin a (2 cos x)(cos x − 1) + cos a (2 sin x)(cos x − 1)<br />

x 2<br />

2(cos x − 1)[sin a cos x +cosa sin x](cos x +1)<br />

= lim<br />

x→0 x 2 (cos x +1)<br />

<br />

−2sin 2 2<br />

x [sin(a + x)]<br />

sin x sin(a + x)<br />

= lim<br />

= −2 lim ·<br />

x→0 x 2 (cos x +1)<br />

x→0 x cos x +1 = sin(a +0)<br />

−2(1)2 cos 0 + 1 = − sin a<br />

23. Let f(x) =e 2x and g(x) =k √ x [k >0]. From the graphs of f and g,<br />

So we must have k √ a =<br />

k<br />

4 √ a<br />

k =2e 1/2 =2 √ e ≈ 3.297.<br />

25. y =<br />

⇒<br />

we see that f will intersect g exactly once when f and g share a tangent<br />

line. Thus, we must have f = g and f 0 = g 0 at x = a.<br />

f(a) =g(a) ⇒ e 2a = k √ a ()<br />

and f 0 (a) =g 0 (a) ⇒ 2e 2a = k<br />

2 √ ⇒ e 2a = k<br />

a<br />

4 √ a .<br />

√ 2 k a = ⇒ a = 1<br />

4k<br />

.From(), 4 e2(1/4) = k 1/4 ⇒<br />

x<br />

√<br />

a2 − 1 − 2<br />

√<br />

a2 − 1 arctan sin x<br />

a + √ a 2 − 1+cosx .Letk = a + √ a 2 − 1. Then<br />

y 0 =<br />

=<br />

1<br />

√<br />

a2 − 1 − 2<br />

√<br />

a2 − 1 ·<br />

1<br />

1+sin 2 x/(k +cosx) · cos x(k +cosx)+sin2 x<br />

2 (k +cosx) 2<br />

1<br />

√<br />

a2 − 1 − 2<br />

√<br />

a2 − 1 · k cos x +cos2 x +sin 2 x<br />

(k +cosx) 2 +sin 2 x = 1<br />

√<br />

a2 − 1 − 2<br />

√<br />

a2 − 1 · k cos x +1<br />

k 2 +2k cos x +1<br />

= k2 +2k cos x +1− 2k cos x − 2<br />

√<br />

a2 − 1(k 2 +2k cos x +1)<br />

=<br />

k 2 − 1<br />

√<br />

a2 − 1(k 2 +2k cos x +1)<br />

But k 2 =2a 2 +2a √ a 2 − 1 − 1=2a a + √ a 2 − 1 − 1=2ak − 1,sok 2 +1=2ak,andk 2 − 1=2(ak − 1).<br />

[continued]

F.<br />

144 ¤ CHAPTER 3 PROBLEMS PLUS<br />

TX.10<br />

So y 0 =<br />

2(ak − 1)<br />

√<br />

a2 − 1(2ak +2k cos x) = ak − 1<br />

√<br />

a2 − 1k (a +cosx) .Butak − 1=a2 + a √ a 2 − 1 − 1=k √ a 2 − 1,<br />

so y 0 =1/(a +cosx).<br />

27. y = x 4 − 2x 2 − x ⇒ y 0 =4x 3 − 4x − 1. The equation of the tangent line at x = a is<br />

y − (a 4 − 2a 2 − a) =(4a 3 − 4a − 1)(x − a) or y =(4a 3 − 4a − 1)x +(−3a 4 +2a 2 ) and similarly for x = b. Soifat<br />

x = a and x = b we have the same tangent line, then 4a 3 − 4a − 1=4b 3 − 4b − 1 and −3a 4 +2a 2 = −3b 4 +2b 2 .Thefirst<br />

equation gives a 3 − b 3 = a − b ⇒ (a − b)(a 2 + ab + b 2 )=(a − b). Assuming a 6=b, wehave1=a 2 + ab + b 2 .<br />

The second equation gives 3(a 4 − b 4 )=2(a 2 − b 2 ) ⇒ 3(a 2 − b 2 )(a 2 + b 2 )=2(a 2 − b 2 ) which is true if a = −b.<br />

Substituting into 1=a 2 + ab + b 2 gives 1=a 2 − a 2 + a 2 ⇒ a = ±1 so that a =1and b = −1 or vice versa. Thus,<br />

the points (1, −2) and (−1, 0) have a common tangent line.<br />

As long as there are only two such points, we are done. So we show that these are in fact the only two such points.<br />

Suppose that a 2 − b 2 6=0.Then3(a 2 − b 2 )(a 2 + b 2 )=2(a 2 − b 2 ) gives 3(a 2 + b 2 )=2 or a 2 + b 2 = 2 3 .<br />

Thus, ab =(a 2 + ab + b 2 ) − (a 2 + b 2 )=1− 2 3 = 1 3 ,sob = 1<br />

3a . Hence, a2 + 1<br />

9a 2 = 2 3 ,so9a4 +1=6a 2<br />

0=9a 4 − 6a 2 +1=(3a 2 − 1) 2 .So3a 2 − 1=0 ⇒ a 2 = 1 3<br />

that a 2 6=b 2 .<br />

⇒ b 2 = 1<br />

9a 2 = 1 3 = a2 , contradicting our assumption<br />

29. Because of the periodic nature of the lattice points, it suffices to consider the points in the 5 × 2 grid shown. We can see that<br />

the minimum value of r occurs when there is a line with slope 2 which touches the circle centered at (3, 1) and the circles<br />

5<br />

centered at (0, 0) and (5, 2).<br />

⇒<br />

To find P , the point at which the line is tangent to the circle at (0, 0), we simultaneously solve x 2 + y 2 = r 2 and<br />

y = − 5 2 x ⇒ x2 + 25 4 x2 = r 2 ⇒ x 2 = 4 29 r2 ⇒ x = 2 √<br />

29<br />

r, y = − 5 √<br />

29<br />

r.Tofind Q, we either use symmetry or<br />

solve (x − 3) 2 +(y − 1) 2 = r 2 and y − 1=− 5 2<br />

2<br />

(x − 3).Asabove,wegetx =3− √<br />

29<br />

r, y =1+ √ 5<br />

29<br />

r. Now the slope of<br />

the line PQis 2 5 ,som PQ =<br />

<br />

1+ √ 5<br />

29<br />

r − − √ 5<br />

29<br />

r<br />

3 − 2 √<br />

29<br />

r − 2 √<br />

29<br />

r<br />

=<br />

10 √<br />

1+ √<br />

29<br />

r 29 + 10r<br />

3 − √ 4<br />

29<br />

r = 3 √ 29 − 4r = 2 5<br />

5 √ 29 + 50r =6 √ 29 − 8r ⇔ 58r = √ 29 ⇔ r = √ 29<br />

58 . So the minimum value of r for which any line with slope 2 5<br />

intersects circles with radius r centered at the lattice points on the plane is r = √ 29<br />

58 ≈ 0.093.<br />

F.<br />

TX.10<br />

31. By similar triangles, r 5 = h 16<br />

⇒<br />

CHAPTER 3 PROBLEMS PLUS ¤ 145<br />

r = 5h . The volume of the cone is<br />

16<br />

2 5h<br />

V = 1 3 πr2 h = 1 π h = 25π<br />

3<br />

16 768 h3 ,so dV<br />

dt = 25π dh<br />

256 h2 dt .Nowtherateof<br />

change of the volume is also equal to the difference of what is being added<br />

(2 cm 3 /min) and what is oozing out (kπrl,whereπrl is the area of the cone and k<br />

Equating the two expressions for dV<br />

dt<br />

is a proportionality constant). Thus, dV<br />

dt<br />

and substituting h =10,<br />

dh<br />

dt<br />

=2− kπrl.<br />

= −0.3, r =<br />

5(10)<br />

16<br />

= 25 8 ,and l<br />

√<br />

281<br />

= 10<br />

16<br />

⇔<br />

l = 5 8<br />

√ 25π<br />

281,weget<br />

256 (10)2 (−0.3) = 2 − kπ 25<br />

8 · 5 √ 125kπ √ 281<br />

281 ⇔<br />

8<br />

64<br />

=2+ 750π . Solving for k gives us<br />

256<br />

256 + 375π<br />

k =<br />

250π √ . To maintain a certain height, the rate of oozing, kπrl, must equal the rate of the liquid being poured in;<br />

281<br />

that is, dV =0. Thus, the rate at which we should pour the liquid into the container is<br />

dt<br />

kπrl =<br />

256 + 375π<br />

250π √ 281 · π · 25 8 · 5 √ 281 256 + 375π<br />

= ≈ 11.204 cm 3 /min<br />

8 128

F.<br />

TX.10

F.<br />

TX.10<br />

4 APPLICATIONS OF DIFFERENTIATION<br />

4.1 Maximum and Minimum Values<br />

1. Afunctionf has an absolute minimum at x = c if f(c) is the smallest function value on the entire domain of f, whereas<br />

f has a local minimum at c if f(c) is the smallest function value when x is near c.<br />

3. Absolute maximum at s, absolute minimum at r,localmaximumatc, local minima at b and r, neither a maximum nor a<br />

minimum at a and d.<br />

5. Absolute maximum value is f(4) = 5; there is no absolute minimum value; local maximum values are f(4) = 5 and<br />

f(6) = 4; local minimum values are f(2) = 2 and f(1) = f(5) = 3.<br />

7. Absolute minimum at 2, absolute maximum at 3,<br />

local minimum at 4<br />

9. Absolute maximum at 5, absolute minimum at 2,<br />

local maximum at 3, local minima at 2 and 4<br />

11. (a) (b) (c)<br />

13. (a) Note: By the Extreme Value Theorem,<br />

f must not be continuous; because if it<br />

were, it would attain an absolute<br />

minimum.<br />

(b)<br />

147

F.<br />

148 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

15. f(x) =8− 3x, x ≥ 1. Absolute maximum f(1) = 5;no<br />

local maximum. No absolute or local minimum.<br />

TX.10<br />

17. f(x) =x 2 , 0

F.<br />

TX.10<br />

SECTION 4.1 MAXIMUM AND MINIMUM VALUES ¤ 149<br />

29. f(x) =5x 2 +4x ⇒ f 0 (x) =10x +4. f 0 (x) =0 ⇒ x = − 2 ,so− 2 5 5<br />

is the only critical number.<br />

31. f(x) =x 3 +3x 2 − 24x ⇒ f 0 (x) =3x 2 +6x − 24 = 3(x 2 +2x − 8).<br />

f 0 (x) =0 ⇒ 3(x +4)(x − 2) = 0 ⇒ x = −4, 2. These are the only critical numbers.<br />

33. s(t) =3t 4 +4t 3 − 6t 2 ⇒ s 0 (t) =12t 3 +12t 2 − 12t. s 0 (t) =0 ⇒ 12t(t 2 + t − 1) ⇒<br />

t =0 or t 2 + t − 1=0. Using the quadratic formula to solve the latter equation gives us<br />

t = −1 ± 1 2 − 4(1)(−1)<br />

2(1)<br />

= −1 ± √ 5<br />

2<br />

≈ 0.618, −1.618. The three critical numbers are 0, −1 ± √ 5<br />

.<br />

2<br />

35. g(y) =<br />

y − 1<br />

y 2 − y +1<br />

⇒<br />

g 0 (y) = (y2 − y +1)(1)− (y − 1)(2y − 1)<br />

= y2 − y +1− (2y 2 − 3y +1)<br />

= −y2 +2y y(2 − y)<br />

=<br />

(y 2 − y +1) 2 (y 2 − y +1) 2 (y 2 − y +1)<br />

2<br />

(y 2 − y +1) . 2<br />

g 0 (y) =0 ⇒ y =0, 2. The expression y 2 − y +1is never equal to 0,sog 0 (y) exists for all real numbers.<br />

The critical numbers are 0 and 2.<br />

37. h(t) =t 3/4 − 2t 1/4 ⇒ h 0 (t) = 3 4 t−1/4 − 2 4 t−3/4 = 1 4 t−3/4 (3t 1/2 − 2) = 3 √ t − 2<br />

4 4√ t 3 .<br />

h 0 (t) =0 ⇒ 3 √ t =2 ⇒ √ t = 2 3<br />

⇒ t = 4 9 . h0 (t) does not exist at t =0, so the critical numbers are 0 and 4 9 .<br />

39. F (x) =x 4/5 (x − 4) 2 ⇒<br />

F 0 (x) =x 4/5 · 2(x − 4) + (x − 4) 2 · 4<br />

5 x−1/5 = 1 5 x−1/5 (x − 4)[5 · x · 2+(x − 4) · 4]<br />

=<br />

(x − 4)(14x − 16) 2(x − 4)(7x − 8)<br />

=<br />

5x 1/5 5x 1/5<br />

F 0 (x) =0 ⇒ x =4, 8 7 . F 0 (0) does not exist. Thus, the three critical numbers are 0, 8 7 ,and4.<br />

41. f(θ) =2cosθ +sin 2 θ ⇒ f 0 (θ) =−2sinθ +2sinθ cos θ. f 0 (θ) =0 ⇒ 2sinθ (cos θ − 1) = 0 ⇒ sin θ =0<br />

or cos θ =1 ⇒ θ = nπ [n an integer] or θ =2nπ. The solutions θ = nπ include the solutions θ =2nπ, so the critical<br />

numbers are θ = nπ.<br />

43. f(x) =x 2 e −3x ⇒ f 0 (x) =x 2 (−3e −3x )+e −3x (2x) =xe −3x (−3x +2). f 0 (x) =0 ⇒ x =0, 2 3<br />

[e −3x is never equal to 0]. f 0 (x) always exists, so the critical numbers are 0 and 2 3 .<br />

45. The graph of f 0 (x) =5e −0.1|x| sin x − 1 has 10 zeros and exists<br />

everywhere, so f has 10 critical numbers.

F.<br />

150 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

47. f(x) =3x 2 − 12x +5, [0, 3]. f 0 (x) =6x − 12 = 0 ⇔ x =2. Applying the Closed Interval Method, we find that<br />

f(0) = 5, f(2) = −7,andf(3) = −4. Sof(0) = 5 is the absolute maximum value and f(2) = −7 is the absolute minimum<br />

value.<br />

49. f(x) =2x 3 − 3x 2 − 12x +1, [−2, 3]. f 0 (x) =6x 2 − 6x − 12 = 6(x 2 − x − 2) = 6(x − 2)(x +1)=0 ⇔<br />

x =2, −1.<br />

f(−2) = −3, f(−1) = 8, f(2) = −19,andf(3) = −8. Sof(−1) = 8 is the absolute maximum value and<br />

f(2) = −19 is the absolute minimum value.<br />

51. f(x) =x 4 − 2x 2 +3, [−2, 3]. f 0 (x) =4x 3 − 4x =4x(x 2 − 1) = 4x(x +1)(x − 1) = 0 ⇔ x = −1, 0, 1.<br />

f(−2) = 11, f(−1) = 2, f(0) = 3, f(1) = 2, f(3) = 66. Sof(3) = 66 is the absolute maximum value and f(±1) = 2 is<br />

the absolute minimum value.<br />

53. f(x) = x , [0, 2].<br />

x 2 +1 f<br />

0 (x) = (x2 +1)− x(2x)<br />

= 1 − x2<br />

=0 ⇔ x = ±1,but−1 is not in [0, 2]. f(0) = 0,<br />

(x 2 +1) 2 (x 2 +1)<br />

2<br />

f(1) = 1 , f(2) = 2 .Sof(1) = 1 2 5 2<br />

is the absolute maximum value and f(0) = 0 is the absolute minimum value.<br />

55. f(t) =t √ 4 − t 2 , [−1, 2].<br />

f 0 (t) =t · 1 (4 − 2 t2 ) −1/2 (−2t)+(4− t 2 ) 1/2 · 1= √ −t2 + √ 4 − t 2 = −t2 +(4− t 2 )<br />

√ = √ 4 − 2t2 .<br />

4 − t<br />

2 4 − t<br />

2 4 − t<br />

2<br />

f 0 (t) =0 ⇒ 4 − 2t 2 =0 ⇒ t 2 =2 ⇒ t = ± √ 2,butt = − √ 2 is not in the given interval, [−1, 2].<br />

f 0 (t) does not exist if 4 − t 2 =0 ⇒ t = ±2, but−2 is not in the given interval. f(−1) = − √ 3, f √ 2 =2,and<br />

f(2) = 0. Sof √ 2 =2is the absolute maximum value and f(−1) = − √ 3 is the absolute minimum value.<br />

57. f(t) =2cost +sin2t, [0, π/2].<br />

f 0 (t) =−2sint +cos2t · 2=−2sint +2(1− 2sin 2 t)=−2(2 sin 2 t +sint − 1) = −2(2 sin t − 1)(sin t +1).<br />

f 0 (t) =0 ⇒ sin t = 1 or sin t = −1 ⇒ t = π . f(0) = 2, f( π )=√ √ √<br />

2 6 6<br />

3+ 1 2 3=<br />

3<br />

2 3 ≈ 2.60,andf(<br />

π<br />

)=0. 2<br />

So f( π )= √ 3<br />

6 2 3 is the absolute maximum value and f(<br />

π<br />

2<br />

)=0is the absolute minimum value.<br />

59. f(x) =xe −x2 /8 , [−1, 4]. f 0 (x) =x · e −x2 /8 · (− x 4 )+e−x2 /8 · 1=e −x2 /8 (− x2<br />

4 +1).Sincee−x2 /8 is never 0,<br />

f 0 (x) =0 ⇒ −x 2 /4+1=0 ⇒ 1=x 2 /4 ⇒ x 2 =4 ⇒ x = ±2,but−2 is not in the given interval, [−1, 4].<br />

f(−1) = −e −1/8 ≈−0.88, f(2) = 2e −1/2 ≈ 1.21,andf(4) = 4e −2 ≈ 0.54. Sof(2) = 2e −1/2 is the absolute maximum<br />

value and f(−1) = −e −1/8 is the absolute minimum value.<br />

61. f(x) =ln(x 2 + x +1), [−1, 1]. f 0 1<br />

(x) =<br />

x 2 + x +1 · (2x +1)=0 ⇔ x = − 1 2 .Sincex2 + x +1> 0 for all x,the<br />

domain of f and f 0 is R. f(−1)=ln1=0, f <br />

− 1 2 =ln<br />

3<br />

≈ −0.29,andf(1) = ln 3 ≈ 1.10. Sof(1) = ln 3 ≈ 1.10 is<br />

4<br />

the absolute maximum value and f <br />

− 1 2 =ln<br />

3<br />

4<br />

≈ −0.29 is the absolute minimum value.

F.<br />

TX.10<br />

SECTION 4.1 MAXIMUM AND MINIMUM VALUES ¤ 151<br />

63. f(x) =x a (1 − x) b , 0 ≤ x ≤ 1, a>0, b>0.<br />

f 0 (x) =x a · b(1 − x) b−1 (−1) + (1 − x) b · ax a−1 = x a−1 (1 − x) b−1 [x · b(−1) + (1 − x) · a]<br />

= x a−1 (1 − x) b−1 (a − ax − bx)<br />

At the endpoints, we have f(0) = f(1) = 0 [the minimum value of f ]. In the interval (0, 1), f 0 (x) =0 ⇔ x = a<br />

a + b .<br />

<br />

f<br />

a<br />

a + b<br />

<br />

=<br />

a<br />

So f<br />

a + b<br />

a<br />

a + b<br />

a <br />

1 − a b<br />

=<br />

a + b<br />

<br />

a a<br />

b a + b − a<br />

=<br />

(a + b) a a + b<br />

a a b b<br />

=<br />

is the absolute maximum value.<br />

a+b<br />

(a + b)<br />

a a<br />

(a + b) a ·<br />

b b<br />

(a + b) = a a b b<br />

b (a + b) . a+b<br />

65. (a) From the graph, it appears that the absolute maximum value is about<br />

f(−0.77) = 2.19, and the absolute minimum value is about<br />

f(0.77) = 1.81.<br />

<br />

(b) f(x) =x 5 − x 3 +2 ⇒ f 0 (x) =5x 4 − 3x 2 = x 2 (5x 2 − 3). Sof 0 3<br />

(x) =0 ⇒ x =0, ± . 5<br />

5 3 <br />

3<br />

3<br />

f − = −<br />

5<br />

5 −<br />

3<br />

−<br />

5 +2=− 3<br />

<br />

2 3<br />

+ 3 3<br />

+2= 3<br />

− <br />

9 3<br />

+2= 6 3<br />

+2(maximum)<br />

5 5 5 5 5 25 5 25 5<br />

<br />

and similarly, f<br />

3<br />

5<br />

<br />

= − 6 25<br />

<br />

3<br />

5 +2(minimum).<br />

67. (a) From the graph, it appears that the absolute maximum value is about<br />

f(0.75) = 0.32, and the absolute minimum value is f(0) = f(1) = 0;<br />

that is, at both endpoints.<br />

(b) f(x) =x √ x − x 2 ⇒ f 0 1 − 2x<br />

(x) =x ·<br />

2 √ x − x + √ x − x 2 = (x − 2x2 )+(2x − 2x 2 )<br />

2 2 √ 3x − 4x2<br />

=<br />

x − x 2 2 √ x − x . 2<br />

So f 0 (x) =0 ⇒ 3x − 4x 2 =0 ⇒ x(3 − 4x) =0 ⇒ x =0or 3 . 4<br />

f(0) = f(1) = 0 (minimum), and f <br />

3<br />

4 =<br />

3<br />

4<br />

<br />

3<br />

4 − 3<br />

4<br />

2<br />

= 3 4<br />

<br />

3<br />

= 3 √ 3<br />

(maximum).<br />

16 16<br />

69. The density is defined as ρ = mass<br />

volume = 1000<br />

V (T ) (in g/cm3 ). But a critical point of ρ will also be a critical point of V<br />

[since dρ<br />

dT = −1000V −2 dV<br />

dT<br />

and V is never 0],andV is easier to differentiate than ρ.<br />

V (T )=999.87 − 0.06426T +0.0085043T 2 − 0.0000679T 3 ⇒ V 0 (T )=−0.06426 + 0.0170086T − 0.0002037T 2 .<br />

Setting this equal to 0 and using the quadratic formula to find T ,weget<br />

T = −0.0170086 ± √ 0.0170086 2 − 4 · 0.0002037 · 0.06426<br />

2(−0.0002037)<br />

≈ 3.9665 ◦ Cor79.5318 ◦ C. Since we are only interested

F.<br />

152 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

in the region 0 ◦ C ≤ T ≤ 30 ◦ C, we check the density ρ at the endpoints and at 3.9665 ◦ C: ρ(0) ≈ 1000<br />

999.87 ≈ 1.00013;<br />

ρ(30) ≈ 1000<br />

1000<br />

≈ 0.99625; ρ(3.9665) ≈ ≈ 1.000255. So water has its maximum density at<br />

1003.7628 999.7447<br />

71. Let a = −0.000 032 37, b =0.000 903 7, c = −0.008 956, d =0.03629, e = −0.04458, andf =0.4074.<br />

Then S(t) =at 5 + bt 4 + ct 3 + dt 2 + et + f and S 0 (t) =5at 4 +4bt 3 +3ct 2 +2dt + e.<br />

We now apply the Closed Interval Method to the continuous function S on the interval 0 ≤ t ≤ 10. SinceS 0 exists for all t,<br />

the only critical numbers of S occur when S 0 (t) =0.Weusearootfinder on a CAS (or a graphing device) to find that<br />

S 0 (t) =0when t 1 ≈ 0.855, t 2 ≈ 4.618, t 3 ≈ 7.292, andt 4 ≈ 9.570. The values of S at these critical numbers are<br />

S(t 1) ≈ 0.39, S(t 2) ≈ 0.43645, S(t 3) ≈ 0.427,andS(t 4) ≈ 0.43641. The values of S at the endpoints of the interval are<br />

S(0) ≈ 0.41 and S(10) ≈ 0.435. Comparing the six numbers, we see that sugar was most expensive at t 2 ≈ 4.618<br />

(corresponding roughly to March 1998) and cheapest at t 1 ≈ 0.855 (June 1994).<br />

73. (a) v(r) =k(r 0 − r)r 2 = kr 0 r 2 − kr 3 ⇒ v 0 (r) =2kr 0 r − 3kr 2 . v 0 (r) =0 ⇒ kr(2r 0 − 3r) =0 ⇒<br />

r =0or 2 r 3 0 (but 0 is not in the interval). Evaluating v at 1 r 2 0, 2 r 3 0,andr 0 ,wegetv 1<br />

r <br />

2 0 =<br />

1<br />

8 kr3 0, v 2<br />

r <br />

3 0 =<br />

4<br />

27 kr3 0,<br />

and v(r 0 )=0.Since 4 27 > 1 8 , v attains its maximum value at r = 2 3 r 0. This supports the statement in the text.<br />

(b) From part (a), the maximum value of v is 4 27 kr3 0.<br />

(c)<br />

75. f(x) =x 101 + x 51 + x +1 ⇒ f 0 (x) =101x 100 +51x 50 +1≥ 1 for all x,sof 0 (x) =0has no solution. Thus, f(x)<br />

has no critical number, so f(x) can have no local maximum or minimum.<br />

77. If f has a local minimum at c,theng(x) =−f(x) has a local maximum at c,sog 0 (c) =0by the case of Fermat’s Theorem<br />

proved in the text. Thus, f 0 (c) =−g 0 (c) =0.<br />

4.2 The Mean Value Theorem<br />

1. f(x) =5− 12x +3x 2 , [1, 3]. Sincef is a polynomial, it is continuous and differentiable on R, so it is continuous on [1, 3]<br />

and differentiable on (1, 3). Alsof(1) = −4 =f(3). f 0 (c) =0 ⇔ −12 + 6c =0 ⇔ c =2, which is in the open<br />

interval (1, 3),soc =2satisfies the conclusion of Rolle’s Theorem.

F.<br />

TX.10<br />

SECTION 4.2 THE MEAN VALUE THEOREM ¤ 153<br />

3. f(x) = √ x − 1 3<br />

x, [0, 9]. f, being the difference of a root function and a polynomial, is continuous and differentiable<br />

on [0, ∞), so it is continuous on [0, 9] and differentiable on (0, 9). Also,f(0) = 0 = f(9). f 0 (c) =0<br />

⇔<br />

1<br />

2 √ c − 1 3 =0 ⇔ 2 √ c =3 ⇔ √ c = 3 2<br />

conclusion of Rolle’s Theorem.<br />

⇒<br />

c = 9 4 , which is in the open interval (0, 9),soc = 9 satisfies the<br />

4<br />

5. f(x) =1− x 2/3 . f(−1) = 1 − (−1) 2/3 =1− 1=0=f(1). f 0 (x) =− 2 3 x−1/3 ,sof 0 (c) =0has no solution. This<br />

does not contradict Rolle’s Theorem, since f 0 (0) does not exist, and so f is not differentiable on (−1, 1).<br />

7.<br />

f(8) − f(0)<br />

8 − 0<br />

= 6 − 4<br />

8<br />

= 1 4 . The values of c which satisfy f 0 (c) = 1 seem to be about c =0.8, 3.2, 4.4,and6.1.<br />

4<br />

9. (a), (b) The equation of the secant line is<br />

(c) f(x) =x +4/x ⇒ f 0 (x) =1− 4/x 2 .<br />

y − 5= 8.5 − 5<br />

8 − 1 (x − 1) ⇔ y = 1 x + 9 . So f 0 (c) = 1 2<br />

⇒ c 2 =8 ⇒ c =2 √ 2,and<br />

2 2<br />

f(c) =2 √ 2+ 4<br />

2 √ =3√ 2. Thus, an equation of the<br />

2<br />

tangent line is y − 3 √ √ <br />

2= 1 2 x − 2 2 ⇔<br />

y = 1 2 x +2√ 2.<br />

11. f(x) =3x 2 +2x +5, [−1, 1]. f is continuous on [−1, 1] and differentiable on (−1, 1) since polynomials are continuous<br />

and differentiable on R.<br />

c =0, which is in (−1, 1).<br />

f 0 (c) =<br />

f(b) − f(a)<br />

b − a<br />

⇔<br />

6c +2=<br />

f(1) − f(−1)<br />

1 − (−1)<br />

= 10 − 6<br />

2<br />

=2 ⇔ 6c =0 ⇔<br />

13. f(x) =e −2x , [0, 3]. f is continuous and differentiable on R, so it is continuous on [0, 3] and differentiable on (0, 3).<br />

f 0 f(b) − f(a)<br />

(c) = ⇔ −2e −2c = e−6 − e 0<br />

⇔ e −2c = 1 − e−6<br />

1 − e<br />

−6<br />

<br />

⇔ −2c =ln<br />

⇔<br />

b − a<br />

3 − 0<br />

6<br />

6<br />

c = − 1 1 − e<br />

−6<br />

2 ln ≈ 0.897, which is in (0, 3).<br />

6

F.<br />

154 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

15. f(x) =(x − 3) −2 ⇒ f 0 (x) =−2(x − 3) −3 . f(4) − f(1) = f 0 (c)(4 − 1) ⇒ 1 1 − 1<br />

2 (−2) = −2<br />

2 (c − 3) · 3 ⇒<br />

3<br />

3<br />

4 = −6 ⇒ (c − 3) 3 = −8 ⇒ c − 3=−2 ⇒ c =1, which is not in the open interval (1, 4). This does not<br />

(c − 3) 3<br />

contradict the Mean Value Theorem since f is not continuous at x =3.<br />

17. Let f(x) =1+2x + x 3 +4x 5 .Thenf(−1) = −6 < 0 and f(0) = 1 > 0. Since f is a polynomial, it is continuous, so the<br />

Intermediate Value Theorem says that there is a number c between −1 and 0 such that f(c) =0. Thus, the given equation has<br />

a real root. Suppose the equation has distinct real roots a and b with a

F.<br />

TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 155<br />

27. We use Exercise 26 with f(x) = √ 1+x, g(x) =1+ 1 2<br />

x,anda =0.Noticethatf(0) = 1 = g(0) and<br />

f 0 (x) =<br />

1<br />

2 √ 1+x < 1 2 = g0 (x) for x>0.SobyExercise26,f(b)

F.<br />

156 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

5. (a) Since f 0 (x) > 0 on (1, 5), f is increasing on this interval. Since f 0 (x) < 0 on (0, 1) and (5, 6), f is decreasing on these<br />

intervals.<br />

(b) Since f 0 (x) =0at x =1and f 0 changes from negative to positive there, f changes from decreasing to increasing and has<br />

a local minimum at x =1.Sincef 0 (x) =0at x =5and f 0 changes from positive to negative there, f changes from<br />

increasing to decreasing and has a local maximum at x =5.<br />

7. There is an inflection point at x =1because f 00 (x) changes from negative to positive there, and so the graph of f changes<br />

from concave downward to concave upward. There is an inflection point at x =7because f 00 (x) changes from positive to<br />

negative there, and so the graph of f changes from concave upward to concave downward.<br />

9. (a) f(x) =2x 3 +3x 2 − 36x ⇒ f 0 (x) =6x 2 +6x − 36 = 6(x 2 + x − 6) = 6(x +3)(x − 2).<br />

We don’t need to include the “6”inthecharttodeterminethesignoff 0 (x).<br />

Interval x +3 x − 2 f 0 (x) f<br />

x− 1 2 ,and<br />

f 00 (x) < 0 ⇔ x

F.<br />

TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 157<br />

13. (a) f(x) =sinx +cosx, 0 ≤ x ≤ 2π. f 0 (x) =cosx − sin x =0 ⇒ cos x =sinx ⇒ 1= sin x<br />

cos x<br />

tan x =1 ⇒ x = π 4 or 5π 4 . Thus, f 0 (x) > 0 ⇔ cos x − sin x>0 ⇔ cos x>sin x ⇔ 0 ln 1 2<br />

⇔<br />

x> 1 (ln 1 − ln 2) ⇔ x>− 1 ln 2 [≈ −0.23] andf 0 3 3<br />

(x) < 0 if x 0 [the sum of two positive terms].<br />

point of inflection.<br />

Thus, f is concave upward on (−∞, ∞) and there is no<br />

17. (a) y = f(x) = ln x √<br />

x<br />

.(Notethatf is only defined for x>0.)<br />

f 0 (x) =<br />

√<br />

x (1/x) − ln x<br />

<br />

1<br />

2 x−1/2 <br />

x<br />

=<br />

1<br />

√ − ln x<br />

x 2 √ x<br />

· 2 √ x<br />

x 2 √ x = 2 − ln x > 0 ⇔ 2 − ln x>0 ⇔<br />

2x 3/2<br />

ln xe 8/3 ,sof is concave upward on (e 8/3 , ∞) and<br />

<br />

concave downward on (0,e 8/3 ).Thereisaninflection point at<br />

e 8/3 , 8 3 e−4/3 ≈ (14.39, 0.70).

F.<br />

158 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

19. f(x) =x 5 − 5x +3 ⇒ f 0 (x) =5x 4 − 5=5(x 2 +1)(x +1)(x − 1).<br />

First Derivative Test: f 0 (x) < 0 ⇒ −1 1 or x 0 ⇒ f(1) = −1 is a local minimum value.<br />

Preference: For this function, the two tests are equally easy.<br />

21. f(x) =x + √ 1 − x ⇒ f 0 (x) =1+ 1 2 (1 − x)−1/2 (−1) = 1 −<br />

1<br />

2 √ .Notethatf is defined for 1 − x ≥ 0;thatis,<br />

1 − x<br />

for x ≤ 1. f 0 (x) =0 ⇒ 2 √ 1 − x =1 ⇒ √ 1 − x = 1 2<br />

⇒ 1 − x = 1 4<br />

⇒ x = 3 4 . f 0 does not exist at x =1,<br />

but we can’t have a local maximum or minimum at an endpoint.<br />

First Derivative Test: f 0 (x) > 0 ⇒ x< 3 and f 0 (x) < 0 ⇒ 3

F.<br />

TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 159<br />

29. The function must be always decreasing (since the first derivative is always negative)<br />

and concave downward (since the second derivative is always negative).<br />

31. (a) f is increasing where f 0 is positive, that is, on (0, 2), (4, 6),and(8, ∞); and decreasing where f 0 is negative, that is,<br />

on (2, 4) and (6, 8).<br />

(b) f has local maxima where f 0 changes from positive to negative, at x =2and at x =6, and local minima where f 0 changes<br />

from negative to positive, at x =4and at x =8.<br />

(c) f is concave upward (CU) where f 0 is increasing, that is, on (3, 6) and (6, ∞),<br />

(e)<br />

and concave downward (CD) where f 0 is decreasing, that is, on (0, 3).<br />

(d) There is a point of inflection where f changes from being CD to being CU, that<br />

is, at x =3.<br />

33. (a) f(x) =2x 3 − 3x 2 − 12x ⇒ f 0 (x) =6x 2 − 6x − 12 = 6(x 2 − x − 2) = 6(x − 2)(x +1).<br />

f 0 (x) > 0 ⇔ x2 and f 0 (x) < 0 ⇔ −1 0 ⇔ x

F.<br />

160 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

(c) f 00 (x) =4− 12x 2 =4(1− 3x 2 ). f 00 (x) =0 ⇔ 1 − 3x 2 =0 ⇔<br />

(d)<br />

x 2 = 1 3<br />

⇔ x = ±1/ √ 3. f 00 (x) > 0 on −1/ √ 3, 1/ √ 3 and f 00 (x) < 0<br />

on −∞, −1/ √ 3 and 1/ √ 3, ∞ .Sof is concave upward on<br />

<br />

−1/<br />

√<br />

3, 1/<br />

√<br />

3<br />

<br />

and f is concave downward on<br />

<br />

−∞, −1/<br />

√<br />

3<br />

<br />

and<br />

√ √ 1/ 3, ∞ . f ±1/ 3 =2+<br />

2<br />

− 1 = 23 . There are points of inflection<br />

3 9 9<br />

at ±1/ √ <br />

3, 23<br />

9 .<br />

37. (a) h(x) =(x +1) 5 − 5x − 2 ⇒ h 0 (x) =5(x +1) 4 − 5. h 0 (x) =0 ⇔ 5(x +1) 4 =5 ⇔ (x +1) 4 =1 ⇒<br />

(x +1) 2 =1 ⇒ x +1=1or x +1=−1 ⇒ x =0or x = −2. h 0 (x) > 0 ⇔ x0 and<br />

h 0 (x) < 0 ⇔ −2 −1 and<br />

h 00 (x) < 0 ⇔ x 0 for x>−2 and A 0 (x) < 0 for −3 −3,soA is concave upward on (−3, ∞). Thereisnoinflection point.<br />

41. (a) C(x) =x 1/3 (x +4)=x 4/3 +4x 1/3 ⇒ C 0 (x) = 4 3 x1/3 + 4 3 x−2/3 = 4 3 x−2/3 (x +1)=<br />

4(x +1)<br />

3 3√ x 2 . C 0 (x) > 0 if<br />

−1

F.<br />

TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 161<br />

(b) C(−1) = −3 is a local minimum value.<br />

(c) C 00 (x) = 4 9 x−2/3 − 8 9 x−5/3 = 4 9 x−5/3 (x − 2) =<br />

4(x − 2)<br />

9 3√ x 5 .<br />

C 00 (x) < 0 for 0 0 ⇔ π

F.<br />

162 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

(d) f 00 (x) = (x2 − 1) 2 (−2) − (−2x) · 2(x 2 − 1)(2x)<br />

[(x 2 − 1) 2 ] 2<br />

= 2(x2 − 1)[−(x 2 − 1) + 4x 2 ]<br />

(x 2 − 1) 4 = 2(3x2 +1)<br />

(x 2 − 1) 3 .<br />

The sign of f 00 (x) is determined by the denominator; that is, f 00 (x) > 0 if<br />

|x| > 1 and f 00 (x) < 0 if |x| < 1. Thus,f is CU on (−∞, −1) and (1, ∞),<br />

and f is CD on (−1, 1). Therearenoinflection points.<br />

(e)<br />

√<br />

47. (a) lim x2 +1− x = ∞ and<br />

x→−∞<br />

√<br />

lim x2 +1− x √<br />

= lim x2 +1− x √ x 2 +1+x<br />

√<br />

x→∞<br />

x→∞<br />

x2 +1+x = lim 1<br />

√ =0,soy =0is a HA.<br />

x→∞ x2 +1+x<br />

(b) f(x) = √ x 2 +1− x ⇒ f 0 x<br />

x<br />

(x) = √ − 1. Since√ x2 +1 x2 +1 < 1 for all x, f 0 (x) < 0,sof is decreasing on R.<br />

(c) No minimum or maximum<br />

(d) f 00 (x) = (x2 +1) 1/2 (1) − x · 1<br />

2 (x2 +1) −1/2 (2x)<br />

√<br />

x2 +1 2<br />

(e)<br />

(x 2 +1) 1/2 x 2<br />

−<br />

(x<br />

=<br />

2 +1) 1/2<br />

x 2 +1<br />

so f is CU on R. NoIP<br />

= (x2 +1)− x 2<br />

(x 2 +1) 3/2 =<br />

1<br />

> 0,<br />

(x 2 +1)<br />

3/2<br />

49. f(x) =ln(1− ln x) is defined when x>0 (so that ln x is defined) and 1 − ln x>0 [so that ln(1 − ln x) is defined].<br />

The second condition is equivalent to 1 > ln x ⇔ x 0 ⇔ ln x

F.<br />

TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 163<br />

(d) f 00 (x) = (x +1)2 e −1/(x+1) 1/(x +1) 2 − e −1/(x+1) [2(x +1)]<br />

[(x +1) 2 ] 2<br />

= e−1/(x+1) [1 − (2x +2)]<br />

(x +1) 4 = − e−1/(x+1) (2x +1)<br />

(x +1) 4 ⇒<br />

(e)<br />

f 00 (x) > 0 ⇔ 2x +1< 0 ⇔ x 0 ⇒ (x − 3) 5 > 0 ⇒ x − 3 > 0 ⇒ x>3. Thus,f is increasing on the interval (3, ∞).<br />

55. (a) From the graph, we get an estimate of f(1) ≈ 1.41 as a local maximum<br />

value, and no local minimum value.<br />

f(x) = x +1 √<br />

x2 +1<br />

⇒ f 0 (x) =<br />

1 − x<br />

(x 2 +1) 3/2 .<br />

f 0 (x) =0 ⇔ x =1. f(1) = 2 √<br />

2<br />

= √ 2 is the exact value.<br />

(b) From the graph in part (a), f increases most rapidly somewhere between x = − 1 and x = − 1 .Tofind the exact value,<br />

2 4<br />

we need to find the maximum value of f 0 , which we can do by finding the critical numbers of f 0 .<br />

f 00 (x) = 2x2 − 3x − 1<br />

=0 ⇔ x = 3 ± √ 17<br />

. x = 3+√ 17<br />

corresponds to the minimum value of f 0 .<br />

(x 2 +1) 5/2 4<br />

4<br />

<br />

Themaximumvalueoff 0 3 −<br />

is at<br />

√ <br />

17 7<br />

4<br />

, − √ 17<br />

≈ (−0.28, 0.69).<br />

6 6<br />

57. f(x) =cosx + 1 2 cos 2x ⇒ f 0 (x) =− sin x − sin 2x ⇒ f 00 (x) =− cos x − 2cos2x<br />

(a)<br />

From the graph of f,itseemsthatf is CD on (0, 1),CUon(1, 2.5),CDon<br />

(2.5, 3.7),CUon(3.7, 5.3), and CD on (5.3, 2π). The points of inflection<br />

appear to be at (1, 0.4), (2.5, −0.6), (3.7, −0.6),and(5.3, 0.4).<br />

(b)<br />

From the graph of f 00 (and zooming in near the zeros), it seems that f is CD<br />

on (0, 0.94),CUon(0.94, 2.57),CDon(2.57, 3.71),CUon(3.71, 5.35),<br />

and CD on (5.35, 2π). Refined estimates of the inflection points are<br />

(0.94, 0.44), (2.57, −0.63), (3.71, −0.63),and(5.35, 0.44).<br />

59. In Maple, we define f andthenusethecommand<br />

plot(diff(diff(f,x),x),x=-2..2);. InMathematica,wedefine f<br />

andthenusePlot[Dt[Dt[f,x],x],{x,-2,2}]. Weseethatf 00 > 0 for<br />

x0.0 [≈ 0.03] andf 00 < 0 for −0.6

F.<br />

164 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

61. (a) The rate of increase of the population is initially very small, then gets larger until it reaches a maximum at about<br />

t =8hours, and decreases toward 0 as the population begins to level off.<br />

(b) The rate of increase has its maximum value at t =8hours.<br />

(c) The population function is concave upward on (0, 8) and concave downward on (8, 18).<br />

(d) At t =8, the population is about 350,sotheinflection point is about (8, 350).<br />

63. Most students learn more in the third hour of studying than in the eighth hour, so K(3) − K(2) is larger than K(8) − K(7).<br />

In other words, as you begin studying for a test, the rate of knowledge gain is large and then starts to taper off, so K 0 (t)<br />

decreases and the graph of K is concave downward.<br />

65. S(t) =At p e −kt with A =0.01, p =4,andk =0.07. Wewillfind the<br />

zeros of f 00 for f(t) =t p e −kt .<br />

f 0 (t) =t p (−ke −kt )+e −kt (pt p−1 )=e −kt (−kt p + pt p−1 )<br />

f 00 (t)=e −kt (−kpt p−1 + p(p − 1)t p−2 )+(−kt p + pt p−1 )(−ke −kt )<br />

= t p−2 e −kt [−kpt + p(p − 1) + k 2 t 2 − kpt]<br />

= t p−2 e −kt (k 2 t 2 − 2kpt + p 2 − p)<br />

Using the given values of p and k gives us f 00 (t) =t 2 e −0.07t (0.0049t 2 − 0.56t +12).SoS 00 (t) =0.01f 00 (t) and its zeros<br />

are t =0and the solutions of 0.0049t 2 − 0.56t +12=0, which are t 1 = 200<br />

7<br />

≈ 28.57 and t 2 = 600<br />

7<br />

≈ 85.71.<br />

At t 1 minutes, the rate of increase of the level of medication in the bloodstream is at its greatest and at t 2 minutes, the rate of<br />

decrease is the greatest.<br />

67. f(x) =ax 3 + bx 2 + cx + d ⇒ f 0 (x) =3ax 2 +2bx + c.<br />

We are given that f(1) = 0 and f(−2) = 3,sof(1) = a + b + c + d =0and<br />

f(−2) = −8a +4b − 2c + d =3.Alsof 0 (1) = 3a +2b + c =0and<br />

f 0 (−2) = 12a − 4b + c =0by Fermat’s Theorem. Solving these four equations, we get<br />

a = 2 , b = 1 , c = − 4 , d = 7 , so the function is f(x) = 1<br />

9 3 3 9 9 2x 3 +3x 2 − 12x +7 .<br />

69. y = 1+x ⇒ y 0 = (1 + x2 )(1) − (1 + x)(2x) 1 − 2x − x2<br />

= ⇒<br />

1+x 2 (1 + x 2 ) 2 (1 + x 2 ) 2<br />

y 00 = (1 + x2 ) 2 (−2 − 2x) − (1 − 2x − x 2 ) · 2(1 + x 2 )(2x)<br />

= 2(1 + x2 )[(1 + x 2 )(−1 − x) − (1 − 2x − x 2 )(2x)]<br />

[(1 + x 2 ) 2 ] 2 (1 + x 2 ) 4<br />

= 2(−1 − x − x2 − x 3 − 2x +4x 2 +2x 3 )<br />

= 2(x3 +3x 2 − 3x − 1)<br />

= 2(x − 1)(x2 +4x +1)<br />

(1 + x 2 ) 3 (1 + x 2 ) 3 (1 + x 2 ) 3<br />

So y 00 =0 ⇒ x =1, −2 ± √ 3.Leta = −2 − √ 3, b = −2+ √ 3,andc =1. We can show that f(a) = 1 4<br />

<br />

1 −<br />

√<br />

3<br />

<br />

,

F.<br />

TX.10 SECTION 4.3 HOW DERIVATIVES AFFECT THE SHAPE OF A GRAPH ¤ 165<br />

f(b) = 1 4<br />

<br />

1+<br />

√<br />

3<br />

<br />

,andf(c) =1. To show that these three points of inflection lie on one straight line, we’ll show that the<br />

slopes m ac and m bc are equal.<br />

m ac =<br />

m bc =<br />

f(c) − f(a)<br />

c − a<br />

f(c) − f(b)<br />

c − b<br />

= 1 − √ <br />

1<br />

4 1 − 3<br />

3<br />

1 − −2 − √ 3 = + √ 1<br />

4 4 3<br />

3+ √ 3 = 1 4<br />

= 1 − √ <br />

1<br />

4 1+ 3<br />

3<br />

1 − −2+ √ 3 = − √ 1<br />

4 4 3<br />

3 − √ 3 = 1 4<br />

71. Suppose that f is differentiable on an interval I and f 0 (x) > 0 for all x in I except x = c. Toshowthatf is increasing on I,<br />

let x 1, x 2 be two numbers in I with x 1

F.<br />

166 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

77. Let the cubic function be f(x) =ax 3 + bx 2 + cx + d ⇒ f 0 (x) =3ax 2 +2bx + c ⇒ f 00 (x) =6ax +2b.<br />

So f is CU when 6ax +2b>0 ⇔ x>−b/(3a),CDwhenx0,so(0, 0) is an inflection point. But g 00 (0) does not<br />

exist.<br />

83. (a) f(x) =x 4 sin 1 x ⇒ f 0 (x) =x 4 cos 1 − 1 <br />

+sin 1 x x 2 x (4x3 )=4x 3 sin 1 x − x2 cos 1 x .<br />

<br />

g(x) =x 4 2+sin 1 <br />

=2x 4 + f(x) ⇒ g 0 (x) =8x 3 + f 0 (x).<br />

x<br />

<br />

h(x) =x 4 −2+sin 1 <br />

= −2x 4 + f(x) ⇒ h 0 (x) =−8x 3 + f 0 (x).<br />

x<br />

It is given that f(0) = 0, sof 0 f(x) − f(0)<br />

x 4 sin 1<br />

(0) = lim<br />

=lim x − 0<br />

=limx 3 sin 1<br />

x→0 x − 0 x→0 x<br />

x→0 x .Since<br />

− x 3 ≤ x 3 sin 1 x ≤ x 3 and lim<br />

x→0<br />

x 3 =0,weseethatf 0 (0) = 0 by the Squeeze Theorem. Also,<br />

g 0 (0) = 8(0) 3 + f 0 (0) = 0 and h 0 (0) = −8(0) 3 + f 0 (0) = 0,so0 is a critical number of f, g,andh.<br />

For x 2n = 1<br />

2nπ<br />

For x 2n+1 =<br />

1<br />

1<br />

[n a nonzero integer], sin =sin2nπ =0and cos =cos2nπ =1,sof 0 (x 2n )=−x 2 2n < 0.<br />

x 2n x 2n<br />

1<br />

(2n +1)π , sin 1<br />

1<br />

=sin(2n +1)π =0and cos =cos(2n +1)π = −1, so<br />

x 2n+1 x 2n+1<br />

f 0 (x 2n+1 )=x 2 2n+1 > 0. Thus,f 0 changes sign infinitely often on both sides of 0.<br />

Next, g 0 (x 2n) =8x 3 2n + f 0 (x 2n) =8x 3 2n − x 2 2n = x 2 2n(8x 2n − 1) < 0 for x 2n < 1 8 ,but<br />

g 0 (x 2n+1 )=8x 3 2n+1 + x 2 2n+1 = x 2 2n+1(8x 2n+1 +1)> 0 for x 2n+1 > − 1 8 ,sog0 changes sign infinitely often on both<br />

sides of 0.<br />

Last, h 0 (x 2n) =−8x 3 2n + f 0 (x 2n) =−8x 3 2n − x 2 2n = −x 2 2n(8x 2n +1)< 0 for x 2n > − 1 8 and<br />

h 0 (x 2n+1 )=−8x 3 2n+1 + x 2 2n+1 = x 2 2n+1(−8x 2n+1 +1)> 0 for x 2n+1 < 1 8 ,soh0 changes sign infinitely often on both<br />

sides of 0.

F.<br />

TX.10SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ¤ 167<br />

(b) f(0) = 0 and since sin 1 x and hence x4 sin 1 is both positive and negative inifinitely often on both sides of 0,and<br />

x<br />

arbitrarily close to 0, f has neither a local maximum nor a local minimum at 0.<br />

Since 2+sin 1 <br />

x ≥ 1, g(x) =x4 2+sin 1 <br />

> 0 for x 6= 0,sog(0) = 0 is a local minimum.<br />

x<br />

Since −2+sin 1 <br />

x ≤−1, h(x) =x4 −2+sin 1 <br />

< 0 for x 6= 0,soh(0) = 0 is a local maximum.<br />

x<br />

4.4 Indeterminate Forms and L'Hospital's Rule<br />

Note: The use of l’Hospital’s Rule is indicated by an H above the equal sign:<br />

1. (a) lim<br />

x→a<br />

f(x)<br />

g(x) is an indeterminate form of type 0 0 .<br />

f(x)<br />

(b) lim =0because the numerator approaches 0 while the denominator becomes large.<br />

x→a p(x)<br />

h(x)<br />

(c) lim =0because the numerator approaches a finite number while the denominator becomes large.<br />

x→a p(x)<br />

H<br />

=<br />

p(x)<br />

(d) If lim p(x) =∞ and f(x) → 0 through positive values, then lim<br />

x→a x→a f(x) = ∞. [Forexample,takea =0, p(x) =1/x2 ,<br />

and f(x) =x 2 p(x)<br />

.] If f(x) → 0 through negative values, then lim<br />

x→a f(x) = −∞. [Forexample,takea =0, p(x) =1/x2 ,<br />

and f(x) =−x 2 .] If f(x) → 0 through both positive and negative values, then the limit might not exist. [For example,<br />

take a =0, p(x) =1/x 2 ,andf(x) =x.]<br />

p(x)<br />

(e) lim<br />

x→a q(x) is an indeterminate form of type ∞ ∞ .<br />

3. (a) When x is near a, f(x) is near 0 and p(x) is large, so f(x) − p(x) is large negative. Thus, lim<br />

x→a<br />

[f(x) − p(x)] = −∞.<br />

(b) lim<br />

x→a<br />

[ p(x) − q(x)] is an indeterminate form of type ∞−∞.<br />

(c) When x is near a, p(x) and q(x) are both large, so p(x)+q(x) is large. Thus, lim<br />

x→a<br />

[ p(x)+q(x)] = ∞.<br />

5. This limit has the form 0 0 .Wecansimplyfactorandsimplifytoevaluatethelimit.<br />

x 2 − 1<br />

lim<br />

x→1 x 2 − x =lim (x +1)(x − 1) x +1<br />

= lim = 1+1 =2<br />

x→1 x(x − 1) x→1 x 1<br />

7. This limit has the form 0 . lim x 9 − 1 H 9x 8<br />

0<br />

=lim<br />

x→1 x 5 − 1 x→1 5x = 9 4 5 lim<br />

x→1 x4 = 9 5 (1) = 9 5<br />

9. This limit has the form 0 0 . lim<br />

x→(π/2) +<br />

11. This limit has the form 0 . lim e t − 1<br />

0 t→0 t 3<br />

cos x<br />

1 − sin x<br />

H<br />

=lim<br />

t→0<br />

H − sin x<br />

= lim<br />

x→(π/2) + − cos x =<br />

lim<br />

x→(π/2)<br />

+<br />

tan x = −∞.<br />

e t<br />

3t 2 = ∞ since et → 1 and 3t 2 → 0 + as t → 0.

F.<br />

168 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

13. This limit has the form 0 . lim tan px H p sec 2 px<br />

=lim<br />

0 x→0 tan qx x→0 q sec 2 qx = p(1)2<br />

q(1) = p 2 q<br />

15. This limit has the form ∞ . lim ln x<br />

√<br />

∞<br />

= H lim<br />

x→∞ x<br />

x→∞<br />

1<br />

2<br />

1/x<br />

= lim<br />

x−1/2 x→∞<br />

2<br />

√<br />

x<br />

=0<br />

17. lim<br />

x→0 + [(ln x)/x] =−∞ since ln x →−∞as x → 0+ anddividingbysmallvaluesofx just increases the magnitude of the<br />

quotient (ln x)/x. L’Hospital’s Rule does not apply.<br />

19. This limit has the form ∞ . lim e x<br />

∞ x→∞ x 3<br />

21. This limit has the form 0 . lim e x − 1 − x<br />

0 x→0 x 2<br />

= H e x<br />

lim<br />

x→∞ 3x 2<br />

= H e x<br />

lim<br />

x→∞ 6x<br />

H e x<br />

= lim<br />

x→∞ 6 = ∞<br />

=lim<br />

H e x − 1 H e x<br />

=lim<br />

x→0 2x x→0 2 = 1 2<br />

23. This limit has the form 0 . lim tanh x H sech 2 x<br />

=lim<br />

0 x→0 tan x x→0 sec 2 x = sech2 0<br />

sec 2 0 = 1 1 =1<br />

25. This limit has the form 0 . lim 5 t − 3 t<br />

0 t→0 t<br />

H 5 t ln 5 − 3 t ln 3<br />

=lim<br />

=ln5− ln 3 = ln 5<br />

t→0 3<br />

1<br />

27. This limit has the form 0 . lim sin −1 x H 1/ √ 1 − x<br />

= lim<br />

2<br />

1<br />

=lim√ 0 x→0 x x→0 1<br />

= 1 x→0 1 − x<br />

2 1 =1<br />

29. This limit has the form 0 . lim 1 − cos x<br />

0<br />

x→0 x 2<br />

=lim<br />

H sin x H cos x<br />

= lim = 1<br />

x→0 2x x→0 2 2<br />

x +sinx<br />

31. lim<br />

x→0 x +cosx = 0+0<br />

0+1 = 0 =0. L’Hospital’s Rule does not apply.<br />

1<br />

33. This limit has the form 0 . lim 1 − x +lnx<br />

0 x→1 1+cosπx<br />

H<br />

=lim<br />

x→1<br />

−1+1/x<br />

−π sin πx<br />

H −1/x 2<br />

=lim<br />

x→1 −π 2 cos πx =<br />

−1<br />

−π 2 (−1) = − 1 π 2<br />

35. This limit has the form 0 . lim x a − ax + a − 1 H ax a−1 − a H a(a − 1)x a−2 a(a − 1)<br />

=lim<br />

= lim<br />

=<br />

0 x→1 (x − 1) 2 x→1 2(x − 1) x→1 2<br />

2<br />

37. This limit has the form 0 . lim cos x − 1+ 1 2 x2<br />

0 x→0 x 4<br />

39. This limit has the form ∞ · 0.<br />

H =lim<br />

x→0<br />

− sin x + x<br />

4x 3<br />

H = lim<br />

x→0<br />

− cos x +1<br />

12x 2<br />

sin(π/x) H cos(π/x)(−π/x 2 )<br />

lim x sin(π/x) = lim = lim<br />

= π lim cos(π/x) =π(1) = π<br />

x→∞ x→∞ 1/x x→∞ −1/x 2<br />

x→∞<br />

41. This limit has the form ∞ · 0. We’ll change it to the form 0 0 .<br />

sin 6x H 6cos6x<br />

lim cot 2x sin 6x =lim =lim<br />

x→0 x→0 tan 2x x→0 2sec 2 2x = 6(1)<br />

2(1) =3 2<br />

=lim<br />

H sin x H cos x<br />

= lim<br />

x→0 24x x→0 24 = 1<br />

24<br />

x 3<br />

43. This limit has the form ∞ · 0. lim<br />

x→∞ x3 e −x2 = lim<br />

x→∞ e x2<br />

= H 3x 2<br />

lim<br />

x→∞ 2xe x2<br />

= lim<br />

x→∞<br />

3x<br />

2e x2<br />

= H 3<br />

lim =0<br />

x→∞ x2<br />

4xe

F.<br />

TX.10SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ¤ 169<br />

45. This limit has the form 0 · (−∞).<br />

lim ln x tan(πx/2) = lim<br />

x→1 + x→1 +<br />

47. This limit has the form ∞−∞.<br />

x<br />

lim<br />

x→1 x − 1 − 1<br />

ln x<br />

ln x<br />

cot(πx/2)<br />

<br />

=lim<br />

x→1<br />

x ln x − (x − 1)<br />

(x − 1) ln x<br />

H<br />

= lim<br />

x→1 + 1/x<br />

(−π/2) csc 2 (πx/2) = 1<br />

(−π/2)(1) 2 = − 2 π<br />

H<br />

= lim<br />

x→1<br />

x(1/x)+lnx − 1<br />

(x − 1)(1/x)+lnx = lim<br />

x→1<br />

H 1/x<br />

=lim<br />

x→1 1/x 2 +1/x · x2<br />

x = lim x<br />

2 x→1 1+x = 1<br />

1+1 = 1 2<br />

ln x<br />

1 − (1/x)+lnx<br />

49. We will multiply and divide by the conjugate of the expression to change the form of the expression.<br />

√<br />

lim x2 + x − x √ √ <br />

x2 + x − x x2 + x + x<br />

x 2 + x − x 2<br />

= lim<br />

· √ = lim √<br />

x→∞<br />

x→∞ 1 x2 + x + x x→∞ x2 + x + x<br />

x<br />

= lim √<br />

x→∞ x2 + x + x = lim 1<br />

1<br />

= √ = 1<br />

x→∞ 1+1/x +1 1+1 2<br />

As an alternate solution, write √ x 2 + x − x as √ x 2 + x − √ x 2 ,factorout √ x 2 ,rewriteas( 1+1/x − 1)/(1/x),and<br />

apply l’Hospital’s Rule.<br />

51. The limit has the form ∞−∞and we will change the form to a product by factoring out x.<br />

<br />

lim (x − ln x) = lim x 1 − ln x <br />

ln x H 1/x<br />

= ∞ since lim = lim<br />

x→∞ x→∞ x<br />

x→∞ x x→∞ 1 =0.<br />

<br />

53. y = x x2 ⇒ ln y = x 2 ln x<br />

ln x,so lim ln y = lim x2 ln x = lim<br />

H 1/x<br />

= lim<br />

x→0 + x→0 + x→0 + 1/x 2 x→0 + −2/x = lim − 1 3 x→0 + 2 x2 =0 ⇒<br />

lim<br />

x→0 + xx2 = lim eln y = e 0 =1.<br />

x→0 +<br />

55. y =(1− 2x) 1/x ⇒ ln y = 1 x<br />

lim<br />

x→0 (1 − 2x)1/x =lim<br />

x→0<br />

e ln y = e −2 .<br />

57. y =<br />

<br />

1+ 3 x + 5 x<br />

⇒ ln y = x ln<br />

1+ 3 x 2 x + 5 <br />

x 2<br />

ln<br />

1+ 3 x + 5 <br />

x<br />

lim ln y = lim<br />

2<br />

x→∞ x→∞ 1/x<br />

ln(1 − 2x) H −2/(1 − 2x)<br />

ln(1 − 2x), solim ln y =lim<br />

= lim<br />

= −2 ⇒<br />

x→0 x→0 x<br />

x→0 1<br />

H<br />

= lim<br />

x→∞<br />

<br />

so lim 1+ 3<br />

x→∞ x + 5 x<br />

= lim<br />

x 2 x→∞ eln y = e 3 .<br />

⇒<br />

<br />

− 3 x 2 − 10<br />

x 3 <br />

1+ 3 x + 5 x 2 <br />

−1/x 2<br />

59. y = x 1/x ln x<br />

⇒ ln y =(1/x) lnx ⇒ lim ln y = lim<br />

x→∞ x→∞ x<br />

lim<br />

x→∞ x1/x = lim<br />

x→∞ eln y = e 0 =1<br />

= lim<br />

x→∞<br />

H 1/x<br />

= lim<br />

x→∞ 1 =0 ⇒<br />

3+ 10 x<br />

1+ 3 x + 5 x 2 =3,

F.<br />

170 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

61. y =(4x +1) cot x ln(4x +1) H<br />

⇒ ln y =cotx ln(4x +1),so lim ln y = lim<br />

= lim<br />

x→0 + x→0 + tan x<br />

x→0 + 4<br />

4x +1<br />

sec 2 x =4<br />

⇒<br />

lim<br />

x→0 +(4x +1)cot x = lim<br />

x→0 + eln y = e 4 .<br />

63. y =(cosx) 1/x2 ⇒ ln y = 1 ln cos x<br />

ln cos x ⇒ lim ln y = lim<br />

x2 x→0 + x→0 + x 2<br />

= H − tan x H − sec 2 x<br />

lim = lim = − 1<br />

x→0 + 2x x→0 + 2 2<br />

⇒<br />

lim<br />

x→0 +(cos x)1/x2 = lim<br />

x→0 + eln y = e −1/2 =1/ √ e<br />

65. From the graph, if x = 500, y ≈ 7.36. The limit has the form 1 ∞ .<br />

<br />

Now y = 1+ 2 x <br />

⇒ ln y = x ln 1+ 2 <br />

⇒<br />

x x<br />

1<br />

− 2 <br />

ln(1 + 2/x) H 1+2/x x<br />

lim ln y = lim<br />

= lim<br />

2<br />

x→∞ x→∞ 1/x x→∞ −1/x 2<br />

=2 lim<br />

x→∞<br />

1<br />

1+2/x =2(1)=2<br />

<br />

lim 1+ 2 x<br />

= lim<br />

x→∞ x x→∞ eln y = e 2 [≈ 7.39]<br />

f(x)<br />

67. From the graph, it appears that lim<br />

x→0 g(x) =lim f 0 (x)<br />

x→0 g 0 (x) =0.25.<br />

We calculate lim<br />

x→0<br />

f(x)<br />

g(x) = lim<br />

x→0<br />

e x − 1<br />

x 3 +4x<br />

⇒<br />

H<br />

= lim<br />

x→0<br />

e x<br />

3x 2 +4 = 1 4 .<br />

e x<br />

69. lim<br />

x→∞ x n<br />

71. lim<br />

x→∞<br />

= H e x<br />

lim<br />

x→∞ nx n−1<br />

x<br />

√<br />

x2 +1<br />

H<br />

= lim<br />

x→∞<br />

H = lim<br />

x→∞<br />

e x<br />

n(n − 1)x n−2<br />

1<br />

1<br />

2 (x2 +1) −1/2 (2x) = lim<br />

x→∞<br />

= H ··· H= e x<br />

lim<br />

x→∞ n! = ∞<br />

√<br />

x2 +1<br />

. Repeated applications of l’Hospital’s Rule result in the<br />

x<br />

original limit or the limit of the reciprocal of the function. Another method is to try dividing the numerator and denominator<br />

by x:<br />

lim<br />

x→∞<br />

x<br />

√<br />

x2 +1 = lim<br />

x→∞<br />

x/x<br />

<br />

x2 /x 2 +1/x 2 = lim<br />

x→∞<br />

1<br />

<br />

1+1/x<br />

2 = 1 1 =1<br />

<br />

73. First we will find lim 1+ r nt, <br />

which is of the form 1<br />

n→∞ n ∞ . y = 1+ r nt <br />

⇒ ln y = nt ln 1+ r <br />

,so<br />

n n<br />

<br />

−r/n<br />

2<br />

lim ln y = lim<br />

1+ nt ln r <br />

ln(1 + r/n)<br />

= t lim<br />

n→∞ n→∞ n n→∞ 1/n<br />

H<br />

= t lim<br />

n→∞<br />

(1 + r/n)(−1/n 2 ) = t lim<br />

n→∞<br />

r<br />

1+i/n = tr<br />

⇒<br />

lim y =<br />

n→∞ ert .Thus,asn →∞, A = A 0<br />

1+ r nt<br />

→ A0 e rt .<br />

n

F.<br />

e E<br />

75. lim P (E) = lim + e −E<br />

E→0 + E→0 + e E − e − 1 <br />

−E E<br />

TX.10SECTION 4.4 INDETERMINATE FORMS AND L’HOSPITAL’S RULE ¤ 171<br />

E e E + e −E − 1 e E − e −E Ee E + Ee −E − e E + e −E <br />

= lim<br />

= lim<br />

form is<br />

0<br />

E→0 + (e E − e −E ) E<br />

E→0 + Ee E − Ee −E 0<br />

H<br />

= lim<br />

E→0 + Ee E + e E · 1+E −e −E + e −E · 1 − e E + −e −E<br />

Ee E + e E · 1 − [E(−e −E )+e −E · 1]<br />

= lim<br />

E→0 +<br />

Ee E − Ee −E<br />

= lim<br />

Ee E + e E + Ee −E − e−E E→0 +<br />

= 0<br />

e E − e −E<br />

, whereL = lim<br />

2+L E→0 + E<br />

Thus, lim P (E) = 0<br />

E→0 + 2+2 =0.<br />

e E − e −E<br />

e E + eE E + e−E − e−E<br />

E<br />

<br />

form is<br />

0<br />

0<br />

H = lim<br />

E→0 + e E + e −E<br />

1<br />

77. We see that both numerator and denominator approach 0, so we can use l’Hospital’s Rule:<br />

√<br />

2a3 x − x<br />

lim<br />

4 − a 3√ aax<br />

x→a a − 4√ ax 3<br />

[divide by E]<br />

= 1+1<br />

1<br />

=lim<br />

H 1<br />

2 (2a3 x − x 4 ) −1/2 (2a 3 − 4x 3 ) − a <br />

1<br />

3 (aax) −2/3 a 2<br />

x→a − 1 4 (ax3 ) −3/4 (3ax 2 )<br />

=<br />

1<br />

2 (2a3 a − a 4 ) −1/2 (2a 3 − 4a 3 ) − 1 3 a3 (a 2 a) −2/3<br />

− 1 4 (aa3 ) −3/4 (3aa 2 )<br />

=2<br />

= (a4 ) −1/2 (−a 3 ) − 1 3 a3 (a 3 ) −2/3<br />

− 3 4 a3(a4 )−3/4<br />

= −a − 1 3 a<br />

− 3 4<br />

= 4 3<br />

4<br />

3 a = 16 9 a<br />

79. Since f(2) = 0, the given limit has the form 0 0 .<br />

f(2 + 3x)+f(2 + 5x) H f 0 (2 + 3x) · 3+f 0 (2 + 5x) · 5<br />

lim<br />

=lim<br />

= f 0 (2) · 3+f 0 (2) · 5=8f 0 (2) = 8 · 7=56<br />

x→0 x<br />

x→0 1<br />

81. Since lim<br />

h→0<br />

[f(x + h) − f(x − h)] = f(x) − f(x) =0 (f is differentiable and hence continuous) and lim<br />

h→0<br />

2h =0,weuse<br />

l’Hospital’s Rule:<br />

f(x + h) − f(x − h) H f 0 (x + h)(1) − f 0 (x − h)(−1)<br />

lim<br />

=lim<br />

= f 0 (x)+f 0 (x)<br />

= 2f 0 (x)<br />

= f 0 (x)<br />

h→0 2h<br />

h→0 2<br />

2<br />

2<br />

f(x + h) − f(x − h)<br />

2h<br />

is the slope of the secant line between<br />

(x − h, f(x − h)) and (x + h, f(x + h)). Ash → 0,thislinegetscloser<br />

to the tangent line and its slope approaches f 0 (x).<br />

f(x)<br />

83. (a) We show that lim<br />

x→0 x =0for every integer n ≥ 0. Lety = 1 n<br />

x .Then 2<br />

f(x)<br />

lim<br />

x→0 x =lim e −1/x2 y n<br />

2n x→0 (x 2 ) n = lim<br />

y→∞ e y<br />

= H ny n−1<br />

lim<br />

y→∞ e y<br />

= H ··· H= n!<br />

lim<br />

y→∞ e =0 ⇒<br />

y<br />

f(x)<br />

lim<br />

x→0 x = lim f(x)<br />

n x→0 xn x =lim f(x)<br />

2n x→0 xn lim<br />

x→0 x =0.Thus,f 0 f(x) − f(0) f(x)<br />

(0) = lim<br />

= lim<br />

2n<br />

x→0 x − 0 x→0 x =0.

F.<br />

172 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

(b) Using the Chain Rule and the Quotient Rule we see that f (n) (x) exists for x 6= 0. In fact, we prove by induction that for<br />

each n ≥ 0, there is a polynomial p n and a non-negative integer k n with f (n) (x) =p n (x)f(x)/x kn for x 6= 0.Thisis<br />

true for n =0; suppose it is true for the nth derivative. Then f 0 (x) =f(x)(2/x 3 ),so<br />

which has the desired form.<br />

f (n+1) (x)= x k n<br />

[p 0 n(x) f(x)+p n(x)f 0 (x)] − k nx k n−1 p n(x) f(x) x −2k n<br />

= x k n<br />

p 0 n(x)+p n(x) 2/x 3 − k nx k n−1 p n(x) f(x)x −2k n<br />

= x kn+3 p 0 n(x)+2p n (x) − k n x kn+2 p n (x) f(x)x −(2kn+3)<br />

Now we show by induction that f (n) (0) = 0 for all n.Bypart(a),f 0 (0) = 0. Suppose that f (n) (0) = 0. Then<br />

f (n+1) f (n) (x) − f (n) (0) f (n) (x) p n(x) f(x)/x k n<br />

(0) = lim<br />

= lim =lim<br />

x→0 x − 0<br />

x→0 x x→0 x<br />

f(x)<br />

=limp n (x) lim<br />

x→0 x→0 x = p n(0) · 0=0<br />

kn+1<br />

= lim<br />

x→0<br />

p n(x) f(x)<br />

x k n+1<br />

4.5 Summary of Curve Sketching<br />

1. y = f(x) =x 3 + x = x(x 2 +1) A. f is a polynomial, so D = R.<br />

H.<br />

B. x-intercept =0, y-intercept = f(0) = 0 C. f(−x) =−f(x),sof is<br />

odd; the curve is symmetric about the origin.<br />

D. f is a polynomial, so there is<br />

no asymptote.<br />

E. f 0 (x) =3x 2 +1> 0,sof is increasing on (−∞, ∞).<br />

F. There is no critical number and hence, no local maximum or minimum value.<br />

G. f 00 (x) =6x>0 on (0, ∞) and f 00 (x) < 0 on (−∞, 0),sof is CU on<br />

(0, ∞) and CD on (−∞, 0). Since the concavity changes at x =0,thereisan<br />

inflection point at (0, 0).<br />

3. y = f(x) =2− 15x +9x 2 − x 3 = −(x − 2) x 2 − 7x +1 A. D = R B. y-intercept: f(0) = 2; x-intercepts:<br />

f(x) =0 ⇒ x =2or (by the quadratic formula) x = 7 ± √ 45<br />

2<br />

≈ 0.15, 6.85 C. No symmetry D. No asymptote<br />

E. f 0 (x) =−15 + 18x − 3x 2 = −3(x 2 − 6x +5)<br />

= −3(x − 1)(x − 5) > 0 ⇔ 1

F.<br />

TX.10<br />

SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 173<br />

5. y = f(x) =x 4 +4x 3 = x 3 (x +4) A. D = R B. y-intercept: f(0) = 0;<br />

H.<br />

x-intercepts: f(x) =0 ⇔ x = −4, 0 C. No symmetry<br />

D. No asymptote E. f 0 (x) =4x 3 +12x 2 =4x 2 (x +3)> 0 ⇔<br />

x>−3,sof is increasing on (−3, ∞) and decreasing on (−∞, −3).<br />

F. Local minimum value f(−3) = −27, no local maximum<br />

G. f 00 (x) =12x 2 +24x =12x(x +2)< 0 ⇔ −2 1,sof is CU on (1, ∞) and<br />

(x − 1)<br />

3<br />

11. y = f(x) =1/(x 2 − 9) A. D = {x | x 6=±3} =(−∞, −3) ∪ (−3, 3) ∪ (3, ∞) B. y-intercept = f(0) = − 1 9 ,no<br />

1<br />

x-intercept C. f(−x) =f(x) ⇒ f is even; the curve is symmetric about the y-axis. D. lim =0,soy =0<br />

x→±∞ x 2 − 9<br />

is a HA.<br />

1<br />

1<br />

lim = −∞, lim<br />

x→3 − x 2 − 9 x→3 + x 2 − 9 = ∞,<br />

lim 1<br />

x→−3 − x 2 − 9 = ∞,<br />

lim 1<br />

= −∞, sox =3and x = −3<br />

x→−3 + x 2 − 9<br />

are VA. E. f 0 2x<br />

(x) =−<br />

(x 2 2<br />

> 0 ⇔ x

F.<br />

174 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

decreasing on (0, 3) and (3, ∞). F. Local maximum value f(0) = − 1 9 .<br />

H.<br />

G. y 00 = −2(x2 − 9) 2 +(2x)2(x 2 − 9)(2x)<br />

(x 2 − 9) 4 = 6(x2 +3)<br />

(x 2 − 9) 3 > 0 ⇔<br />

x 2 > 9 ⇔ x>3 or x 0 ⇔ −3

F.<br />

TX.10<br />

SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 175<br />

17. y = f(x) = x2<br />

x 2 +3 = (x2 +3)− 3<br />

=1− 3<br />

x 2 +3 x 2 +3<br />

A. D = R B. y-intercept: f(0) = 0;<br />

x-intercepts: f(x) =0 ⇔ x =0 C. f(−x) =f(x),sof is even; the graph is symmetric about the y-axis.<br />

D. lim<br />

x→±∞<br />

x 2<br />

x 2 +3 =1,soy =1is a HA. No VA. E. Using the Reciprocal Rule, f 0 (x) =−3 ·<br />

f 0 (x) > 0 ⇔ x>0 and f 0 (x) < 0 ⇔ x 0 ⇒ x>− 1 and f 0 (x) < 0 ⇒ x

F.<br />

176 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

23. y = f(x) =x/ √ x 2 +1 A. D = R B. y-intercept: f(0) = 0; x-intercepts: f(x) =0 ⇒ x =0<br />

C. f(−x) =−f(x),sof is odd; the graph is symmetric about the origin.<br />

D. lim f(x) = lim<br />

x→∞ x→∞<br />

and<br />

lim f(x) =<br />

x→−∞<br />

=<br />

lim<br />

x→−∞<br />

x<br />

√<br />

x2 +1 = lim<br />

x→∞<br />

x<br />

√<br />

x2 +1 =<br />

lim<br />

x→−∞<br />

x/x<br />

√<br />

x2 +1/x = lim<br />

x→∞<br />

x/x<br />

√<br />

x2 +1/x =<br />

1<br />

− √ = −1 so y = ±1 are HA.<br />

1+0<br />

lim<br />

x→−∞<br />

x/x<br />

√<br />

x2 +1/ √ x 2 = lim<br />

x→∞<br />

x/x<br />

√<br />

x2 +1/− √ x 2 = lim<br />

x→−∞<br />

1<br />

<br />

1+1/x<br />

2 = 1<br />

√ 1+0<br />

=1<br />

1<br />

− 1+1/x 2<br />

No VA.<br />

√ 2x<br />

x2 +1− x ·<br />

E. f 0 2 √ x<br />

(x) =<br />

2 +1<br />

= x2 +1− x 2<br />

[(x 2 +1) 1/2 ] 2 (x 2 +1) = 1<br />

> 0 for all x,sof is increasing on R.<br />

3/2 (x 2 3/2<br />

+1)<br />

F. No extreme values<br />

G. f 00 (x) =− 3 2 (x2 +1) −5/2 · 2x =<br />

−3x<br />

(x 2 +1) 5/2 ,sof 00 (x) > 0 for x0. Thus,f is CU on (−∞, 0) and CD on (0, ∞).<br />

IP at (0, 0)<br />

25. y = f(x) = √ 1 − x 2 /x A. D = {x ||x| ≤ 1, x 6= 0} =[−1, 0) ∪ (0, 1] B. x-intercepts ±1, noy-intercept<br />

√<br />

1 − x<br />

2<br />

C. f(−x) =−f(x), so the curve is symmetric about (0, 0) . D. lim<br />

x→0 + x<br />

<br />

−x 2 / √ 1 − x 2 − √ 1 − x 2<br />

so x =0is a VA. E. f 0 (x) =<br />

on (−1, 0) and (0, 1).<br />

G. f 00 (x) =<br />

<br />

f is CU on −1, −<br />

<br />

2<br />

IP at ± , ± √ 1<br />

3 2<br />

F. No extreme values<br />

2 − 3x 2<br />

x 3 (1 − x 2 ) > 0 3/2 ⇔ <br />

<br />

−1 1 and f 0 (x) < 0 when 0 < |x| < 1,sof is increasing on (−∞, −1) and (1, ∞),and<br />

decreasing on (−1, 0) and (0, 1) [hence decreasing on (−1, 1) since f is<br />

H.<br />

H.<br />

continuous on (−1, 1)].<br />

F. Local maximum value f(−1) = 2, local minimum<br />

value f(1) = −2 G. f 00 (x) = 2 3 x−5/3 < 0 when x 0<br />

when x>0,sof is CD on (−∞, 0) and CU on (0, ∞). IPat(0, 0)

F.<br />

TX.10<br />

SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 177<br />

29. y = f(x) = 3√ x 2 − 1 A. D = R B. y-intercept: f(0) = −1; x-intercepts: f(x) =0 ⇔ x 2 − 1=0 ⇔<br />

x = ±1 C. f(−x) =f(x), so the curve is symmetric about the y-axis. D. No asymptote<br />

E. f 0 (x) = 1 3 (x2 − 1) −2/3 (2x) =<br />

increasing on (0, ∞) and decreasing on (−∞, 0).<br />

G. f 00 (x)= 2 3 · (x2 − 1) 2/3 (1) − x · 2<br />

3 (x2 − 1) −1/3 (2x)<br />

[(x 2 − 1) 2/3 ] 2<br />

2x<br />

3 3 (x 2 − 1) 2 . f 0 (x) > 0 ⇔ x>0 and f 0 (x) < 0 ⇔ x 0 ⇔ −1 0 ⇔ x ∈ 2nπ − π , 2nπ + π<br />

2 2 for each integer n, andf 0 (x) < 0 ⇔ cos x0 and sin x 6=±1 ⇔ x ∈ <br />

2nπ, 2nπ + π 2 ∪ 2nπ +<br />

π<br />

, 2nπ + π for some integer n.<br />

2<br />

f 00 (x) > 0 ⇔ sin x

F.<br />

178 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

35. y = f(x) = 1 2<br />

x − sin x, 0 0 on (0,α) and (β,2π) [ f is CU].<br />

The inflection points occur when x = α, β.<br />

H.

F.<br />

TX.10<br />

SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 179<br />

41. y =1/(1 + e −x ) A. D = R B. No x-intercept; y-intercept = f(0) = 1 2<br />

. C. No symmetry<br />

D. lim 1/(1 +<br />

x→∞ e−x )= 1 =1and lim 1/(1 + 1+0 x→−∞ e−x )=0 since<br />

lim<br />

x→−∞ e−x = ∞], so f has horizontal asymptotes<br />

y =0and y =1. E. f 0 (x) =−(1 + e −x ) −2 (−e −x )=e −x /(1 + e −x ) 2 . This is positive for all x,sof is increasing on R.<br />

F. No extreme values G. f 00 (x) = (1 + e−x ) 2 (−e −x ) − e −x (2)(1 + e −x )(−e −x )<br />

= e−x (e −x − 1)<br />

(1 + e −x ) 4 (1 + e −x ) 3<br />

The second factor in the numerator is negative for x>0 and positive for x 0 ⇒ 1 > 1/x ⇒ x>1 and<br />

H.<br />

f 0 (x) < 0 ⇒ 0 0]<br />

x→−∞<br />

f(x) =1,soy =0and y =1are HA; no VA<br />

E. f 0 (x) =−2(1 + e x ) −3 e x = −2ex < 0,sof is decreasing on R F. No local extrema<br />

(1 + e x )<br />

3<br />

G. f 00 (x)=(1+e x ) −3 (−2e x )+(−2e x )(−3)(1 + e x ) −4 e x<br />

H.<br />

= −2e x (1 + e x ) −4 [(1 + e x ) − 3e x ]= −2ex (1 − 2e x )<br />

(1 + e x ) 4 .<br />

f 00 (x) > 0 ⇔ 1 − 2e x < 0 ⇔ e x > 1 2<br />

⇔ x>ln 1 2 and<br />

f 00 (x) < 0 ⇔ x0} = ∞ <br />

n=−∞<br />

(2nπ, (2n +1)π) =···∪ (−4π, −3π) ∪ (−2π, −π) ∪ (0,π) ∪ (2π, 3π) ∪ ···<br />

B. No y-intercept; x-intercepts: f(x) =0 ⇔ ln(sin x) =0 ⇔ sin x = e 0 =1 ⇔ x =2nπ + π for each<br />

2<br />

integer n. C. f is periodic with period 2π. D. lim f(x) =−∞ and lim<br />

x→(2nπ) +<br />

x = nπ are VAs for all integers n.<br />

x→[(2n+1)π]<br />

−<br />

f(x) =−∞, so the lines<br />

E. f 0 (x) = cos x<br />

sin x =cotx,sof 0 (x) > 0 when 2nπ < x < 2nπ + π for each<br />

2<br />

integer n,andf 0 (x) < 0 when 2nπ + π 2

F.<br />

180 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

decreasing on 2nπ + π 2 , (2n +1)π for each integer n.<br />

TX.10<br />

H.<br />

F. Local maximum values f 2nπ + π 2<br />

<br />

=0, no local minimum.<br />

G. f 00 (x) =− csc 2 x 0 ⇔ x 2 < 1 2<br />

⇔ |x| < 1 √<br />

2<br />

,sof is increasing on<br />

<br />

and decreasing on −∞, − √ 1<br />

2<br />

and<br />

<br />

<br />

√<br />

1<br />

1<br />

2<br />

, ∞ . F. Local maximum value f √<br />

2<br />

=1/ √ 2e, local minimum<br />

<br />

value f − √ 1<br />

2<br />

= −1/ √ 2e G. f 00 (x) =−2xe −x2 (1 − 2x 2 ) − 4xe −x2 =2xe −x2 (2x 2 − 3) > 0 ⇔<br />

<br />

<br />

3<br />

x> or − 3<br />

2e −2x [multiply by e 2x ] ⇔<br />

H.<br />

e 5x > 2 3<br />

⇔ 5x >ln 2 3<br />

⇔ x> 1 5 ln 2 3 ≈−0.081. Similarly, f 0 (x) < 0 ⇔<br />

x< 1 5 ln 2 3 . f is decreasing on −∞, 1 5 ln 2 3<br />

<br />

and increasing on<br />

1<br />

5 ln 2 3 , ∞ .<br />

F. Local minimum value f 1<br />

ln 2<br />

5 3 = 2<br />

3/5<br />

+ <br />

2 −2/5<br />

≈ 1.96; no local maximum.<br />

3<br />

3<br />

G. f 00 (x) =9e 3x +4e −2x ,so f 00 (x) > 0 for all x,andf is CU on (−∞, ∞). NoIP<br />

53. m = f(v) =<br />

m 0<br />

<br />

1 − v2 /c .Them-intercept is f(0) = m 0.Therearenov-intercepts. lim f(v) =∞,sov = c is a VA.<br />

2 v→c− f 0 (v) =− 1 2 m0(1 − v2 /c 2 ) −3/2 (−2v/c 2 )=<br />

increasing on (0,c). There are no local extreme values.<br />

m 0 v<br />

c 2 (1 − v 2 /c 2 ) 3/2 =<br />

f 00 (v)= (c2 − v 2 ) 3/2 (m 0 c) − m 0 cv · 3<br />

2 (c2 − v 2 ) 1/2 (−2v)<br />

[(c 2 − v 2 ) 3/2 ] 2<br />

= m 0c(c 2 − v 2 ) 1/2 [(c 2 − v 2 )+3v 2 ]<br />

(c 2 − v 2 ) 3 = m 0c(c 2 +2v 2 )<br />

(c 2 − v 2 ) 5/2 > 0,<br />

so f is CU on (0,c). Therearenoinflection points.<br />

m 0 v<br />

c 2 (c 2 − v 2 ) 3/2<br />

c 3 =<br />

m 0 cv<br />

> 0,sof is<br />

(c 2 − v 2 )<br />

3/2

F.<br />

TX.10<br />

SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 181<br />

55. y = − W<br />

24EI x4 + WL<br />

12EI x3 − WL2<br />

24EI x2 = − W<br />

24EI x2 x 2 − 2Lx + L 2<br />

= −W<br />

24EI x2 (x − L) 2 = cx 2 (x − L) 2<br />

where c = − W isanegativeconstantand0 ≤ x ≤ L. Wesketch<br />

24EI<br />

f(x) =cx 2 (x − L) 2 for c = −1. f(0) = f(L) =0.<br />

f 0 (x) =cx 2 [2(x − L)] + (x − L) 2 (2cx) =2cx(x − L)[x +(x − L)] = 2cx(x − L)(2x − L). Sofor0

F.<br />

182 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION<br />

TX.10<br />

C. No symmetry D. lim f(x) =−∞ and lim<br />

x→(1/2) −<br />

lim<br />

x→±∞<br />

[f(x) − (−x +2)]= lim<br />

E. f 0 (x) =−1 −<br />

f(x) =∞,sox = 1 x→(1/2) + 2<br />

is a VA.<br />

1<br />

=0,sotheliney = −x +2is a SA.<br />

x→±∞ 2x − 1<br />

2<br />

(2x − 1) < 0 for x 6= 1 ,sof is decreasing on <br />

−∞, 1 2 2 2<br />

and 1<br />

2 , ∞ . F. No extreme values G. f 0 (x) =−1 − 2(2x − 1) −2 ⇒<br />

H.<br />

f 00 (x) =−2(−2)(2x − 1) −3 8<br />

(2) =<br />

(2x − 1) 3 ,sof 00 (x) > 0 when x> 1 and 2<br />

f 00 (x) < 0 when x< 1 . Thus, f is CU on 1<br />

, ∞ and CD on <br />

−∞, 1 2 2 2 .NoIP<br />

63. y = f(x) =(x 2 +4)/x = x +4/x A. D = {x | x 6= 0} =(−∞, 0) ∪ (0, ∞) B. No intercept<br />

C. f(−x) =−f(x) ⇒ symmetry about the origin D. lim (x +4/x) =∞ but f(x) − x =4/x → 0 as x → ±∞,<br />

x→∞<br />

so y = x is a slant asymptote.<br />

lim (x +4/x) =∞ and<br />

x→0 +<br />

lim<br />

x→0 − (x +4/x) =−∞,sox =0 is a VA. E. f 0 (x) =1− 4/x 2 > 0 ⇔<br />

x 2 > 4 ⇔ x>2 or x 0 ⇔ x>0 so f is CU on<br />

(0, ∞) and CD on (−∞, 0). NoIP<br />

65. y = f(x) = 2x3 + x 2 +1<br />

x 2 +1<br />

=2x +1+ −2x<br />

x 2 +1<br />

A. D = R B. y-intercept: f(0) = 1; x-intercept: f(x) =0 ⇒<br />

0=2x 3 + x 2 +1=(x + 1)(2x 2 − x +1) ⇒ x = −1 C. No symmetry D. No VA<br />

lim<br />

x→±∞<br />

[f(x) − (2x + 1)] = lim<br />

x→±∞<br />

−2x<br />

x 2 +1 =<br />

lim<br />

x→±∞<br />

−2/x<br />

=0, so the line y =2x +1is a slant asymptote.<br />

1+1/x2 E. f 0 (x) =2+ (x2 +1)(−2) − (−2x)(2x)<br />

= 2(x4 +2x 2 +1)− 2x 2 − 2+4x 2<br />

= 2x4 +6x 2<br />

(x 2 +1) 2 (x 2 +1) 2 (x 2 +1) = 2x2 (x 2 +3)<br />

2 (x 2 +1) 2<br />

so f 0 (x) > 0 if x 6= 0.Thus,f is increasing on (−∞, 0) and (0, ∞). Sincef is continuous at 0, f is increasing on R.<br />

F. No extreme values<br />

G. f 00 (x) = (x2 +1) 2 · (8x 3 +12x) − (2x 4 +6x 2 ) · 2(x 2 +1)(2x)<br />

[(x 2 +1) 2 ] 2<br />

= 4x(x2 +1)[(x 2 + 1)(2x 2 +3)− 2x 4 − 6x 2 ]<br />

= 4x(−x2 +3)<br />

(x 2 +1) 4 (x 2 +1) 3<br />

so f 00 (x) > 0 for x

F.<br />

TX.10<br />

SECTION 4.5 SUMMARY OF CURVE SKETCHING ¤ 183<br />

67. y = f(x) =x − tan −1 x, f 0 (x) =1− 1<br />

1+x 2 = 1+x2 − 1<br />

1+x 2 = x2<br />

1+x 2 ,<br />

f 00 (x) = (1 + x2 )(2x) − x 2 (2x)<br />

= 2x(1 + x2 − x 2 ) 2x<br />

=<br />

(1 + x 2 ) 2 (1 + x 2 ) 2 (1 + x 2 ) . 2<br />

<br />

lim f(x) − x −<br />

π<br />

x→∞<br />

2 = lim π −<br />

x→∞<br />

2 tan−1 x = π − π =0,soy = x − π is a SA.<br />

2 2 2<br />

Also,<br />

lim<br />

x→−∞<br />

<br />

f(x) −<br />

<br />

x +<