Answer key to review lab

REVIEW LAB ANSWER KEY1. Let fx 2x . x−1a. Find the average rate of change of fx between x 2 and x 5.To find average rate of change, we need the y −coords of points on the graph of fwith x coordinates 2 and 5.f5 5 2f2 4Av r.o.c f5−f2 − 1 5−2 2b. Fill in the five blanks: ___Your answer to (a)_ −1 is the slope of the _secant__2line between the points 2,_4_ and _5___,_ 5 ___. 2c. Find the instantaneous rate of change of fx at x 2.f ′ x 2 −2 x−1x.x−1 2f ′ 2 − 2d. Fill in the three blanks: ___Your answer to (c)_-2___ is the slope of the __tangent____ line at the point2,_4_REVIEW LAB, PAGE22. Given is the graph of a function fx.y54321-4 -3 -2 -1 0 1 2 3The graph of fx.a. Find the following limits and values (if they exist). No work needed.xf−3 f−2 f−1 f0 f1 f2DNE 0 4 2 1 31

lim fx DNE () limx→−3− fx 0x→−2 −lim fx 1x→−1 −lim fx 1x→1 −lim fxDNE () limx→−3 fx 0x→−2 lim fx 1x→−1 lim fx 3x→1 limx→−3fx DNE limx→−2 fx 0lim fx 1x→−1lim fx DNEx→1b. Assume that the domain of this function includes all x − values to the left and rightof the viewing window. With that information, determine:limx→fx 3 limx→− fx 0notice that on the left hand side of the picture, the graph is swooping down andapproaching the x-axis. That is why I determined the limit at - is 0, since thex-axis is the line y0.On the right hand side the function has become horizontal with height 3 so thelimit at is 3.c. Is the function f continuous at x −1? Justify your answer by explaining whetheror not f satisfies each of the conditions in the formal definition of continuity.The function has value 4 at -1 but the limit at -1 was 1. Since the value of thefunction is not equal to the limit, the third condition of the definition of continuity isnot met. (Visually we have a hole in the graph and a point elsewhere) so thefunction f is not continuous at x-1.d. Use the definition of the derivative to determine whether or not f isdifferentiable at x 0.Notice visually that the function has a "sharp corner" at 0. But to use thedefinition of the derivative I need to turn "sharp corner" into a mathematicalargument. It should go something like this:From the left, the slope of the tangent lines at 0 should be positive (heading up).fxh−fxhThis suggests the left hand limit of should be positive. But on the right of0, the function is heading sharply down, which means that the right hand limit offxh−fxwould be negative. Since these limits aren’t the same,hfxh−fxlim h→0 DNE so the derivative does not exist there. That means f is nothdifferentiable at x0.REVIEW LAB, PAGE 33. (Tan 3.1 #73)The body mass index (BMI) measures body weight in relation to heigh. ABMI of 25 to 29.9 is considered overweight, a BMI of 30 or more is considered obese, anda BMI of 40 or more is morbidly obese. The percent of the U.S. population that is obese isapproximated by the functionPt 0.0004t 3 0.0036t 2 0.8t 12 0 ≤ t ≤ 13where t is measured in years, with t 0 corresponding to the beginning of 1991.a. What percentage of the U.S. population was deemed obese at the beginning of1991?2

P0 0.00040 3 0.00360 2 0.80 12 12 percentb. What percentage of the U.S population was deemed obese at the beginning of2004?2004-1991t13P13 23. 887percentc. What was the average rate of change of the percent of the population deemedobese from 1991 to 2004?P13−P013−0 0.9144percent per yeard. How fast was the percent of the U.S. population that is deemed obese changingat the beginning of 1991?IusedSNtocalculateformethatP ′ t 0.0012t 2 0.0072t 0.8.P ′ 0 0.8 so the percent of obese people was changing by 0.8percent/year atthe beginning of 1991.e. How fast was the percent of the U.S. population that is that is deemed obesechanging at the beginning of 2004?2004 corresponds to t13. P ′ 13 1. 0964So the percent of obese people waschanging by 1.0694 percent/year at the beginning of 2004.f. Based on the derivative of the function Pt, will tne percentage of the populationdeemed obese be increasing or decreasing in 2008?2008 corresponds to t17. P ′ 17 1. 2692 0 so the percentage deemedobese will still be increasing because the derivative is positive.REVIEW LAB, PAGE 44. Let fx 2x 4 − x 2 − 4. UseSNtogiveacarefulsketchoffx for the following twowindows. Notice that the one on the left is zoomed in more.-1.0 -0.5 0.0 0.5 1.0y-3-4xy40302010-5-3 -2 -1 1 2 3x-10a. Use the limit definition of the derivative to find f ′ x by doing the following:fxh−fxi. Find the unsimplified difference quotient for fx.Make sure tohuse enough parentheses.fxh−fxh 2xh4 −xh 2 −4−2x 4 −x 2 −4hii. Use SN to simplify your difference quotient.2xh 4 −xh 2 −4−2x 4 −x 2 −4 − 1 h x 2 − 2h x 4 − x 2 2x 4 h h2h 3 8h 2 x 12hx 2 −h 8x 3 −2xiii. By hand (versus using SN),take the limit as h → 0. lim h→0 2h 3 8h 2 x 12hx 2 − h 8x 3 − 2x 20 3 80 2 x 120x 2 − 0 8x 3 − 2x 8x 3 − 2xlim h→0fxh−fxh3

So,f ′ x ________8x 3 − 2x_________b. For what x −value(s) is the tangent line to fx horizontal? Draw those line(s) inthe graph on the left.8x 3 − 2x 0, Solution is: 1 ,0,− 1 So draw the horizontal lines tangent to the2 2graph through the points with x coords .5, 0, -.5 See them on the picture?c. Find the equation of the tangent line to fx at x 2. Draw your tangent line inthe graph on the right.f ′ 2 60slope of tangent line. f2 24.0equation: y 60x − 2 24d. By using your answer to part (b) and looking at the graph of fx, over whatinterval(s) is f ′ x 0 and over what intervals is f ′ x 0?The derivative is positive when the function is increasing, so between - 1 and 0 21and to the right of . The derivative is negative when f is decreasing, left of − 1 ,2 20to 1 . 2In interval notation the correct answers are:−,− 1 0, 1 2 2 f′ 0.− 1 ,0 1 2 2 ,,f′ 0.REVIEW LAB, PAGE 55. Evaluate each limit algebraically and show your steps. If you use SN to simplify anexpression (and you are encouraged to do so) please indicate at which step. You mayuse SN to check your answer, but solely giving the answer is not enough work in general.a. lim −2x 2 − 4x 70x→1 3x 2 − 30x 75 4 3First try plugging in on top and bottom.−21 2 − 41 7031 2 − 301 75 I evaluate in SN and get −212 − 41 70 64 and31 2 − 301 75 48. since these are numbers and the bottom is nonzero, their64quotient is the limit, 4 48 3b. lim −2x 2 − 4x 70x→5 3x 2 − 30x 75 DNEFirst try plugging in on top and bottom.−25 2 − 45 70 035 2 − 305 75 0 this is a 0/0 situation so I need to have SN factor the topand bottom.SN factors the top: −2x 2 − 4x 70 − 2x 7x − 5SN factors the bottom: 3x 2 − 30x 75 3x − 5 2Now I look at my new quotient: −2x 2 − 4x 703x 2 − 30x 75 −2x7x−5 − 2x7 plug 5 into3x−5 23x−5top and bottom again.25 7 24 35 − 5 0 we have 24/0 so that means the expression isundefined and the limit DNE,c. lim −2x 2 − 4x 70x→−7 3x 2 − 30x 75 0First try plugging in on top and bottom.−2−7 2 − 4−7 70 03−7 2 − 30−7 75 4324

since these are numbers and the bottom is nonzero, their quotient is the limit,0432 0d. limx 4 10x 3 35x 2 50x 24 DNEx→−3 x 4 6x 3 5x 2 − 24x − 36First try plugging in on top and bottom. −3 4 10−3 3 35−3 2 50−3 24 0−3 4 6−3 3 5−3 2 − 24−3 − 36 0this is a 0/0 situation so I need to haveSN factor the top and bottom.x 4 10x 3 35x 2 50x 24 x 4x 3x 2x 1x 4 6x 3 5x 2 − 24x − 36 x − 2x 2x 3 2So now we divide and cancel like terms,x4x3x2x1x−2x2x3 2 x1x4x−2x3Note now if you plug in -3 to the top, you get (-2)( 1)-2 and in the bottom you get(-5)(0)0.Since this is in the form (nonzero number)/0, the limit will not be defined and thusyou write DNE (SN will evaluate it as "undefined").x 4 10x 3 35x 2 50x 24 1x→−2 x 4 6x 3 5x 2 2− 24x − 36e. limNotice this is the same function as in part d, but we’re taking the limit at adifferent point. So to start with I’m just going to use the simplified, factored formof x4 10x 3 35x 2 50x 24x1x4that we got above: . Notice when you plugx 4 6x 3 5x 2 − 24x − 36x−2x3in -2 everywhere, you getso this is the limit.f. lim x→−21−24−2−2−23 1 22x 3 2xx 4 − x 3 1 0This is a limit at infinity. So to do it remember how I taught you in class: divide out the topand bottom by the largest power of x present anywhere. In this case, we divide the top andbottom by x 4 since that’s the largest present. Note that SN won’t be able to help you withthis sort of dividing; you’ll have to use your own knowledge of exponents and division to doit.2x 3 2xx 4x 4 −x 3 1x 42x 3x 4 2xx 4x 4x 4 − x3x 4 1x 42x 2x 31x 4 − 1 x 1Now we know that:2 2 1x , , ,− 1 x 3 x 4 x allgoto0whenx → while 1 is still 1 at infinity. so we’ve got:00 0 is the limit at infinity.001REVIEW LAB, PAGE 6Below this line are problems that you should be able to do WITHOUT SN at all, and in anexam you should be prepared to solve problems like this without technology.6. You are given the following table of values for fx.x 4.9 4.99 4.999 4.9999 5 5.0001 5.001 5.01 5.1fx 47. 61 48. 8601 48. 9860 48. 9986 10 36. 001 36. 0120 36. 1201 37. 21a. Evaluate f5105

. Evaluate lim fxx→536c. Evaluate lim fxx→5−49d. Evaluate limfxx→5Does not exist, since left and right limits are different.e. Is fx continuous at x 5? Explain your answer by referring to the formal (limit)definition of continuity.No, since the limit doesn’t exist at 5 and therefore the limit and the function can’tbe equal there.7. Without using SN, find the derivative of each of the following (you do not need to simplifyyour answers):a. fx 3x 3 − 5x −1 x − 6f ′ x 33x 2 − 5−1x −2 1 0b. gx 4 x 2 x 1x −21notice the trick here! x 2x −2g ′ x 4 1 2 x 1 2 −1 2− 1 2 x− 1 2 −1 2x 2−1 c. hx 3 x 10 − 6x −5/3 16x 4we’ll do the last 4 of these in class....d. fx x 7x −5 3 1 x − 5xwe’ll do the last 4 of these in class....e. fx 7x2 − 7x −2 3x 4 − 10we’ll do the last 4 of these in class....f. fx 1 x − 73x 3x4 3x −6 − 7 5xwe’ll do the last 4 of these in class....6