Thomas Calculus 13th [Solutions]

320 Chapter 4 Applications of Derivatives
15. (a) g( t) 2
sin t 3t g ( t) 2sin t cost 3 sin(2 t ) 3 g 0 g( t ) is always falling and hence must
decrease on every interval in its domain.
2
(b) One, since sin t 3t 5
2
0 and sin t 3t 5 have the same solutions: f ( t) 2
sin t 3t 5 has the same
derivative as g( t ) in part (a) and is always decreasing with f ( 3) 0 and f (0) 0. The Intermediate Value
Theorem guarantees the continuous function f has a root in [ 3, 0].
16. (a)
(b)
17. (a)
(b)
18. (a)
(b)
dy 2
y tan sec 0 y tan is always rising on its domain y tan increases on every
d
interval in its domain
The interval
4 , is not in the tangents domain because tan is undefined at . Thus the tangent
2
need not increase on this interval.
4 2
3
f ( x) x 2x
2 f ( x) 4x 4 x . Since f (0) 2 0, f (1) 1 0 and f ( x) 0 for 0 x 1, we
may conclude from the Intermediate Value Theorem that f ( x ) has exactly one solution when 0 x 1.
2 2 4 8 2
x
0 x 3 1 and x 0 x .7320508076 .8555996772
2
y x
x 1
y 1 0, for all x in the domain of x y x is increasing in every interval in its
( x 1)
2
x 1 x 1
domain.
3
2
3
y x 2x
y 3x 2 0 for all x the graph of y x 2x is always increasing and can never
have a local maximum or minimum
19. Let V ( t ) represent the volume of the water in the reservoir at time t, in minutes, let V (0) a 0 be the initial amount
and V (1440) a 0 (1400)(43,560)(7.58) gallons be the amount of water contained in the reservoir after the rain,
where 24 hr 1440 min. Assume that V ( t ) is continuous on [0, 1440] and differentiable on (0, 1440). The Mean
V (1400) V (0) a0 (1440)(43,560)(7.48) a0
Value Theorem says that for some t 0 in (0, 1440) we have V ( t0)
1440 0 1440
456,160,320 gal
1440 min
225,000 gal/min.
316,778 gal/min. Therefore at t 0 the reservoirs volume was increasing at a rate in excess of
20. Yes, all differentiable functions g( x ) having 3 as a derivative differ by only a constant. Consequently, the
difference 3 x g( x ) is a constant K because g ( x) 3 d (3 x ). Thus g( x) 3 x K , the same form as F( x).
dx
21. No, x 1 1 x differs from 1 by the constant 1. Both functions have the same derivative
x 1 x 1 x 1
x 1
d x ( x 1) x(1)
1 d 1 .
dx x 1 2
2
( x 1) ( x 1) dx x 1
22. f ( x) g ( x ) 2 x f ( x) g( x)
C for some constant C the graphs differ by a vertical shift.
2 2
( x 1)
23. The global minimum value of 1 2
occurs at x 2.
24. (a) The function is increasing on the intervals [ 3, 2] and [1, 2].
(b) The function is decreasing on the intervals [ 2, 0) and (0, 1].
(c) The local maximum values occur only at x 2, and at x 2; local minimum values occur at x 3 and
at x 1 provided f is continuous at x 0.
25. (a) t 0, 6, 12
(b) t 3, 9 (c) 6 t 12
(d) 0 t 6, 12 t 14
26. (a) t 4
(b) at no time (c) 0 t 4
(d) 4 t 8
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