Solução_Calculo_Stewart_6e

F.
152 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
in the region 0 ◦ C ≤ T ≤ 30 ◦ C, we check the density ρ at the endpoints and at 3.9665 ◦ C: ρ(0) ≈ 1000
999.87 ≈ 1.00013;
ρ(30) ≈ 1000
1000
≈ 0.99625; ρ(3.9665) ≈ ≈ 1.000255. So water has its maximum density at
1003.7628 999.7447
about 3.9665 ◦ C.
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.
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.
We now apply the Closed Interval Method to the continuous function S on the interval 0 ≤ t ≤ 10. SinceS 0 exists for all t,
the only critical numbers of S occur when S 0 (t) =0.Weusearootfinder on a CAS (or a graphing device) to find that
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
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
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
(corresponding roughly to March 1998) and cheapest at t 1 ≈ 0.855 (June 1994).
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 ⇒
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
r
2 0 =
1
8 kr3 0, v 2
r
3 0 =
4
27 kr3 0,
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.
(b) From part (a), the maximum value of v is 4 27 kr3 0.
(c)
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)
has no critical number, so f(x) can have no local maximum or minimum.
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
proved in the text. Thus, f 0 (c) =−g 0 (c) =0.
4.2 The Mean Value Theorem
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]
and differentiable on (1, 3). Alsof(1) = −4 =f(3). f 0 (c) =0 ⇔ −12 + 6c =0 ⇔ c =2, which is in the open
interval (1, 3),soc =2satisfies the conclusion of Rolle’s Theorem.