Solução_Calculo_Stewart_6e

F.
164 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
61. (a) The rate of increase of the population is initially very small, then gets larger until it reaches a maximum at about
t =8hours, and decreases toward 0 as the population begins to level off.
(b) The rate of increase has its maximum value at t =8hours.
(c) The population function is concave upward on (0, 8) and concave downward on (8, 18).
(d) At t =8, the population is about 350,sotheinflection point is about (8, 350).
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).
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)
decreases and the graph of K is concave downward.
65. S(t) =At p e −kt with A =0.01, p =4,andk =0.07. Wewillfind the
zeros of f 00 for f(t) =t p e −kt .
f 0 (t) =t p (−ke −kt )+e −kt (pt p−1 )=e −kt (−kt p + pt p−1 )
f 00 (t)=e −kt (−kpt p−1 + p(p − 1)t p−2 )+(−kt p + pt p−1 )(−ke −kt )
= t p−2 e −kt [−kpt + p(p − 1) + k 2 t 2 − kpt]
= t p−2 e −kt (k 2 t 2 − 2kpt + p 2 − p)
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
are t =0and the solutions of 0.0049t 2 − 0.56t +12=0, which are t 1 = 200
7
≈ 28.57 and t 2 = 600
7
≈ 85.71.
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
decrease is the greatest.
67. f(x) =ax 3 + bx 2 + cx + d ⇒ f 0 (x) =3ax 2 +2bx + c.
We are given that f(1) = 0 and f(−2) = 3,sof(1) = a + b + c + d =0and
f(−2) = −8a +4b − 2c + d =3.Alsof 0 (1) = 3a +2b + c =0and
f 0 (−2) = 12a − 4b + c =0by Fermat’s Theorem. Solving these four equations, we get
a = 2 , b = 1 , c = − 4 , d = 7 , so the function is f(x) = 1
9 3 3 9 9 2x 3 +3x 2 − 12x +7 .
69. y = 1+x ⇒ y 0 = (1 + x2 )(1) − (1 + x)(2x) 1 − 2x − x2
= ⇒
1+x 2 (1 + x 2 ) 2 (1 + x 2 ) 2
y 00 = (1 + x2 ) 2 (−2 − 2x) − (1 − 2x − x 2 ) · 2(1 + x 2 )(2x)
= 2(1 + x2 )[(1 + x 2 )(−1 − x) − (1 − 2x − x 2 )(2x)]
[(1 + x 2 ) 2 ] 2 (1 + x 2 ) 4
= 2(−1 − x − x2 − x 3 − 2x +4x 2 +2x 3 )
= 2(x3 +3x 2 − 3x − 1)
= 2(x − 1)(x2 +4x +1)
(1 + x 2 ) 3 (1 + x 2 ) 3 (1 + x 2 ) 3
So y 00 =0 ⇒ x =1, −2 ± √ 3.Leta = −2 − √ 3, b = −2+ √ 3,andc =1. We can show that f(a) = 1 4
1 −
√
3
,