Solução_Calculo_Stewart_6e

F.
188 ¤ CHAPTER 4 APPLICATIONS OF DIFFERENTIATION
TX.10
on (−∞, −35.3), (−5.0, −1) and (−0.5, −0.1). We check back on the graphs of f to find the y-coordinates of the
inflection points, and find that these points are approximately (−35.3, −0.015), (−5.0, −0.005), (−1, 0), (−0.5, 0.00001),
and (−0.1, 0.0000066).
17. y = f(x) =
√
x
x 2 + x +1 . FromaCAS,y0 = − 3x2 + x − 1
2 √ x (x 2 + x +1) 2 and y00 = 15x4 +10x 3 − 15x 2 − 6x − 1
4x 3/2 (x 2 + x +1) 3 .
f 0 (x) =0 ⇔ x ≈ 0.43,sof is increasing on (0, 0.43) and decreasing on (0.43, ∞). There is a local maximum value of
f(0.43) ≈ 0.41. f 00 (x) =0 ⇔ x ≈ 0.94,sof is CD on (0, 0.94) and CU on (0.94, ∞). There is an inflection point at
(0.94, 0.34).
19. y = f(x) = √ x +5sinx, x ≤ 20.
From a CAS, y 0 =
5cosx +1
2 √ x +5sinx and y00 = − 10 cos x +25sin2 x +10x sin x +26
.
4(x +5sinx) 3/2
We’ll start with a graph of g(x) =x +5sinx. Notethatf(x) = g(x) is only defined if g(x) ≥ 0. g(x) =0 ⇔ x =0
or x ≈−4.91, −4.10, 4.10,and4.91. Thus, the domain of f is [−4.91, −4.10] ∪ [0, 4.10] ∪ [4.91, 20].
From the expression for y 0 , we see that y 0 =0 ⇔ 5cosx +1=0 ⇒ x 1 =cos −1
− 1 5 ≈ 1.77 and
x 2 =2π − x 1 ≈−4.51 (not in the domain of f ). The leftmost zero of f 0 is x 1 − 2π ≈−4.51. Moving to the right, the zeros
of f 0 are x 1 , x 1 +2π, x 2 +2π, x 1 +4π, andx 2 +4π. Thus, f is increasing on (−4.91, −4.51), decreasing on
(−4.51, −4.10),increasingon(0, 1.77), decreasing on (1.77, 4.10),increasingon(4.91, 8.06), decreasing on (8.06, 10.79),
increasing on (10.79, 14.34), decreasing on (14.34, 17.08), and increasing on (17.08, 20). The local maximum values are
f(−4.51) ≈ 0.62, f(1.77) ≈ 2.58, f(8.06) ≈ 3.60,andf(14.34) ≈ 4.39. The local minimum values are f(10.79) ≈ 2.43
and f(17.08) ≈ 3.49.
f is CD on (−4.91, −4.10), (0, 4.10), (4.91, 9.60),CUon(9.60, 12.25),CD
on (12.25, 15.81),CUon(15.81, 18.65), and CD on (18.65, 20). Thereare
inflection points at (9.60, 2.95), (12.25, 3.27), (15.81, 3.91),and(18.65, 4.20).