Solução_Calculo_Stewart_6e

F.
SECTION TX.10 3.1 DERIVATIVES OF POLYNOMIALS AND EXPONENTIAL FUNCTIONS ¤ 89
43. (a) (b) From the graph in part (a), it appears that f 0 is zero at x 1 ≈−1.25, x 2 ≈ 0.5,
and x 3 ≈ 3. The slopes are negative (so f 0 is negative) on (−∞,x 1) and
(x 2,x 3). The slopes are positive (so f 0 is positive) on (x 1,x 2) and (x 3, ∞).
(c) f(x) =x 4 − 3x 3 − 6x 2 +7x +30
⇒
f 0 (x) =4x 3 − 9x 2 − 12x +7
45. f(x) =x 4 − 3x 3 +16x ⇒ f 0 (x) =4x 3 − 9x 2 +16 ⇒ f 00 (x) =12x 2 − 18x
47. f(x) =2x − 5x 3/4 ⇒ f 0 (x) =2− 15
4 x−1/4 ⇒ f 00 (x) = 15
16 x−5/4
Note that f 0 is negative when f is decreasing and positive when f is
increasing. f 00 is always positive since f 0 is always increasing.
49. (a) s = t 3 − 3t ⇒ v(t) =s 0 (t) =3t 2 − 3 ⇒ a(t) =v 0 (t) =6t
(b) a(2) = 6(2) = 12 m/s 2
(c) v(t) =3t 2 − 3=0when t 2 =1,thatis,t =1and a(1) = 6 m/s 2 .
51. The curve y =2x 3 +3x 2 − 12x +1has a horizontal tangent when y 0 =6x 2 +6x − 12 = 0 ⇔ 6(x 2 + x − 2) = 0 ⇔
6(x +2)(x − 1) = 0 ⇔ x = −2 or x =1. The points on the curve are (−2, 21) and (1, −6).
53. y =6x 3 +5x − 3 ⇒ m = y 0 =18x 2 +5,butx 2 ≥ 0 for all x,som ≥ 5 for all x.
55. The slope of the line 12x − y =1(or y =12x − 1)is12, so the slope of both lines tangent to the curve is 12.
y =1+x 3 ⇒ y 0 =3x 2 . Thus, 3x 2 =12 ⇒ x 2 =4 ⇒ x = ±2, which are the x-coordinates at which the tangent
lines have slope 12. The points on the curve are (2, 9) and (−2, −7), so the tangent line equations are y − 9=12(x − 2)
or y =12x − 15 and y + 7 = 12(x +2)or y =12x +17.