Solução_Calculo_Stewart_6e

F.
90 ¤ CHAPTER 3 DIFFERENTIATION RULES
TX.10
57. The slope of y = x 2 − 5x +4is given by m = y 0 =2x − 5. The slope of x − 3y =5 ⇔ y = 1 3 x − 5 3 is 1 3 ,
so the desired normal line must have slope 1 , and hence, the tangent line to the parabola must have slope −3. This occurs if
3
2x − 5=−3 ⇒ 2x =2 ⇒ x =1.Whenx =1, y =1 2 − 5(1) + 4 = 0, and an equation of the normal line is
y − 0= 1 3 (x − 1) or y = 1 3 x − 1 3 .
59. Let a, a 2 be a point on the parabola at which the tangent line passes through the
point (0, −4). The tangent line has slope 2a and equation y − (−4) = 2a(x − 0)
⇔
y =2ax − 4. Since a, a 2 also lies on the line, a 2 =2a(a) − 4,ora 2 =4.So
a = ±2 and the points are (2, 4) and (−2, 4).
1
61. f 0 f(x + h) − f(x)
(x) = lim
= lim
x + h − 1 x
h→0 h
h→0 h
= lim
h→0
x − (x + h)
hx(x + h) =lim
h→0
−h
hx(x + h) =lim
h→0
63. Let P (x) =ax 2 + bx + c. ThenP 0 (x) =2ax + b and P 00 (x) =2a. P 00 (2) = 2 ⇒ 2a =2 ⇒ a =1.
P 0 (2) = 3 ⇒ 2(1)(2) + b =3 ⇒ 4+b =3 ⇒ b = −1.
P (2) = 5 ⇒ 1(2) 2 +(−1)(2) + c =5 ⇒ 2+c =5 ⇒ c =3.SoP (x) =x 2 − x +3.
−1
x(x + h) = − 1 x 2
65. y = f(x) =ax 3 + bx 2 + cx + d ⇒ f 0 (x) =3ax 2 +2bx + c. Thepoint(−2, 6) is on f, sof(−2) = 6 ⇒
−8a +4b − 2c + d =6 (1). The point (2, 0) is on f, sof(2) = 0 ⇒ 8a +4b +2c + d =0 (2). Since there are
horizontal tangents at (−2, 6) and (2, 0), f 0 (±2) = 0. f 0 (−2) = 0 ⇒ 12a − 4b + c =0 (3) and f 0 (2) = 0 ⇒
12a +4b + c =0 (4). Subtracting equation (3) from (4) gives 8b =0 ⇒ b =0. Adding (1) and (2) gives 8b +2d =6,
so d =3since b =0.From(3) we have c = −12a,so(2) becomes 8a +4(0)+2(−12a)+3=0 ⇒ 3=16a ⇒
a = 3
3
16 16 = −
9
3
and the desired cubic function is y =
4 16 x3 − 9 x +3. 4
67. f(x) =2− x if x ≤ 1 and f(x) =x 2 − 2x +2if x>1. Now we compute the right- and left-hand derivatives defined in
Exercise 2.8.54:
f 0 −(1) =
f(1 + h) − f(1) 2 − (1 + h) − 1 −h
lim
= lim
= lim
h→0 − h
h→0 − h
h→0 − h = lim −1 =−1 and
h→0− f+(1) 0 f(1 + h) − f(1) (1 + h) 2 − 2(1 + h)+2− 1 h 2
= lim
= lim
= lim
h→0 + h
h→0 + h
h→0 + h = lim h =0.
h→0 +
Thus, f 0 (1) does not exist since f−(1) 0 6=f+(1),sof
0
is not differentiable at 1. Butf 0 (x) =−1 for x1.