Solução_Calculo_Stewart_6e

F.
TX.10
SECTION 4.1 MAXIMUM AND MINIMUM VALUES ¤ 151
63. f(x) =x a (1 − x) b , 0 ≤ x ≤ 1, a>0, b>0.
f 0 (x) =x a · b(1 − x) b−1 (−1) + (1 − x) b · ax a−1 = x a−1 (1 − x) b−1 [x · b(−1) + (1 − x) · a]
= x a−1 (1 − x) b−1 (a − ax − bx)
At the endpoints, we have f(0) = f(1) = 0 [the minimum value of f ]. In the interval (0, 1), f 0 (x) =0 ⇔ x = a
a + b .
f
a
a + b
=
a
So f
a + b
a
a + b
a
1 − a b
=
a + b
a a
b a + b − a
=
(a + b) a a + b
a a b b
=
is the absolute maximum value.
a+b
(a + b)
a a
(a + b) a ·
b b
(a + b) = a a b b
b (a + b) . a+b
65. (a) From the graph, it appears that the absolute maximum value is about
f(−0.77) = 2.19, and the absolute minimum value is about
f(0.77) = 1.81.
(b) f(x) =x 5 − x 3 +2 ⇒ f 0 (x) =5x 4 − 3x 2 = x 2 (5x 2 − 3). Sof 0 3
(x) =0 ⇒ x =0, ± . 5
5 3
3
3
f − = −
5
5 −
3
−
5 +2=− 3
2 3
+ 3 3
+2= 3
−
9 3
+2= 6 3
+2(maximum)
5 5 5 5 5 25 5 25 5
and similarly, f
3
5
= − 6 25
3
5 +2(minimum).
67. (a) From the graph, it appears that the absolute maximum value is about
f(0.75) = 0.32, and the absolute minimum value is f(0) = f(1) = 0;
that is, at both endpoints.
(b) f(x) =x √ x − x 2 ⇒ f 0 1 − 2x
(x) =x ·
2 √ x − x + √ x − x 2 = (x − 2x2 )+(2x − 2x 2 )
2 2 √ 3x − 4x2
=
x − x 2 2 √ x − x . 2
So f 0 (x) =0 ⇒ 3x − 4x 2 =0 ⇒ x(3 − 4x) =0 ⇒ x =0or 3 . 4
f(0) = f(1) = 0 (minimum), and f
3
4 =
3
4
3
4 − 3
4
2
= 3 4
3
= 3 √ 3
(maximum).
16 16
69. The density is defined as ρ = mass
volume = 1000
V (T ) (in g/cm3 ). But a critical point of ρ will also be a critical point of V
[since dρ
dT = −1000V −2 dV
dT
and V is never 0],andV is easier to differentiate than ρ.
V (T )=999.87 − 0.06426T +0.0085043T 2 − 0.0000679T 3 ⇒ V 0 (T )=−0.06426 + 0.0170086T − 0.0002037T 2 .
Setting this equal to 0 and using the quadratic formula to find T ,weget
T = −0.0170086 ± √ 0.0170086 2 − 4 · 0.0002037 · 0.06426
2(−0.0002037)
≈ 3.9665 ◦ Cor79.5318 ◦ C. Since we are only interested