Solução_Calculo_Stewart_6e

F.
TX.10SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS ¤ 123
3.10 Linear Approximations and Differentials
1. f(x) =x 4 +3x 2 ⇒ f 0 (x) =4x 3 +6x,sof(−1) = 4 and f 0 (−1) = −10.
Thus, L(x) =f(−1) + f 0 (−1)(x − (−1)) = 4 + (−10)(x +1)=−10x − 6.
3. f(x) =cosx ⇒ f 0 (x) =− sin x, sof π
2 =0and f
0 π
Thus, L(x) =f π
2
+ f
0 π
2
5. f(x) = √ 1 − x ⇒ f 0 (x) =
x −
π
2
=0− 1
x −
π
2
2
= −1.
= −x +
π
2 .
−1
2 √ 1 − x ,sof(0) = 1 and f 0 (0) = − 1 2 .
Therefore,
√ 1 − x = f(x) ≈ f(0) + f 0 (0)(x − 0) = 1 + − 1 2
(x − 0) = 1 −
1
2 x.
So √ 0.9 = √ 1 − 0.1 ≈ 1 − 1 (0.1) = 0.95
2
and √ 0.99 = √ 1 − 0.01 ≈ 1 − 1 (0.01) = 0.995.
2
7. f(x) = 3√ 1 − x =(1− x) 1/3 ⇒ f 0 (x) =− 1 3 (1 − x)−2/3 ,sof(0) = 1
and f 0 (0) = − 1 .Thus,f(x) ≈ f(0) + f 0 3
(0)(x − 0) = 1 − 1 x.Weneed
3
3√ 1 − x − 0.1 < 1 −
1
x< 3√ 3
1 − x +0.1, which is true when
−1.204