   Chapter 3.10, Problem 6E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Find the linear approximation of the function g ( x ) = 1 + x 3 at a = 0 and use it to approximate the numbers 0.95 3 and 1.1 3 . Illustrate by graphing g and the tangent line.

To determine

To find: The linear approximation of the function f(x)=1x at a=0 and use it to approximate the numbers 0.953 and 1.13.

Explanation

Given:

The function is g(x)=1+x3 and the point a=0.

Result used:

The linear approximation of the function at x=a is g(x)g(a)+g(a)(xa).

Derivative rules: Chain rule

If y=f(u) and u=g(x) are both differentiable function, then dydx=dydududx.

Calculation:

Consider g(x)=1+x3.

Differentiate with respect to x,

g(x)=ddx(1+x3)=ddx((1+x)13)

Substitute u=1+x,

g(x)=ddx((u)13)

Apply the chain rule and simplify the terms,

g(x)=ddu((u)13)dudx=13u131dudx=13u23dudx

Substitute u=1+x in g(x),

g(x)=13(1+x)23ddx(1+x)=13(1+x)23[ddx(1)+ddx(x)]=13(1+x)23[0+1]=13(1+x)23

Thus, the derivative is g(x)=13(1+x)23.

Substitute x=0 in g(x)=13(1+x)23,

g(0)=13(1+0)23=13

Thus, the value is g(0)=13

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 