# Find the linearization L(x) of the function at a. 1. f(x) = x^3 - x^2 + 3, a = -2 2. f(x) = sin...

## Question:

Find the linearization {eq}\, L(x) \, {/eq} of the function at {eq}\, a {/eq}.

**1.** {eq}f(x) \, = \, x^{3} \, - \, x^{2} \, + \, 3, \quad a \, = \, -2
{/eq}

**2.** {eq}f(x) \, = \, \sin \, x, \quad a \, = \, \dfrac{\pi}{6}
{/eq}

## Linearization:

The linearization of {eq}f(x) {/eq} at {eq}x=a {/eq} is defined to be {eq}L(x)=f'(a)(x-a)+f(a) {/eq} provided that {eq}f'(a) {/eq} exists.

The linearization also called the linear approximation is really just the equation of the tangent line and we use it to approximate the values of the function near {eq}a {/eq}.

## Answer and Explanation: 1

1. Let {eq}f(x)=x^3-x^2+3 {/eq}. We then have {eq}f'(x)=3x^2-2x {/eq}. Evaluating at {eq}a=-2 {/eq} gives us

{eq}f(-2)=(-2)^3-(-2)^2+3=-9\\ f'(-2)=3(-2)^2-2(-2)=16 {/eq}

This gives a linearization of {eq}L(x)=f'(-2)(x-(-2))+f(-2)=16(x+2)-9 {/eq}

2. Let {eq}f(x)=\sin x {/eq}. We then have {eq}f'(x)=\cos x {/eq}

Evaluating at {eq}a=\frac{\pi}{6} {/eq} we have

{eq}f(\pi/6)=\sin \pi/6=\frac{1}{2}\\ f'(\pi/6)=\cos \pi/6=\frac{\sqrt{3}}{2} {/eq}

This gives a linearization of

{eq}L(x)=f'(\pi/6)(x-\frac{\pi}{6})+f(\pi/6)=\frac{\sqrt{3}}{2}(x-\frac{\pi}{6})+\frac{1}{2} {/eq}

#### Ask a question

Our experts can answer your tough homework and study questions.

Ask a question Ask a question#### Search Answers

#### Learn more about this topic:

from

Chapter 10 / Lesson 1Over the river and through the woods to Grandmother's house we go ... Are we there yet? In this lesson, apply linearization to estimate when we will finally get to Grandma's house!