# Given f(x) = 2x^2 + 3x + 2, look the linearization of f at a = 2. a. Compare l(x) and f(x) for x...

## Question:

Given {eq}f(x) = 2x^2 + 3x + 2 {/eq}, look the linearization of f at a = 2.

a. Compare l(x) and f(x) for x = 2.02.

b. Compare y and dy when x changes from 2 to 2.01.

## Function Linearization:

We consider a function of one real variable {eq}f(x). {/eq}

The linearization of the function at the point {eq}x_0 {/eq} is found by means of the following formula:

{eq}L(x) = f(x_0) + f'(x)(x-x_0). {/eq}

## Answer and Explanation:

We are given the function

{eq}f(x) = 2x^2 + 3x + 2. {/eq}

having its derivative equal to

{eq}f'(x) = 4x + 3. {/eq}

The linearization of the function at {eq}x=2 {/eq} is calculated as

{eq}f(2) = 16 \\ f'(2) = 11 \\ L(x) = f(2) + f'(2)(x-2) \\ = 16 + 11(x-2) \\ = -6 + 11x. {/eq}

a. The comparison between f(x) and L(x) when {eq}x=2.02 {/eq} provides

{eq}f(2.02) = 2(2.02)^2 + 3(2.02) + 2 = 16.2208 \\ L(2.02) = 16.22. {/eq}

b. The comparison between the finite increment {eq}\Delta y {/eq} and {eq}dy {/eq} when x changes from 2 to 2.01 is reported as follows

{eq}dx = 2.01-2=0.01 \\ \Delta y = f(2.02) - f(2) = 16.2208-16 = 0.2208 \\ dy = f'(2)dx = 11(0.01) =0.11. {/eq}

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from Math 104: Calculus

Chapter 10 / Lesson 2