# Find the linearization of f(x) = xe^{-x^2} at a = 0. a. L(x) = x b. L(x) = -x c. L(x) = 1 d. L(x)...

## Question:

Find the linearization of {eq}f(x) = xe^{-x^2} {/eq} at {eq}a = 0 {/eq}.

a. {eq}L(x) = x {/eq}

b. {eq}L(x) = -x {/eq}

c. {eq}L(x) = 1 {/eq}

d. {eq}L(x) = 0 {/eq}

## Linearization

A linearization of a function near a given point is the tangent line to the graph at the given point.

To linearly approximate a function {eq}\displaystyle f(x) {/eq} near {eq}\displaystyle a {/eq}

we will use the tangent line to the graph of {eq}\displaystyle f(x) {/eq} at the point {eq}\displaystyle a {/eq} , i.e. {eq}\displaystyle L(x)=f(a)+f'(a)(x-a). {/eq}

Linearizations of functions are used in approximate the value of the function in a point near a point of tangency,

which is easier to evaluate it on a tangent line than on a complex function.

## Answer and Explanation:

To find the linearization of the function {eq}\displaystyle xe^{-x^2}, {/eq} at {eq}\displaystyle a=0, {/eq} we need to find the equation of the tangent line to the graph at that point.

For this, we need the derivative function, first.

{eq}\displaystyle f'(x)=\frac{d}{dx}\left(xe^{-x^2}\right) =e^{-x^2}-2x^2e^{-x^2} \implies f'(0)=e^0=1. {/eq}

So, the tangent line is now

{eq}\displaystyle y=f(0)+f'(0)x=x\implies \\ \displaystyle \text{ the linearization of } f(x) \text{ at } a=0 \text{ is }\boxed{ L(x)=x \text{ (- option (a)}}. {/eq}

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