# Find the linearization at x = a. f ( x ) = 1 / x^ 3 , a = 7

## Question:

Find the linearization at x = a.

{eq}f(x)=\frac{1}{x^{3}},a=7 {/eq}

## Linearization:

In some cases, it may be easy to calculate *f(a)* of a function, but difficult or impossible to compute nearby values of *f*. In such a case, we can use the easily computed value *f(a)* as an approximation.

We use the tangent line at *(a, f(a))* as an approximation to the curve *y =f(x)* when *x* is near *a*.

An equation of this tangent line *y=L(x)* is: {eq}L(x) = f(a) + f'(a)(x-a) {/eq}

The linearization *L(x)* is given by the following formula:

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

## Answer and Explanation:

We have

{eq}f(x) =x^{-3} \\ a = 7 {/eq}

So

{eq}f(a) =a^{-3} = 7^{-3} = 1/7^3 \\ f'(x) =-3x^{-4} f'(a) =3/7^4 {/eq}

Hence the linearization...

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View this answerWe have

{eq}f(x) =x^{-3} \\ a = 7 {/eq}

So

{eq}f(a) =a^{-3} = 7^{-3} = 1/7^3 \\ f'(x) =-3x^{-4} f'(a) =3/7^4 {/eq}

Hence the linearization is:

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