# 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}

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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