# Given that f(3)= 5 and f '(x)= \frac{x}{x^3 + 3}, find the linear approximation of f(x) at x= 3.

## Question:

Given that f(3)= 5 and {eq}f '(x)= \frac{x}{x^3 + 3}, {/eq} find the linear approximation of f(x) at x= 3.

## Linearization of the Function:

We have to find the linear approximation of f(x) at x= 3. We have given the value of f(3) and with help of f'(x) we will find the value of f'(3), then plug the values into the formula of the linear approximation to get the desired results.

Plug the value of x=3 into f'(x) and we have \begin{align*} f'\left( x \right) &= \frac{x}{{{x^3} + 3}}\\ f'\left( 3 \right) &= \frac{3}{{{{\left( 3 \right)}^3} + 3}}\\ f'\left( 3 \right) &= \frac{3}{{30}}\\ f'\left( 3 \right) &= \frac{1}{{10}}. \end{align*}

The linear approximation of f(x) at a= 3 is \begin{align*} y &= f\left( a \right) + f'\left( a \right)\left( {x - a} \right)\\ y &= f\left( 3 \right) + f'\left( 3 \right)\left( {x - 3} \right)\\ y &= 5 + \frac{1}{{10}}\left( {x - 3} \right)\\ y &= \frac{x}{{10}} + 5 - \frac{3}{{10}}\\ y &= \frac{1}{{10}}\left( {x + 47} \right). \end{align*}