# Use the linearization approximation (1+x)^k=1+kx to find an approximation for the function...

## Question:

Use the linearization approximation {eq}(1+x)^k=1+kx {/eq} to find an approximation for the function {eq}f(x)=\frac1{\sqrt{4+x}} {/eq} for values of x near zero.

## Linearization of a Function:

Let {eq}f {/eq} be a differentiable function at {eq}x=a {/eq}. Then the approximating function {eq}L\left ( x \right )= f\left ( a \right )+f{}'\left ( a \right )\left ( x-a \right ) {/eq} is the linearization of {eq}f {/eq} at {eq}x=a {/eq}. The point {eq}x=a {/eq} is called the center of the approximation.

{eq}1+kx {/eq} is an approximation to the function {eq}g\left ( x \right )=\left ( 1+x \right )^{k} {/eq} at {eq}x = 0 {/eq}.

Given function is

{eq}\begin{align*} f\left ( x \right )&=\frac{1}{\sqrt{4+x}} \\ &=\left ( 4+x \right )^{\frac{-1}{2}} \\ &=\left ( 4\left ( 1+\frac{x}{4} \right ) \right )^{\frac{-1}{2}} \\ &=\frac{1}{2}\left ( 1+\frac{x}{4} \right )^{\frac{-1}{2}} \end{align*} {/eq}

Therefore, the approximation for the function {eq}f\left ( x \right )=\frac{1}{\sqrt{4+x}} {/eq} is

{eq}\frac{1}{2}\left ( 1+\left (-\frac{1}{2} \right )\left ( \frac{x}{4} \right )\right )=\frac{1}{2}\left ( 1-\frac{x}{8} \right ) {/eq} 