# Find the linearization L(x) of the function at a.

– $f(x) \space = \space \sqrt ( x ) \space , \space a \space = \space 4$

The main objective of this question is to find the linearization of the given function.

This question uses the concept of linearization of a function. Determining the linear approximation of a function at a specific location is referred to as linearization.

Derivative of function

The very first level Taylor expansion around at the point of interest is the linear approximations of a function.

Taylor expansion

We have to find the linearization of the given function.

We are given:

$\space f(x) \space = \space \sqrt ( x ) \space , \space a \space = \space 4$

So:

$\space f(x) \space = \space \sqrt (x)$

By putting value, we get:

$\space f(4) \space = \space \sqrt (4)$

$\space = \space 2$

Now taking the derivative will result in:

$\space f”(x) \space = \space \frac{1}{2 \sqrt (4)}$

$\space = \space \frac{1}{4}$

Thus, $L(x)$ at the value of $4$.

$\space L(x) \space = \space f(a) \space + \space f'(a) (x \space – \space a )$

$\space L(x) \space = \space 2 \space + \space \frac{1}{4} (x \space – \space 4)$

$\space L(x) \space = \space 2 \space + \space \frac{1}{4} (x \space – \space 4)$

## Numerical Results

The linearization of the given function is:

$\space L(x) \space = \space 2 \space + \space \frac{1}{4} (x \space – \space 4)$

## Example

Find the linearization of the given  two functions.

• $\space f(x) \space = \space \sqrt ( x ) \space , \space a \space = \space 9$
• $\space f(x) \space = \space \sqrt ( x ) \space , \space a \space = \space 16$

We have to find the linearization of the given function.

We are given that:

$\space f(x) \space = \space \sqrt ( x ) \space , \space a \space = \space 9$

So:

$\space f(x) \space = \space \sqrt (x)$

By putting value, we get:

$\space f(4) \space = \space \sqrt (9)$

$\space = \space 3$

Now taking the derivative will result in:

$\space f”(x) \space = \space \frac{1}{2 \sqrt (9)}$

$\space = \space \frac{1}{6}$

Thus, $L(x)$ at the value of $9$.

$\space L(x) \space = \space f(a) \space + \space f'(a) (x \space – \space a )$

$\space L(x) \space = \space 3 \space + \space \frac{1}{6} (x \space – \space 9)$

$\space L(x) \space = \space 3 \space + \space \frac{1}{6} (x \space – \space 9)$

Now for the second expression. We have to find the linearization of the given function.

We are given that:

$\space f(x) \space = \space \sqrt ( x ) \space , \space a \space = \space 16$

So:

$\space f(x) \space = \space \sqrt (x)$

By putting value, we get:

$\space f(4) \space = \space \sqrt (16)$

$\space = \space 4$

Now taking the derivative will result in:

$\space f”(x) \space = \space \frac{1}{2 \sqrt (16)}$

$\space = \space \frac{1}{8}$

Thus, $L(x)$ at the value of $9$.

$\space L(x) \space = \space f(a) \space + \space f'(a) (x \space – \space a )$

$\space L(x) \space = \space 4 \space + \space \frac{1}{8} (x \space – \space 16)$

$\space L(x) \space = \space 4 \space + \space \frac{1}{8} (x \space – \space 16)$