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

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

Expert Answer

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 $.

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

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

The answer is:

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

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 $.

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

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

The answer is:

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

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 $.

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

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

The answer is:

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

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