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