Answer :
Sure, let's go through the problem step by step to find the linearization of the given function at a certain point.
Given the function:
[tex]\[ f(x) = \sqrt{\cos\left(8x + \frac{\pi}{2}\right) + 4} \][/tex]
To find the linearization [tex]\( L(x) \)[/tex] at [tex]\( x_0 = 0 \)[/tex]:
### Step 1: Evaluate [tex]\( f(x_0) \)[/tex]
First, we need to compute the value of the function at [tex]\( x_0 = 0 \)[/tex]:
[tex]\[ f(0) = \sqrt{\cos\left(8 \cdot 0 + \frac{\pi}{2}\right) + 4} \][/tex]
[tex]\[ f(0) = \sqrt{\cos\left(\frac{\pi}{2}\right) + 4} \][/tex]
Considering [tex]\( \cos\left(\frac{\pi}{2}\right) = 0 \)[/tex]:
[tex]\[ f(0) = \sqrt{0 + 4} = \sqrt{4} = 2 \][/tex]
### Step 2: Compute the derivative [tex]\( f'(x) \)[/tex]
We need to find the derivative of the function. The general form of the derivative for [tex]\( f(x) = \sqrt{g(x)} \)[/tex] is:
[tex]\[ f'(x) = \frac{g'(x)}{2 \sqrt{g(x)}} \][/tex]
In our case, [tex]\( g(x) = \cos\left(8x + \frac{\pi}{2}\right) + 4 \)[/tex], thus:
[tex]\[ g'(x) = -8 \sin\left(8x + \frac{\pi}{2}\right) \][/tex]
So the derivative [tex]\( f'(x) \)[/tex] is:
[tex]\[ f'(x) = \frac{-8 \sin\left(8x + \frac{\pi}{2}\right)}{2 \sqrt{\cos\left(8x + \frac{\pi}{2}\right) + 4}} \][/tex]
Evaluate the derivative at [tex]\( x_0 = 0 \)[/tex]:
[tex]\[ f'(0) = \frac{-8 \sin\left(8 \cdot 0 + \frac{\pi}{2}\right)}{2 \sqrt{\cos\left(8 \cdot 0 + \frac{\pi}{2}\right) + 4}} \][/tex]
[tex]\[ f'(0) = \frac{-8 \sin\left(\frac{\pi}{2}\right)}{2 \sqrt{\cos\left(\frac{\pi}{2}\right) + 4}} \][/tex]
Considering [tex]\( \sin\left(\frac{\pi}{2}\right) = 1 \)[/tex]:
[tex]\[ f'(0) = \frac{-8 \cdot 1}{2 \cdot 2} = \frac{-8}{4} = -2 \][/tex]
### Step 3: Formulate the linearization [tex]\( L(x) \)[/tex]
Linearization of a function at [tex]\( x_0 \)[/tex] is given by:
[tex]\[ L(x) = f(x_0) + f'(x_0) \cdot (x - x_0) \][/tex]
Using [tex]\( x_0 = 0 \)[/tex], [tex]\( f(0) = 2 \)[/tex], and [tex]\( f'(0) = -2 \)[/tex]:
[tex]\[ L(x) = 2 + (-2)(x - 0) \][/tex]
[tex]\[ L(x) = 2 - 2x \][/tex]
Thus, the linearization [tex]\( L(x) \)[/tex] of [tex]\( f \)[/tex] at [tex]\( x_0 = 0 \)[/tex] is:
[tex]\[ L(x) = 2 - 2x \][/tex]
### Summary of Values
- [tex]\( f(0) = 2 \)[/tex]
- [tex]\( f'(0) = -2 \)[/tex]
- Linearization [tex]\( L(x) = 2 - 2x \)[/tex]
Given the function:
[tex]\[ f(x) = \sqrt{\cos\left(8x + \frac{\pi}{2}\right) + 4} \][/tex]
To find the linearization [tex]\( L(x) \)[/tex] at [tex]\( x_0 = 0 \)[/tex]:
### Step 1: Evaluate [tex]\( f(x_0) \)[/tex]
First, we need to compute the value of the function at [tex]\( x_0 = 0 \)[/tex]:
[tex]\[ f(0) = \sqrt{\cos\left(8 \cdot 0 + \frac{\pi}{2}\right) + 4} \][/tex]
[tex]\[ f(0) = \sqrt{\cos\left(\frac{\pi}{2}\right) + 4} \][/tex]
Considering [tex]\( \cos\left(\frac{\pi}{2}\right) = 0 \)[/tex]:
[tex]\[ f(0) = \sqrt{0 + 4} = \sqrt{4} = 2 \][/tex]
### Step 2: Compute the derivative [tex]\( f'(x) \)[/tex]
We need to find the derivative of the function. The general form of the derivative for [tex]\( f(x) = \sqrt{g(x)} \)[/tex] is:
[tex]\[ f'(x) = \frac{g'(x)}{2 \sqrt{g(x)}} \][/tex]
In our case, [tex]\( g(x) = \cos\left(8x + \frac{\pi}{2}\right) + 4 \)[/tex], thus:
[tex]\[ g'(x) = -8 \sin\left(8x + \frac{\pi}{2}\right) \][/tex]
So the derivative [tex]\( f'(x) \)[/tex] is:
[tex]\[ f'(x) = \frac{-8 \sin\left(8x + \frac{\pi}{2}\right)}{2 \sqrt{\cos\left(8x + \frac{\pi}{2}\right) + 4}} \][/tex]
Evaluate the derivative at [tex]\( x_0 = 0 \)[/tex]:
[tex]\[ f'(0) = \frac{-8 \sin\left(8 \cdot 0 + \frac{\pi}{2}\right)}{2 \sqrt{\cos\left(8 \cdot 0 + \frac{\pi}{2}\right) + 4}} \][/tex]
[tex]\[ f'(0) = \frac{-8 \sin\left(\frac{\pi}{2}\right)}{2 \sqrt{\cos\left(\frac{\pi}{2}\right) + 4}} \][/tex]
Considering [tex]\( \sin\left(\frac{\pi}{2}\right) = 1 \)[/tex]:
[tex]\[ f'(0) = \frac{-8 \cdot 1}{2 \cdot 2} = \frac{-8}{4} = -2 \][/tex]
### Step 3: Formulate the linearization [tex]\( L(x) \)[/tex]
Linearization of a function at [tex]\( x_0 \)[/tex] is given by:
[tex]\[ L(x) = f(x_0) + f'(x_0) \cdot (x - x_0) \][/tex]
Using [tex]\( x_0 = 0 \)[/tex], [tex]\( f(0) = 2 \)[/tex], and [tex]\( f'(0) = -2 \)[/tex]:
[tex]\[ L(x) = 2 + (-2)(x - 0) \][/tex]
[tex]\[ L(x) = 2 - 2x \][/tex]
Thus, the linearization [tex]\( L(x) \)[/tex] of [tex]\( f \)[/tex] at [tex]\( x_0 = 0 \)[/tex] is:
[tex]\[ L(x) = 2 - 2x \][/tex]
### Summary of Values
- [tex]\( f(0) = 2 \)[/tex]
- [tex]\( f'(0) = -2 \)[/tex]
- Linearization [tex]\( L(x) = 2 - 2x \)[/tex]
