Westonci.ca offers fast, accurate answers to your questions. Join our community and get the insights you need now. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
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]
We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
