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To find the absolute maximum value of the function [tex]\( f(x) = 6x \cos(x) \)[/tex] on the interval [tex]\( [0, \pi] \)[/tex] correct to six decimal places, we can follow these steps:
### Step 1: Identify Critical Points
To identify the critical points in the interval, we need to find where the first derivative of the function [tex]\( f(x) \)[/tex] equals zero.
#### First Derivative
The first derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx}(6x \cos(x)) = 6 \cos(x) - 6x \sin(x) \][/tex]
Set the first derivative to zero to find the critical points:
[tex]\[ 6 \cos(x) - 6x \sin(x) = 0 \][/tex]
[tex]\[ \cos(x) = x \sin(x) \][/tex]
To solve this equation, we often use numerical methods such as Newton's method.
### Step 2: Use Newton's Method
Newton's method iteratively finds roots of functions. Given an initial guess [tex]\( x_0 \)[/tex], the method uses the update formula:
[tex]\[ x_{n+1} = x_n - \frac{f'(x_n)}{f''(x_n)} \][/tex]
#### Second Derivative
We need the second derivative of [tex]\( f(x) \)[/tex] for Newton's method:
[tex]\[ f''(x) = \frac{d}{dx}(6 \cos(x) - 6x \sin(x)) = -6 \sin(x) - 6 \sin(x) - 6x \cos(x) = -12 \sin(x) - 6x \cos(x) \][/tex]
Choose an initial guess; let's start with [tex]\( x_0 = \frac{\pi}{2} \)[/tex].
Applying the Newton's method:
[tex]\[ x_{n+1} = x_n - \frac{6 \cos(x_n) - 6x_n \sin(x_n)}{-12 \sin(x_n) - 6x_n \cos(x_n)} \][/tex]
After iterating, one critical point found in the interval [tex]\( [0, \pi] \)[/tex] is approximately:
[tex]\[ x \approx 0.860334 \][/tex]
### Step 3: Determine the Nature of the Critical Point
To determine if this critical point is a maximum, evaluate the second derivative at this point:
[tex]\[ f''(0.860334) = -12 \sin(0.860334) - 6 \cdot 0.860334 \cos(0.860334) \][/tex]
If [tex]\( f''(0.860334) < 0 \)[/tex], the point is a local maximum.
### Step 4: Evaluate the Function at Key Points
We need to evaluate [tex]\( f(x) \)[/tex] at the critical points and at the endpoints of the interval:
#### Critical Point
[tex]\[ f(0.860334) \approx 3.366578 \][/tex]
#### Endpoints
[tex]\[ f(0) = 6 \cdot 0 \cdot \cos(0) = 0 \][/tex]
[tex]\[ f(\pi) = 6 \cdot \pi \cdot \cos(\pi) = 6 \cdot \pi \cdot (-1) = -6\pi \approx -18.849556 \][/tex]
### Step 5: Determine the Absolute Maximum
Compare the function values at the critical point and the endpoints:
[tex]\[ f(0.860334) \approx 3.366578, \quad f(0) = 0, \quad f(\pi) \approx -18.849556 \][/tex]
The absolute maximum value of the function [tex]\( f(x) \)[/tex] on the interval [tex]\( [0, \pi] \)[/tex] is therefore:
[tex]\[ \boxed{3.366578} \][/tex]
### Step 1: Identify Critical Points
To identify the critical points in the interval, we need to find where the first derivative of the function [tex]\( f(x) \)[/tex] equals zero.
#### First Derivative
The first derivative of [tex]\( f(x) \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx}(6x \cos(x)) = 6 \cos(x) - 6x \sin(x) \][/tex]
Set the first derivative to zero to find the critical points:
[tex]\[ 6 \cos(x) - 6x \sin(x) = 0 \][/tex]
[tex]\[ \cos(x) = x \sin(x) \][/tex]
To solve this equation, we often use numerical methods such as Newton's method.
### Step 2: Use Newton's Method
Newton's method iteratively finds roots of functions. Given an initial guess [tex]\( x_0 \)[/tex], the method uses the update formula:
[tex]\[ x_{n+1} = x_n - \frac{f'(x_n)}{f''(x_n)} \][/tex]
#### Second Derivative
We need the second derivative of [tex]\( f(x) \)[/tex] for Newton's method:
[tex]\[ f''(x) = \frac{d}{dx}(6 \cos(x) - 6x \sin(x)) = -6 \sin(x) - 6 \sin(x) - 6x \cos(x) = -12 \sin(x) - 6x \cos(x) \][/tex]
Choose an initial guess; let's start with [tex]\( x_0 = \frac{\pi}{2} \)[/tex].
Applying the Newton's method:
[tex]\[ x_{n+1} = x_n - \frac{6 \cos(x_n) - 6x_n \sin(x_n)}{-12 \sin(x_n) - 6x_n \cos(x_n)} \][/tex]
After iterating, one critical point found in the interval [tex]\( [0, \pi] \)[/tex] is approximately:
[tex]\[ x \approx 0.860334 \][/tex]
### Step 3: Determine the Nature of the Critical Point
To determine if this critical point is a maximum, evaluate the second derivative at this point:
[tex]\[ f''(0.860334) = -12 \sin(0.860334) - 6 \cdot 0.860334 \cos(0.860334) \][/tex]
If [tex]\( f''(0.860334) < 0 \)[/tex], the point is a local maximum.
### Step 4: Evaluate the Function at Key Points
We need to evaluate [tex]\( f(x) \)[/tex] at the critical points and at the endpoints of the interval:
#### Critical Point
[tex]\[ f(0.860334) \approx 3.366578 \][/tex]
#### Endpoints
[tex]\[ f(0) = 6 \cdot 0 \cdot \cos(0) = 0 \][/tex]
[tex]\[ f(\pi) = 6 \cdot \pi \cdot \cos(\pi) = 6 \cdot \pi \cdot (-1) = -6\pi \approx -18.849556 \][/tex]
### Step 5: Determine the Absolute Maximum
Compare the function values at the critical point and the endpoints:
[tex]\[ f(0.860334) \approx 3.366578, \quad f(0) = 0, \quad f(\pi) \approx -18.849556 \][/tex]
The absolute maximum value of the function [tex]\( f(x) \)[/tex] on the interval [tex]\( [0, \pi] \)[/tex] is therefore:
[tex]\[ \boxed{3.366578} \][/tex]
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