Answer :
To find the global maximum and minimum points of the function [tex]\( f(x) = 2x^3 - 3x^2 - 12x - 5 \)[/tex] on the interval [tex]\([-2, 4]\)[/tex], we need to follow these steps:
1. Find the critical points of the function on the given interval:
- First, determine the derivative of the function [tex]\( f'(x) \)[/tex].
- Solve the equation [tex]\( f'(x) = 0 \)[/tex] to find the critical points within the interval [tex]\([-2, 4]\)[/tex].
2. Evaluate the function at critical points and at the endpoints of the interval:
- Calculate the value of [tex]\( f(x) \)[/tex] at each critical point.
- Also, evaluate the function at the interval's endpoints [tex]\([-2, 4]\)[/tex].
3. Determine the global maximum and minimum values:
- Compare the values of [tex]\( f(x) \)[/tex] at the critical points and the endpoints to find the global maximum and minimum.
### Step-by-Step Solution
1. Find the derivative of the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 2x^3 - 3x^2 - 12x - 5 \][/tex]
The derivative [tex]\( f'(x) \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x - 5) = 6x^2 - 6x - 12 \][/tex]
2. Solve for critical points by setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 6x^2 - 6x - 12 = 0 \][/tex]
Simplify the equation:
[tex]\[ x^2 - x - 2 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x - 2)(x + 1) = 0 \][/tex]
So, the critical points are:
[tex]\[ x = 2 \quad \text{and} \quad x = -1 \][/tex]
3. Evaluate [tex]\( f(x) \)[/tex] at the critical points and the endpoints:
- At [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 2(-2)^3 - 3(-2)^2 - 12(-2) - 5 = -16 - 12 + 24 - 5 = -9 \][/tex]
- At [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 2(4)^3 - 3(4)^2 - 12(4) - 5 = 128 - 48 - 48 - 5 = 27 \][/tex]
- At [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2(2)^3 - 3(2)^2 - 12(2) - 5 = 16 - 12 - 24 - 5 = -25 \][/tex]
- At [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) - 5 = -2 - 3 + 12 - 5 = 2 \][/tex]
4. Determine the global maximum and minimum values:
- The evaluated values are:
[tex]\[ \begin{aligned} f(-2) & = -9, \\ f(4) & = 27, \\ f(2) & = -25, \\ f(-1) & = 2. \end{aligned} \][/tex]
- The highest value is [tex]\( 27 \)[/tex] at [tex]\( x = 4 \)[/tex], which is the global maximum.
- The lowest value is [tex]\( -25 \)[/tex] at [tex]\( x = 2 \)[/tex], which is the global minimum.
### Conclusion
- Global Maximum: [tex]\( f(4) = 27 \)[/tex] at [tex]\( x = 4 \)[/tex]
- Global Minimum: [tex]\( f(2) = -25 \)[/tex] at [tex]\( x = 2 \)[/tex]
1. Find the critical points of the function on the given interval:
- First, determine the derivative of the function [tex]\( f'(x) \)[/tex].
- Solve the equation [tex]\( f'(x) = 0 \)[/tex] to find the critical points within the interval [tex]\([-2, 4]\)[/tex].
2. Evaluate the function at critical points and at the endpoints of the interval:
- Calculate the value of [tex]\( f(x) \)[/tex] at each critical point.
- Also, evaluate the function at the interval's endpoints [tex]\([-2, 4]\)[/tex].
3. Determine the global maximum and minimum values:
- Compare the values of [tex]\( f(x) \)[/tex] at the critical points and the endpoints to find the global maximum and minimum.
### Step-by-Step Solution
1. Find the derivative of the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = 2x^3 - 3x^2 - 12x - 5 \][/tex]
The derivative [tex]\( f'(x) \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 - 12x - 5) = 6x^2 - 6x - 12 \][/tex]
2. Solve for critical points by setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 6x^2 - 6x - 12 = 0 \][/tex]
Simplify the equation:
[tex]\[ x^2 - x - 2 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x - 2)(x + 1) = 0 \][/tex]
So, the critical points are:
[tex]\[ x = 2 \quad \text{and} \quad x = -1 \][/tex]
3. Evaluate [tex]\( f(x) \)[/tex] at the critical points and the endpoints:
- At [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = 2(-2)^3 - 3(-2)^2 - 12(-2) - 5 = -16 - 12 + 24 - 5 = -9 \][/tex]
- At [tex]\( x = 4 \)[/tex]:
[tex]\[ f(4) = 2(4)^3 - 3(4)^2 - 12(4) - 5 = 128 - 48 - 48 - 5 = 27 \][/tex]
- At [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = 2(2)^3 - 3(2)^2 - 12(2) - 5 = 16 - 12 - 24 - 5 = -25 \][/tex]
- At [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = 2(-1)^3 - 3(-1)^2 - 12(-1) - 5 = -2 - 3 + 12 - 5 = 2 \][/tex]
4. Determine the global maximum and minimum values:
- The evaluated values are:
[tex]\[ \begin{aligned} f(-2) & = -9, \\ f(4) & = 27, \\ f(2) & = -25, \\ f(-1) & = 2. \end{aligned} \][/tex]
- The highest value is [tex]\( 27 \)[/tex] at [tex]\( x = 4 \)[/tex], which is the global maximum.
- The lowest value is [tex]\( -25 \)[/tex] at [tex]\( x = 2 \)[/tex], which is the global minimum.
### Conclusion
- Global Maximum: [tex]\( f(4) = 27 \)[/tex] at [tex]\( x = 4 \)[/tex]
- Global Minimum: [tex]\( f(2) = -25 \)[/tex] at [tex]\( x = 2 \)[/tex]
