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To find the absolute minimum and maximum values of the function [tex]\( f(x) = x^4 - 32x^2 + 5 \)[/tex] on the interval [tex]\([-3, 9]\)[/tex], we need to follow these steps:
1. Find the derivative of the function:
To locate the critical points, we first find the derivative of the function [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^4 - 32x^2 + 5) = 4x^3 - 64x \][/tex]
2. Solve for critical points:
Set the derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 4x^3 - 64x = 0 \][/tex]
Factor out the common term [tex]\( 4x \)[/tex]:
[tex]\[ 4x(x^2 - 16) = 0 \][/tex]
This results in:
[tex]\[ 4x = 0 \quad \text{or} \quad x^2 - 16 = 0 \][/tex]
Solving these, we get:
[tex]\[ x = 0, \quad x^2 = 16 \implies x = 4 \text{ or } x = -4 \][/tex]
So, the critical points are [tex]\( x = 0, x = -4, \)[/tex] and [tex]\( x = 4 \)[/tex].
3. Evaluate the function at the endpoints of the interval:
The endpoints of the interval are [tex]\( x = -3 \)[/tex] and [tex]\( x = 9 \)[/tex]. Calculate the function values at these points:
[tex]\[ f(-3) = (-3)^4 - 32(-3)^2 + 5 = 81 - 288 + 5 = -202 \][/tex]
[tex]\[ f(9) = 9^4 - 32(9^2) + 5 = 6561 - 2592 + 5 = 3974 \][/tex]
4. Evaluate the function at the critical points within the interval:
The critical points within the interval [tex]\([-3, 9]\)[/tex] are [tex]\( x = 0 \)[/tex] and [tex]\( x = 4 \)[/tex] (since [tex]\( x = -4 \)[/tex] is outside the interval). Calculate the function values at these points:
[tex]\[ f(0) = 0^4 - 32(0)^2 + 5 = 5 \][/tex]
[tex]\[ f(4) = 4^4 - 32(4^2) + 5 = 256 - 512 + 5 = -251 \][/tex]
5. Identify the absolute minimum and maximum values:
We now compare all the function values obtained from the critical points and endpoints:
[tex]\[ f(-3) = -202, \quad f(9) = 3974, \quad f(0) = 5, \quad f(4) = -251 \][/tex]
- The absolute minimum value is the smallest of these values:
[tex]\[ \min\{-202, 3974, 5, -251\} = -251 \][/tex]
- The absolute maximum value is the largest of these values:
[tex]\[ \max\{-202, 3974, 5, -251\} = 3974 \][/tex]
Therefore, the absolute minimum value of the function on the interval [tex]\([-3, 9]\)[/tex] is [tex]\(-251\)[/tex] and the absolute maximum value is [tex]\(3974\)[/tex].
1. Find the derivative of the function:
To locate the critical points, we first find the derivative of the function [tex]\( f(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx}(x^4 - 32x^2 + 5) = 4x^3 - 64x \][/tex]
2. Solve for critical points:
Set the derivative equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[ 4x^3 - 64x = 0 \][/tex]
Factor out the common term [tex]\( 4x \)[/tex]:
[tex]\[ 4x(x^2 - 16) = 0 \][/tex]
This results in:
[tex]\[ 4x = 0 \quad \text{or} \quad x^2 - 16 = 0 \][/tex]
Solving these, we get:
[tex]\[ x = 0, \quad x^2 = 16 \implies x = 4 \text{ or } x = -4 \][/tex]
So, the critical points are [tex]\( x = 0, x = -4, \)[/tex] and [tex]\( x = 4 \)[/tex].
3. Evaluate the function at the endpoints of the interval:
The endpoints of the interval are [tex]\( x = -3 \)[/tex] and [tex]\( x = 9 \)[/tex]. Calculate the function values at these points:
[tex]\[ f(-3) = (-3)^4 - 32(-3)^2 + 5 = 81 - 288 + 5 = -202 \][/tex]
[tex]\[ f(9) = 9^4 - 32(9^2) + 5 = 6561 - 2592 + 5 = 3974 \][/tex]
4. Evaluate the function at the critical points within the interval:
The critical points within the interval [tex]\([-3, 9]\)[/tex] are [tex]\( x = 0 \)[/tex] and [tex]\( x = 4 \)[/tex] (since [tex]\( x = -4 \)[/tex] is outside the interval). Calculate the function values at these points:
[tex]\[ f(0) = 0^4 - 32(0)^2 + 5 = 5 \][/tex]
[tex]\[ f(4) = 4^4 - 32(4^2) + 5 = 256 - 512 + 5 = -251 \][/tex]
5. Identify the absolute minimum and maximum values:
We now compare all the function values obtained from the critical points and endpoints:
[tex]\[ f(-3) = -202, \quad f(9) = 3974, \quad f(0) = 5, \quad f(4) = -251 \][/tex]
- The absolute minimum value is the smallest of these values:
[tex]\[ \min\{-202, 3974, 5, -251\} = -251 \][/tex]
- The absolute maximum value is the largest of these values:
[tex]\[ \max\{-202, 3974, 5, -251\} = 3974 \][/tex]
Therefore, the absolute minimum value of the function on the interval [tex]\([-3, 9]\)[/tex] is [tex]\(-251\)[/tex] and the absolute maximum value is [tex]\(3974\)[/tex].
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