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To find the absolute minimum and maximum values of the function [tex]\( f(x) = x^4 - 50x^2 + 6 \)[/tex] on the interval [tex]\([-4, 11]\)[/tex], we must follow these steps:
1. Calculate the derivative of [tex]\( f(x) \)[/tex]:
The first derivative of [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx}(x^4 - 50x^2 + 6) = 4x^3 - 100x \][/tex]
2. Find the critical points by setting the first derivative to zero:
[tex]\[ 4x^3 - 100x = 0 \][/tex]
Factor out the common term:
[tex]\[ 4x(x^2 - 25) = 0 \][/tex]
This gives the solutions:
[tex]\[ x = 0, \quad x^2 - 25 = 0 \implies x = \pm 5 \][/tex]
So, the critical points are [tex]\( x = 0, x = 5, \text{ and } x = -5 \)[/tex].
3. Evaluate the function at the critical points and at the endpoints of the interval:
The interval given is [tex]\([-4, 11]\)[/tex].
- Endpoint [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = (-4)^4 - 50(-4)^2 + 6 = 256 - 800 + 6 = -538 \][/tex]
- Endpoint [tex]\( x = 11 \)[/tex]:
[tex]\[ f(11) = 11^4 - 50(11)^2 + 6 = 14641 - 6050 + 6 = 8597 \][/tex]
- Critical point [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^4 - 50(0)^2 + 6 = 6 \][/tex]
- Critical point [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = 5^4 - 50(5)^2 + 6 = 625 - 1250 + 6 = -619 \][/tex]
- Critical point [tex]\( x = -5 \)[/tex] (note this is outside the interval [tex]\([-4, 11]\)[/tex], so we do not consider it).
4. Compare the function values at these points to determine the absolute minimum and maximum:
- At [tex]\( x = -4 \)[/tex], [tex]\( f(-4) = -538 \)[/tex]
- At [tex]\( x = 11 \)[/tex], [tex]\( f(11) = 8597 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( f(0) = 6 \)[/tex]
- At [tex]\( x = 5 \)[/tex], [tex]\( f(5) = -619 \)[/tex]
Thus, we have the following function values:
- [tex]\( f(-4) = -538 \)[/tex]
- [tex]\( f(11) = 8597 \)[/tex]
- [tex]\( f(0) = 6 \)[/tex]
- [tex]\( f(5) = -619 \)[/tex]
The absolute minimum value is [tex]\( -619 \)[/tex] and occurs at [tex]\( x = 5 \)[/tex].
The absolute maximum value is [tex]\( 8597 \)[/tex] and occurs at [tex]\( x = 11 \)[/tex].
Therefore, the absolute minimum value is [tex]\(-619\)[/tex] and the absolute maximum value is [tex]\(8597\)[/tex].
1. Calculate the derivative of [tex]\( f(x) \)[/tex]:
The first derivative of [tex]\( f(x) \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ f'(x) = \frac{d}{dx}(x^4 - 50x^2 + 6) = 4x^3 - 100x \][/tex]
2. Find the critical points by setting the first derivative to zero:
[tex]\[ 4x^3 - 100x = 0 \][/tex]
Factor out the common term:
[tex]\[ 4x(x^2 - 25) = 0 \][/tex]
This gives the solutions:
[tex]\[ x = 0, \quad x^2 - 25 = 0 \implies x = \pm 5 \][/tex]
So, the critical points are [tex]\( x = 0, x = 5, \text{ and } x = -5 \)[/tex].
3. Evaluate the function at the critical points and at the endpoints of the interval:
The interval given is [tex]\([-4, 11]\)[/tex].
- Endpoint [tex]\( x = -4 \)[/tex]:
[tex]\[ f(-4) = (-4)^4 - 50(-4)^2 + 6 = 256 - 800 + 6 = -538 \][/tex]
- Endpoint [tex]\( x = 11 \)[/tex]:
[tex]\[ f(11) = 11^4 - 50(11)^2 + 6 = 14641 - 6050 + 6 = 8597 \][/tex]
- Critical point [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^4 - 50(0)^2 + 6 = 6 \][/tex]
- Critical point [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = 5^4 - 50(5)^2 + 6 = 625 - 1250 + 6 = -619 \][/tex]
- Critical point [tex]\( x = -5 \)[/tex] (note this is outside the interval [tex]\([-4, 11]\)[/tex], so we do not consider it).
4. Compare the function values at these points to determine the absolute minimum and maximum:
- At [tex]\( x = -4 \)[/tex], [tex]\( f(-4) = -538 \)[/tex]
- At [tex]\( x = 11 \)[/tex], [tex]\( f(11) = 8597 \)[/tex]
- At [tex]\( x = 0 \)[/tex], [tex]\( f(0) = 6 \)[/tex]
- At [tex]\( x = 5 \)[/tex], [tex]\( f(5) = -619 \)[/tex]
Thus, we have the following function values:
- [tex]\( f(-4) = -538 \)[/tex]
- [tex]\( f(11) = 8597 \)[/tex]
- [tex]\( f(0) = 6 \)[/tex]
- [tex]\( f(5) = -619 \)[/tex]
The absolute minimum value is [tex]\( -619 \)[/tex] and occurs at [tex]\( x = 5 \)[/tex].
The absolute maximum value is [tex]\( 8597 \)[/tex] and occurs at [tex]\( x = 11 \)[/tex].
Therefore, the absolute minimum value is [tex]\(-619\)[/tex] and the absolute maximum value is [tex]\(8597\)[/tex].
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