To find the absolute maximum and minimum values of the function [tex]\( f(x) = x^2 - 8x - 3 \)[/tex] over the interval [tex]\([0, 6]\)[/tex], we need to follow these steps:
1. Find the critical points of the function within the given interval by setting the first derivative to zero and solving for [tex]\( x \)[/tex].
2. Evaluate the function at the critical points.
3. Evaluate the function at the endpoints of the interval.
4. Compare all these values to determine the absolute maximum and minimum values and the corresponding [tex]\( x \)[/tex]-values.
### Step-by-Step Solution:
1. Find the critical points:
First, we need to find the derivative of [tex]\( f(x) \)[/tex]:
[tex]\[
f'(x) = \frac{d}{dx} (x^2 - 8x - 3) = 2x - 8
\][/tex]
Set the derivative equal to zero to find the critical points:
[tex]\[
2x - 8 = 0 \implies x = 4
\][/tex]
Thus, the critical point is [tex]\( x = 4 \)[/tex].
2. Evaluate the function at the critical point:
[tex]\[
f(4) = 4^2 - 8(4) - 3 = 16 - 32 - 3 = -19
\][/tex]
3. Evaluate the function at the endpoints of the interval [tex]\([0, 6]\)[/tex]:
For [tex]\( x = 0 \)[/tex]:
[tex]\[
f(0) = 0^2 - 8(0) - 3 = -3
\][/tex]
For [tex]\( x = 6 \)[/tex]:
[tex]\[
f(6) = 6^2 - 8(6) - 3 = 36 - 48 - 3 = -15
\][/tex]
4. Determine the absolute maximum and minimum values:
- At [tex]\( x = 0 \)[/tex], [tex]\( f(x) = -3 \)[/tex]
- At [tex]\( x = 4 \)[/tex], [tex]\( f(x) = -19 \)[/tex]
- At [tex]\( x = 6 \)[/tex], [tex]\( f(x) = -15 \)[/tex]
Compare these values:
- The absolute minimum value is [tex]\(-19\)[/tex] at [tex]\( x = 4 \)[/tex].
- The absolute maximum value is [tex]\(-3\)[/tex] at [tex]\( x = 0 \)[/tex].
### Conclusion:
The absolute maximum value is [tex]\(-3\)[/tex] at [tex]\( x = 0 \)[/tex].
The absolute minimum value is [tex]\(-19\)[/tex] at [tex]\( x = 4 \)[/tex].
