To determine the absolute maximum value of $f(x) = x^4 - 8x^2 + 8$ on the closed interval $[-1, 3]$, we follow these steps:

Step 1: Find the critical points

To locate the critical points of $f(x)$, we first take the derivative of $f(x)$ and set it equal to zero.

$f'(x) = \frac{d}{dx} \left( x^4 - 8x^2 + 8 \right)$ $f'(x) = 4x^3 - 16x$

Set $f'(x) = 0$:

$4x^3 - 16x = 0$

Factor out $4x$:

$4x(x^2 - 4) = 0$

Simplify:

$4x(x-2)(x+2) = 0$

Thus, the critical points are:

$x = 0, \, x = 2, \, x = -2$

Step 2: Evaluate $f(x)$ at the endpoints and critical points

The closed interval is $[-1, 3]$. We evaluate $f(x)$ at:

1. The endpoints $x = -1$ and $x = 3$,
2. The critical points $x = 0$, $x = 2$, and $x = -2$, but only those that fall in the interval $[-1, 3]$.

Check critical points in $[-1, 3]$:

• $x = 0$ → included.
• $x = 2$ → included.
• $x = -2$ → excluded since it lies outside $[-1, 3]$.

Now, calculate $f(x)$ at the relevant points:

1. At $x = -1$: $f(-1) = (-1)^4 - 8(-1)^2 + 8$ $f(-1) = 1 - 8 + 8 = 1$

2. At $x = 0$: $f(0) = (0)^4 - 8(0)^2 + 8$ $f(0) = 8$

3. At $x = 2$: $f(2) = (2)^4 - 8(2)^2 + 8$ $f(2) = 16 - 8(4) + 8$ $f(2) = 16 - 32 + 8 = -8$

4. At $x = 3$: $f(3) = (3)^4 - 8(3)^2 + 8$ $f(3) = 81 - 8(9) + 8$ $f(3) = 81 - 72 + 8 = 17$

Step 3: Compare values

The values of $f(x)$ at the endpoints and critical points are:

• $f(-1) = 1$
• $f(0) = 8$
• $f(2) = -8$
• $f(3) = 17$

The absolute maximum value is:

$\boxed{17 \, \text{at} \, x = 3}$

Summary

The absolute maximum value of $f(x) = x^4 - 8x^2 + 8$ on the interval $[-1, 3]$ is $17$, which occurs at $x = 3$.

Let me know if you'd like further explanations or step-by-step clarifications!

Related Questions:

1. How do you determine the absolute minimum value of $f(x)$ on the same interval?
2. How does the behavior of $x^4$ impact the shape of the graph?
3. What happens to the maximum value if the interval changes to $[-2, 2]$?
4. Can you find the inflection points of $f(x)$?
5. How does the derivative test confirm whether critical points are maxima or minima?

Tip: Always evaluate the function at both critical points and endpoints when working on closed intervals.