We are tasked with finding the absolute maximum and minimum values of the function:

$f(x) = x^4 - 16x^2 + 64$

on the interval $[0, 5]$.

Steps to solve:

1. Find the derivative of $f(x)$ to determine critical points: $f'(x) = \frac{d}{dx}(x^4 - 16x^2 + 64) = 4x^3 - 32x$

2. Set the derivative equal to zero to find critical points: $4x^3 - 32x = 0$ Factor the equation: $4x(x^2 - 8) = 0$ This gives: $x = 0 \quad \text{or} \quad x^2 = 8 \quad \Rightarrow \quad x = \pm \sqrt{8} = \pm 2\sqrt{2}$

3. Check the critical points within the interval $[0, 5]$:

   • $x = 0$ is within the interval.
   • $x = 2\sqrt{2} \approx 2.828$ is also within the interval.
   • $x = -2\sqrt{2}$ is not within the interval, so we discard it.

4. Evaluate the function at the critical points and at the endpoints of the interval:

   • $f(0) = 0^4 - 16(0^2) + 64 = 64$
   • $f(2\sqrt{2}) = (2\sqrt{2})^4 - 16(2\sqrt{2})^2 + 64 = 64 - 128 + 64 = 0$
   • $f(5) = 5^4 - 16(5^2) + 64 = 625 - 400 + 64 = 289$

5. Conclusion:

   • Absolute minimum value occurs at $x = 2\sqrt{2}$, where $f(2\sqrt{2}) = 0$.
   • Absolute maximum value occurs at $x = 5$, where $f(5) = 289$.

Final Answer:

• $\text{min} = 0$
• $\text{max} = 289$

Let me know if you need more details or explanations! Here are 5 related questions to further your understanding:

1. What is the difference between absolute and relative extrema?
2. How do you determine if a critical point is a minimum or maximum?
3. How would this method change if the interval was different?
4. Can a function have more than one absolute minimum or maximum?
5. Why is it important to check both endpoints of the interval?

Tip: Always check both critical points and endpoints when looking for absolute extrema!