The given function is $f(x) = 4x - \frac{1}{2}x^2$ on the interval $x \in [-2, \frac{9}{2}]$.

Step 1: Find the derivative

To find the critical points, we first need to differentiate $f(x)$.

$f'(x) = \frac{d}{dx}\left(4x - \frac{1}{2}x^2\right) = 4 - x$

Step 2: Set the derivative equal to zero

To find the critical points, set $f'(x) = 0$:

$4 - x = 0 \implies x = 4$

Step 3: Evaluate the function at the critical point and endpoints

Now, we evaluate the function $f(x)$ at the critical point $x = 4$, as well as at the endpoints $x = -2$ and $x = \frac{9}{2}$.

• At $x = -2$: $f(-2) = 4(-2) - \frac{1}{2}(-2)^2 = -8 - \frac{1}{2}(4) = -8 - 2 = -10$

• At $x = 4$: $f(4) = 4(4) - \frac{1}{2}(4)^2 = 16 - \frac{1}{2}(16) = 16 - 8 = 8$

• At $x = \frac{9}{2}$: $f\left(\frac{9}{2}\right) = 4\left(\frac{9}{2}\right) - \frac{1}{2}\left(\frac{9}{2}\right)^2 = 18 - \frac{1}{2}\left(\frac{81}{4}\right) = 18 - \frac{81}{8} = 18 - 10.125 = 7.875$

Step 4: Determine the absolute maximum and minimum

• $f(-2) = -10$
• $f(4) = 8$
• $f\left(\frac{9}{2}\right) = 7.875$

Thus, the absolute maximum value is 8 at $x = 4$, and the absolute minimum value is -10 at $x = -2$.

Would you like further details or explanations on any step?

Related questions:

1. How would the result change if the interval was different?
2. Can you explain why we check the endpoints in this process?
3. What would happen if we added constraints to this optimization problem?
4. How do we handle similar problems when the function involves higher powers of $x$?
5. What is the significance of the second derivative in these types of problems?

Tip: Always check the derivative at critical points and endpoints when finding absolute extrema on a closed interval.