The problem asks to determine the value of $x$ where the function $f(x) = \frac{1}{2}x^2 - 2x + 3$ reaches its maximum in the interval $[0, 4]$.

Let's proceed by analyzing the function step by step:

Step 1: Find the critical points

To find critical points, we first take the derivative of $f(x)$ with respect to $x$:

$f'(x) = \frac{d}{dx} \left( \frac{1}{2}x^2 - 2x + 3 \right) = x - 2$

Set $f'(x) = 0$ to find the critical points:

$x - 2 = 0 \implies x = 2$

Step 2: Evaluate at the critical point and endpoints

Since the interval is $[0, 4]$, we need to check the function values at the critical point $x = 2$ and the endpoints $x = 0$ and $x = 4$.

• $f(0) = \frac{1}{2}(0)^2 - 2(0) + 3 = 3$
• $f(2) = \frac{1}{2}(2)^2 - 2(2) + 3 = 2 - 4 + 3 = 1$
• $f(4) = \frac{1}{2}(4)^2 - 2(4) + 3 = 8 - 8 + 3 = 3$

Step 3: Determine the maximum

Comparing the values:

• $f(0) = 3$
• $f(2) = 1$
• $f(4) = 3$

The maximum value occurs at $x = 0$ and $x = 4$, where the function value is 3. Therefore, the maximum is achieved at $x = 0$ and $x = 4$.

Given the options, the correct answer is A. 4.

Would you like more details or have any questions? Here are some additional questions for practice:

1. How do you find the critical points of a function?
2. What is the importance of evaluating a function at its endpoints in a closed interval?
3. How can you distinguish between local maximum and global maximum?
4. Why do you need to set the derivative equal to zero when finding critical points?
5. What happens if a function does not have critical points within the interval?

Tip: Always remember to check the endpoints when working with closed intervals, as maxima and minima can occur there too!