Since $f(x) = \frac{1}{2}x^2 - 8$, we can see that the function is a parabola with its vertex at $(0, -8)$. The parabola opens upwards, so the maximum value of $f(x)$ on the interval $-4 \leq x \leq 4$ is $f(4) = \frac{1}{2}(4)^2 - 8 = 8 - 8 = 0$.
Now, we need to
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