The maximum value of $f(x) = x^2 – x – 2$ in the closed interval $[-4, 4]$ is $\boxed{10}$.
To find the maximum value, we can first find the critical points of $f$. The derivative of $f$ is $f'(x) = 2x – 1$. Setting $f'(x) = 0$, we find that the critical point is $x = \frac{1}{2}$.
Since $f$ is a polynomial, it is continuous for all real numbers. Therefore, according to the Extreme Value Theorem, $f$ must have an absolute maximum (or minimum) over any closed interval.
To find the absolute maximum value of $f$, we can evaluate $f$ at the critical point and at the endpoints of the interval. The critical point is $x = \frac{1}{2}$, and the endpoints of the interval are $x = -4$ and $x = 4$. Therefore, the possible maximum values of $f$ are $f\left(\frac{1}{2}\right), f(-4)$, and $f(4)$.
We can evaluate $f$ at each of these points to find that $f\left(\frac{1}{2}\right) = \frac{9}{4} < 10$, $f(-4) = 16 – 4 – 2 = 10$, and $f(4) = 16 – 4 – 2 = 10$. Therefore, the maximum value of $f$ is $\boxed{10}$.
The other options are incorrect because they are not the maximum value of $f$. Option A is incorrect because $18 > 10$. Option B is incorrect because $10$ is the maximum value of $f$. Option C is incorrect because $-2.25 < 0$. Option D is incorrect because the maximum value of $f$ is not indeterminate.