Based on the extreme value theorem, what is the maximum value of f(x) = –x2 + 6x over the interval [1, 4]?

Because f(x) = −x² + 6x is continuous on the closed interval [1, 4], its maximum occurs at an endpoint or a critical point. Its derivative is f′(x) = −2x + 6, which is zero at x = 3; evaluating gives f(3) = 9. The endpoint values are f(1) = 5 and f(4) = 8, so the maximum value is 9, choice D.

AI-written and checked against the verified answer.