The absolute minimum value of \(f(x) = 2x^{3} - 3x^{2} - 12x + 5\) on the interval [−3, 3] is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(-40\)
On a closed interval the absolute extreme values occur at critical points or at the end points. The critical points:
\( f'(x) = 6x^{2} - 6x - 12 \)
\( = 6(x - 2)(x + 1) \)
So x = −1 and x = 2, both inside [−3, 3]. The values of f:
\( f(-3) = -54 - 27 + 36 + 5 = -40 \)
\( f(-1) = -2 - 3 + 12 + 5 = 12 \)
\( f(2) = 16 - 12 - 24 + 5 = -15 \)
\( f(3) = 54 - 27 - 36 + 5 = -4 \)
The least of these is −40, at the end point x = −3.
Why the other options are incorrect
Option (2) \(-15\) — Incorrect. This is the local minimum value f(2); the end point x = −3 gives a smaller value.
Option (3) \(-4\) — Incorrect. This is f(3), the value at the right end point.
Option (4) 12 — Incorrect. This is the local maximum value f(−1), which is also the absolute maximum on [−3, 3].