The function f(x) = 2x^3 - 3x^2 - 12x + 4 has:

The correct answer is Option (3) → one maximum and one minimum

The critical points are where f'(x) = 0:

    \(f'(x) = 6x^2 - 6x - 12\)
    \(= 6(x^2 - x - 2)\)
    \(= 6(x - 2)(x + 1)\)

So x = −1 or x = 2. By the second derivative test, with \(f''(x) = 12x - 6\):

    \(f''(-1) = -18 \lt 0\): local maximum at x = −1
    \(f''(2) = 18 \gt 0\): local minimum at x = 2

So f has one maximum and one minimum.

NCERT Exemplar:
Application of Derivatives