For the function f(x) = 2x^3-3x^2 - 12x+5 , the difference of maximum…

The correct answer is Option (3) → 27

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

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

So the critical points are \(x = -1\) and \(x = 2\). By the second derivative test:

    \(f''(x) = 12x - 6\)
    \(f''(-1) = -18 \lt 0\), a local maximum at \(x = -1\)
    \(f''(2) = 18 \gt 0\), a local minimum at \(x = 2\)

    \(f(-1) = -2 - 3 + 12 + 5 = 12\)
    \(f(2) = 16 - 12 - 24 + 5 = -15\)

A cubic takes every real value, so it has no greatest or least value on R. Here “maximum and minimum value” means the local maximum value 12 and the local minimum value −15:

    \(12 - (-15) = 27\)