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

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

$f(x)= 2x^3-3x^2-12x+4$

$f'(x)= 6x^2-6x-12=0$

$x^2-x-2=0$

$⇒x=-1,2$

$f''(x)=12x-6$

$f''(-1)=-18$ point of maxima

$f''(2)=18$ point of minima