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

The correct answer is Option (3) → One local maxima and one local minima

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

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

$⇒6(x^2-x-2)=0$

$6(x-2)(x+1)=0$

$x=2,-1$

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

at $x=2$ point of local minima

$f''(2)=18>0$

at $x=-1$

$f''(-1)=-18<0$ point of local maxima

f(x) has one local maxima and one local minima