The function $f(x)=2x^3-3x^2-12x+4$ has :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → one maximum and one minima
$f(x)=2x^3-3x^2-12x+4$
$f'(x)=6x^2-6x-12=0$
$x^2-x-2=0$
$x=2,-1$
$f''(x)=12x$ so $f''(2)=24>0$
$f''(-1)=-12$
2 → point of local minima
-1 → point of local maxima