The function $f(x)=2x^3-3x^2-12x+4$ has : |
Two points of local maximum Two points of local minimum One local maxima and one local minima No maxima or minima |
One local maxima and one local minima |
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 |