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