Mathematics - Application of Derivative Question with Solution | TestHub

$\begin{array}{rl}& f(x)=2x^3-3x^2-12x+4\\ & \Rightarrow f^{\prime }(x)=6x^2-6x-12=6\left(x^2-x-2\right)\\ & =6(x-2)(x+1)\end{array}$ For maxima and minima ${\mathrm{f}}^{\prime }(\mathrm{x})=0$ $\therefore 6(\mathrm{x}-2)(\mathrm{x}+1)=0\Rightarrow \mathrm{x}=2,-1$ Now, ${\mathrm{f}}^{\prime \prime }(\mathrm{x})=12\mathrm{x}-6$ $\begin{array}{rl}& \text{ At }\mathrm{x}=2;{\mathrm{f}}^{\prime \prime }(\mathrm{x})=24-6=18>0\\ & \therefore \mathrm{x}=2,\text{ local min. point }\\ & \text{ At }\mathrm{x}=-1;{\mathrm{f}}^{\prime \prime }(\mathrm{x})=12(-1)-6=-18<0\\ & \therefore \mathrm{x}=-1\text{ local max. point }\end{array}$