#### The function $f (x) = 2x^3 - 3x^2 - 12x + 4$, has A. two points of local maximum B. two points of local minimum C. one maxima and one minima D. no maxima or minima

Let $f (x) = 2x^3 - 3x^2 - 12x + 4$

Apply first derivative and get

\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(2 \mathrm{x}^{3}-3 \mathrm{x}^{2}-12 \mathrm{x}+4\right)}{\mathrm{dx}}\\ &\text { Apply sum rule and get }\\ &\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}\left(2 \mathrm{x}^{3}\right)}{\mathrm{dx}}-3 \frac{\mathrm{d}\left(\mathrm{x}^{2}\right)}{\mathrm{d} \mathrm{x}}-12 \frac{\mathrm{d}(\mathrm{x})}{\mathrm{dx}}+0 \end{aligned}

Apply derivative,
$\\ { \Rightarrow f' (x)=6x\textsuperscript{2}-6x-12}\\$

Put f'(x)=0, and get

$\\{6x\textsuperscript{2}-6x-12=0}\\ { \Rightarrow 6(x\textsuperscript{2}-x-2)=80}\\ { \Rightarrow x\textsuperscript{2}-x-2=0}\\$
Split middle term and get

$\\ { \Rightarrow x\textsuperscript{2}-2x+x-2=0}\\ { \Rightarrow x(x-2)+1(x-2)=0}\\ { \Rightarrow (x-2)(x+1)=0}\\ { \Rightarrow x-2=0 or x+1=0}\\ { \Rightarrow x=2 or x=-1}\\$

Now we find the values of f(x) at x=-1, 2

$\\ {f(x)= 2x\textsuperscript{3} - 3x\textsuperscript{2} - 12x + 4}\\ {f(-1)= 2(-1)\textsuperscript{3} - 3(-1)\textsuperscript{2} - 12(-1) + 4=-2-3+12+4=11}\\ {f(2)= 2(2)\textsuperscript{3} - 3(2)\textsuperscript{2} - 12(2) + 4 =16-12-24+4=-16}\\$

Hence from above we find that the point of local maxima is x=-1 and 11 is the maximum value of f(x).

Whereas the point of local minima is x=2 and -16 is the minimum value of f(x).

So, the correct answer is option C.

Hence, the given function has 1 minima and 1 maxima.