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# The function $f(x)=2x^3-3x^2-12x+4$,has

$\begin{array}{1 1}(A)\;\text{two points of local maximum} & (B)\;\text{two points of local minimum}\\(C)\;\text{one maxima and one minima} & (D)\;\text{no maxima or no minima}\end{array}$

Toolbox:
• To obtain the absolute maxima or minima for the function $f(x)$
• (i) Find $f'(x)$ and put $f'(x)=0$
• (ii) Obtain the points from $f'(x)=0$
• (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$f(x)=2x^3-3x^2-12x+4$
Differentiating w.r.t $x$ we get,
$f'(x)=6x^2-6x-12$
Again differentiating w.r.t we get,
$f''(x)=12x-6$
when $f'(x)=0$
$\Rightarrow 6x^2-6x-12=0$
$\Rightarrow 6(x^2-x-2)=0$
$\therefore 6(x-2)(x+1)=0$
$x = 2, -1$
Step 2
when $x=2$
$f''(x)=6(2)-6$
$= 12-6$
$= 6 >0$
Hence $x = 2$ is a point of local minimum.
Step 3
when $x = -1$
$f''(x)=6(-1)-6$
$= -6-6$
$= -12 <0$
Hence $x=-1$ is a point of local maximum.
Hence it has one point of maximum and one point of minimum.
Hence the correct answer is C.