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

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

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  • 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
Differentiating w.r.t $x$ we get,
Again differentiating w.r.t we get,
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$
$ = 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.
answered Aug 12, 2013 by thanvigandhi_1
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