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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Maximum slope of the curve $y=x^3+3x^2+9x-27$ is


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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
$ y=-x^3+3x^2+9x-27 \: \: \: \: (1)$
Differentiating w.r.t $x$ we get,
$ \large\frac{dy}{dx}=-3x^2+6x+9$
Again differentiating w.r.t $x$ we get,
$ \large\frac{d^2y}{dx^2}=-6x+6$
$ = 6(-x+1)$
Step 2
when $ \large\frac{dy}{dx}=0$
$ -3x^2+6x+9=0$
$ \Rightarrow -3(x^2-2x-3)=0$
$ \Rightarrow -3(x-3)(x+1)=0$
$ \Rightarrow x=3, -1$
when $x=3$
$ \large\frac{d^2y}{dx^2}=-12$ which is $ <0$
when $x=-1$
$ \large\frac{d^2y}{dx^2}=6(2)$
$ = 12$
Step 3
Hence the slope is maximum when $x=3$
when $x=3$,
$ = -27+27+27-27$
$ = 0$
Hence the maximum slope is 0
The correct option is A
answered Aug 12, 2013 by thanvigandhi_1
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