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# Maximum slope of the curve $y=x^3+3x^2+9x-27$ is

$(A)\;0\quad(B)\;12\quad(C)\;16\quad(D)\;32$

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## 1 Answer

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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
$y=-x^3+3x^2+9x-27 \: \: \: \: (1)$
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 $x=-1$
$\large\frac{d^2y}{dx^2}=6(2)$
$= 12$
$>0$
Step 3
Hence the slope is maximum when $x=3$
when $x=3$,
$y=(-3)^3+3(3)^2+9(3)-27$
$= -27+27+27-27$
$= 0$
Hence the maximum slope is 0
The correct option is A
answered Aug 12, 2013

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