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# At what point,the slope of the curve $y=-x^3+3x^2+9x-27$ is maximum?Also find the maximum slope.

Toolbox:
• Let $f(x)$ be a function with domain $D\: \subset \: R$. Then $f(x)$ is said to attain the maximum value at a point, $a \in D,$ if $f'(x) \leq f(a)$ for all $x \in D$.
Step 1
$f(r)=-x^3+3x^2+9x-27$
Differentiating w.r.t $x$ we get
$f'(x)=-3x^2+6x+9$
If $f'(x)=0$
$\Rightarrow -3x^2+6x+9=0$
$3(-x^2+2x+3)=0$
factorising this we get,
$f'(x)=(x+3)(x-1)$
when $x=-3$
$f(x)=-(3)^3+3(3)^2+9(3)-27$
$= -27+27+27-27$
$=0$
when $x=+1$
$f(1)=-(+1)^3+3(+1)^2+9(+1)-27$
$= -16$
Hence the point is $(1, -16)$
The maximum slope =
$f'(1)=-3(1)+6(1)+9$
$= 12$