Find the maximum slope of the curve

Solution:

Given : $y=-x^{3}+3 x^{2}+2 x-27$             .....(1)

Slope $=\frac{d y}{d x}=-3 x^{2}+6 x+2$

Now,

$M=-3 x^{2}+6 x+2$

$\Rightarrow \frac{d M}{d x}=-6 x+6$

For maximum or minimum values of $M$, we must have

$\frac{d M}{d x}=0$

$\Rightarrow-6 x+6=0$

$\Rightarrow 6 x=6$

$\Rightarrow x=1$

Substituing the value of $x$ in eq. (1), we get

$y=-1^{3}+3 \times 1^{2}+2 \times 1-27=-23$

$\frac{d^{2} M}{d x^{2}}=-6<0$

So, the slope is maximum when $x=1$ and $y=-23$.

$\therefore \operatorname{At}(1,-23):$

Maximum slope $=-3(1)^{2}+6(1)+2=-3+6+2=5$