Question:
Find the maximum slope of the curve $y=-x^{3}+3 x^{2}+2 x-27 .$
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 $=-3(1)^{2}+6(1)+2=-3+6+2=5$