Question

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

Answer 1

We have, $y=-x^3+3x^2+9x-27\therefore \frac{dy}{dx}=-3x^2+6x+9=SlopeoftangenttothecurveNow,\frac{d^{2}y}{d{x}^2}=-6x+6for\frac{d}{dx}\left(\frac{dy}{dx}\right)=0.-6x+6=0\Rightarrow x=\frac{-6}{-6}=1\therefore \frac{d}{dx}\left(\frac{d^{2}y}{d{x}^2}\right)=-6<0$
So, the slope of tangent to the curve is maximum when x = 1
for $x=1,{\left(\frac{dy}{dx}\right)}_{x=1}=-3\left(1^2\right)+6\left(1\right)+9=12.$
Which is the maximum slope.

Also, for $x=1,y=-1^3+{3.1}^2+9.1-27=-1+3+9-27=-16$