Question

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

• A. $(-1,16)$
• B. $(1,-16)$
• C. $(16,-1)$
• D. $(-16,1)$

Answer 1

Answer: (B)

$(1,-16)$

$y=-x^3+3x^2+9x-27$

Therefore, $\frac{dy}{dx}=-3x^2+6x+9$ = Slope of tangent to the curve

Now, $\frac{d^{2}y}{d{x}^2}=-6x+6$

For $\frac{d}{dx}\left(\frac{dy}{dx}\right)=0,x=1$

Also, $\frac{d}{dx}\left(\frac{d^{2}y}{d{x}^2}\right)=-6<0$

So, the slope of tangent to the curve is maximum, when $x=1$

For $x=1,\frac{dy}{dx}=12$, which is the maximum slope.

Also, for $x=1,y=-16$

So, the required point is $(1,-16)$.