The maximum slope of curve y = -x^3 + 3x^2 + 9x - 27 is

The correct answer is Option (2) → 12 ##

We have, $y = -x^3 + 3x^2 + 9x - 27$

$∴\frac{dy}{dx} = -3x^2 + 6x + 9 = \text{Slope of the curve}$

and $\frac{d^2y}{dx^2} = -6x + 6 = -6(x - 1)$

$∴\frac{d^2y}{dx^2} = 0$

$\Rightarrow -6(x - 1) = 0 \Rightarrow x = 1 > 0$

Now, $\frac{d^3y}{dx^3} = -6 < 0$

So, the maximum slope of given curve is at $x = 1$.

$∴\left( \frac{dy}{dx} \right)_{(x=1)} = -3 \cdot 1^2 + 6 \cdot 1 + 9 = 12$