Question

Maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$ is

• A. $0$
• B. $12$
• C. $16$
• D. $32$

Answer 1

Answer: (B)

$12$

Answer 2

$y = -x^3 + 3x^2 + 9x -27$ Slope $= \frac{dy}{dx} = m = - 3 x^{ 2} + 6 x + 9$ Now, $\frac{dm}{dx} = - 6x + 6$ Now, $\frac{dm}{dx} = 0$ $\Rightarrow -6x + 6 = 0$ $\Rightarrow x = 1$ Now, $\frac{d^{2}m}{dx^{2}} =-6 < 0 \,\forall\, x$ $\therefore x = 1$ is a point of local maximum. $\therefore$ Maximum slope $= -3\left(1\right)^{2 }+ 6\left(1\right) + 9 = 12$