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)

$y = -x^3 + 3x^2 + 9x -27$
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 $= -3\left(1\right)^{2 }+ 6\left(1\right) + 9 = 12$