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