Question

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

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

Answer 1

The correct option is C $12$

$y=-x^3+3x^2+9x-27$
$\Rightarrow \frac{dy}{dx}=-3x^2+6x+9$
Let Slope of tangent to given curve is
$f\left(x\right)=\frac{dy}{dx}=-3x^2+6x+9$
$\Rightarrow f^{\prime }\left(x\right)=-6x+6$
Substituting $f^{\prime }\left(x\right)=0$,
$\Rightarrow -6x+6=0$
$\Rightarrow x=1$
$\Rightarrow f^{\prime }\left(x\right)=-6x+6$
$\Rightarrow f"\left(x\right)=-6$
$\Rightarrow f"\left(x\right){|}_{x=1}=-6<0$
Hence, $x=1$ is the point of maxima
$\Rightarrow f\left(x\right){|}_{x=1}=15-3=12$
Thus, maximum slope of the curve
$y=-x^3+3x^2+9x-27$ is $12$
Hence, option (b) is correct.