Question

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

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

Answer 1

The correct option is B

$12$

Explanation for the correct option:

Step:1 Finding first and second derivative of $y=-x^3+3x^2+9x-27$

Given: $y=-x^3+3x^2+9x-27$

Differentiating w.r.t.$x$
$\frac{dy}{dx}=-3x^2+6x+9=$ the slope of the curve

Again differentiating w.r.t.$x$

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

$$\begin{gathered} \therefore \frac{d^{2}y}{d{x}^2}=0 \\ \Rightarrow -(x-1)=0 \\ \Rightarrow x=1>0 \end{gathered}$$

Step:2 Finding thrid derivative and checking maxima condition

Now $\frac{d^{3}y}{d{x}^3}=-6<0$

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

Step:3 Finding maximum value of $\frac{dy}{dx}=-3x^2+6x+9=$

Hence we find maxima at $x=1.$ in the equation of the slope.

Hence the maximum value of the slope is $\frac{dy}{dx}$​ which is,

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

Hence, option (B) is the correct answer.