Question

The point on the curve $y=12x-x^3$ at which the gradient is zero are

• A. $(0,2),(2,16)$
• B. $(0,-2),(2,-16)$
• C. $(2,-16),(-2,16)$
• D. $(2,16),(-2,-16)$

Answer 1

The correct option is D

$(2,16),(-2,-16)$

Compute the required points:

Given: $y=12x-x^3$………………$\left(1\right)$

Since the gradient is zero

So, $\frac{dy}{dx}=0$

$\Rightarrow$$12-3x^2=0$

$\Rightarrow$ $3x^2=12$

$\Rightarrow$ $x^2=4$

$\Rightarrow$ $x=\pm 2$

Substitute the value of $x$ in equation $\left(1\right)$,

$$\begin{gathered} \left(\mathrm{i}\right)x=2 \\ \Rightarrow y=24-8 \\ \Rightarrow y=16 \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right)x=-2 \\ \Rightarrow y=-24+8 \\ \Rightarrow y=-16 \end{gathered}$$

Therefore, the points are $\left(2,16\right)$ and $\left(-2,-16\right)$.

Hence option $\left(D\right)$ is the correct answer.