#### Explain solution RD Sharma class 12 Chapter 15 Tangents and Normals exercise Fill in the blanks question 3

$\left ( 2,16 \right )\left ( -2,-16 \right )$

Hint:

Gradient is zero. i.e. $\frac{dy}{dx }=0$

Given:

Given curve, $y=12x-x^{3}$

To find:

We have to find the points on the given curve at which the gradient is zero.

Solution:

Here      $y=12x-x^{3}$                                                                                                                                      … (i)

i.e. $\frac{dy}{dx }=0$

Differentiating equation (i), we get

$\Rightarrow \quad \frac{d y}{d x}=12-3 x^{2}$

Since  $\frac{dy}{dx }=0$

$\begin{array}{lrl} \Rightarrow & & 12-3 x^{2}=0 \\\\ \Rightarrow & & -3 x^{2}=-12 \\\\ \Rightarrow & & x^{2}=4 \\\\ \Rightarrow & & x=\pm 2 \end{array}$

Putting the value $x=\pm 2$ in equation (i), we get

\begin{aligned} &y=12(2)-(2)^{3} \text { or } y=12(-2)-(-2)^{3} \\\\ &y=16 \text { or } y=-16 \end{aligned}

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