#### Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 13

$\inline x=\frac{1}{4}$  is the point of local minima and the values of local minima at $x=\frac{1}{4} \ \ \ is \ \ \ \ \frac{-1}{512}$

Hint

Use first derivative test to find the value and point of local maxima and local minima.

Given:

$\inline f(x)=x^{3}(2 x-1)^{3}$

Solution: -

$\inline f(x)=x^{3}(2 x-1)^{3}$

Differentiating $\inline f(x) w \cdot r . t^{\prime} x^{\prime} then$

\inline \begin{aligned} f^{\prime}(x) &=\frac{d}{d x}\left\{x^{3}(2 x-1)^{3}\right\} \\ &=x^{3} \frac{d}{d x}(2 x-1)^{3}+(2 x-1)^{3} \frac{d}{d x} x^{3} \quad\left\{\because \frac{d}{d x}(y x)=y \frac{d}{d x} x+x \frac{d}{d x} y\right\} \\ &=x^{3} 3(2 x-1)^{3-1} \frac{d}{d x}(2 x-1)+(2 x-1)^{3} 3 x^{3-1} \quad\left\{\because \frac{d}{d x} x^{n}=n x^{n-1} \& \frac{d}{d x}(a x+b)^{n}=n(a x+b)^{n-1} \frac{d}{d x}(a x+b)\right\} \\ &=x^{3} 3(2 x-1)^{2}\left\{\frac{d}{d x}(2 x)-\frac{d}{d x}(1)^{2}\right\}+(2 x-1)^{3} 3 x^{2} \quad\left\{\because \frac{d}{d x}(a x+b)=\frac{d}{d x}(a x)+\frac{d}{d x}(b)\right\} \end{aligned}

$\inline =x^{3} 3(2 x-1)^{2}\left\{2 \frac{d}{d x}(x)-0\right\}+(2 x-1)^{3} 3 x^{2} \quad\left\{\because \frac{d}{d x}(a x)=a \frac{d}{d x}(x), \frac{d}{d x}(x)=1, \frac{d}{d x} \text { constant }=0\right\}$

$\inline \\ =3 x^{3}(2 x-1)^{2}(2 \cdot 1)+3 x^{2}(2 x-1)^{3} \\ =6 x^{3}(2 x-1)^{2}+3 x^{2}(2 x-1)^{3} \\ =3 x^{2}(2 x-1)^{2}(2 x+2 x-1) \\$

$\inline f^{\prime}(x) =3 x^{2}(2 x-1)^{2}(4 x-1)$

By first derivative test, for local maxima and local minima, we have

$\inline \Rightarrow f^{\prime}(x)=0$

$\inline \Rightarrow 3 x^{2}(2 x-1)^{2}(4 x-1)=0$

$\inline \Rightarrow x^{2}(2 x-1)^{2}(4 x-1)=0 \quad[\because 3 \neq 0]$

$\inline \Rightarrow x^{2}=0 \quad or (2 x-1)^{2}=0 \Rightarrow(4 x-1)=0$

$\inline \Rightarrow x=0 \quad or (2 x-1)=0 \Rightarrow 2 x=1 or \Rightarrow 4 x=1 \Rightarrow x=1 / 4$

$\inline \Rightarrow x=1 / 2$

$\inline \therefore x=0, \frac{1}{2}, \frac{1}{4}$

-                       -                              +                      +

-∞                    0                     1/4                  1/2

Since  f(x) changes from –ve to +ve when x increases through  $\frac{1}{4}$   so   $x=\frac{1}{4}$   is the point of local minima

The value of the local minima of $f\left ( x \right )$ at $x=\frac{1}{4}$

\inline \begin{aligned} f\left(\frac{1}{4}\right) &=\left(\frac{1}{4}\right)^{3}\left(2 \cdot \frac{1}{4}-1\right)^{3} \\ &=\frac{1}{4^{3}}\left(\frac{1}{2}-1\right)^{3}=\frac{1}{64}\left(\frac{1-2}{2}\right)^{3}=\frac{1}{64}\left(\frac{-1}{2}\right)^{3} \\ &=\frac{1}{64} \cdot \frac{-1}{8}=\frac{-1}{512} \end{aligned}