Need Solution for RD Sharma Maths Class 12 Chapter Maxima and Minima Exercise 17.2 Question 3

$x= 1$,$x= \frac{3}{5}$ is the point of local minima and local maxima respectively. The value of local minima and local maxima is 0 and $\frac{108}{3125}$respectively.

Hint:

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

Given:

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

Solution:

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

Differentiating f(x) with respect to ‘x’ then,

$\frac{d}{d x}(f(x))=\frac{d}{d x}\left\{x^{3}(x-1)^{2}\right\}$

$=x^{3} \frac{d}{d x}(x-1)^{2}+(x-1)^{2} \frac{d}{d x}\left(x^{3}\right) \quad\left[\because \frac{d}{d x}(u v)=u \frac{d v}{d x}+v \frac{d u}{d x}\right]$

$=x^{3} 2(x-1)^{2-1} \frac{d}{d x}(x-1)+(x-1)^{2} .3 x^{3-1}\left[\because \frac{d}{d x}(x+a)^{n}=n(x+a)^{n-1} \frac{d}{d x}(x+a), \frac{d x^{n}}{d x}=n x^{n-1}\right]$

$=x^{3} \cdot 2(x-1)\left\{\frac{d x}{d x}-\frac{d(1)}{d x}\right\}+(x-1)^{2} \cdot 3 x^{2} \quad\left[\because \frac{d}{d x}(a x+b)=\frac{d(a x)}{d x}+\frac{d(b)}{d x}\right]$

\begin{aligned} &=2 x^{3}(x-1)+3 x^{2}(x-1)^{2} \\ &=x^{2}(x-1)\{2 x+3(x-1)\}=x^{2}(x-1)(2 x+3 x-3) \\ &\therefore f^{\prime}(x)=+x^{2}(x-1)(5 x-3) \end{aligned}

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

$f^{\prime}(x)$

\begin{aligned} &\Rightarrow x^{2}(x-1)(5 x-3)=0 \\ &\Rightarrow x^{2}=0 \quad \text { or } \quad(x-1)=0 \quad \text { or } 5 x-3=0 \\ &\Rightarrow x=0 \quad \text { or } \quad x=1 \quad \text { or } \quad 5 x=3 \Rightarrow x=\frac{3}{5} \\ &\quad \therefore x=0,1, \frac{3}{5} \end{aligned}

−                                +                   -               +

-∞                    0                        3/5               1            ∞

Since $f^{\prime}(x)$ changes from negative to positive when $x$ increases through 1.

So, $x= 1$ is the point of local minima.

The value of local minima of $f\left ( x \right )$ at x=1 is

$f\left ( 1 \right )= 1^{3}\left ( 1-1 \right )=0$

Again, since $f^{\prime}(x)$  changes from +ve to –ve when $x$ increases through 3/5.

So, $x= \frac{3}{5}$ is the point of local maxima.

The value of local maxima of $f\left ( x \right )$ at $x= \frac{3}{5}$ is

\begin{aligned} f\left(\frac{3}{5}\right) &=\left(\frac{3}{5}\right)^{3}\left(\frac{3}{5}-1\right)^{2}=\frac{27}{125}\left(\frac{3-5}{5}\right)^{2} \\ &=\frac{27}{125}\left(\frac{-2}{5}\right)^{2}=\frac{27}{125} \times \frac{4}{25} \\ &=\frac{108}{3125} \end{aligned}

NOTE: Answer given in the book is incorrect.