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

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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}


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


 f(x)=x^{3}(2 x-1)^{3}   

Solution: -   

 f(x)=x^{3}(2 x-1)^{3}   

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

\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}

            =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\}

            \\ =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) \\

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

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

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

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

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

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

                                                            \Rightarrow x=1 / 2

                              \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}

\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}


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