#### Please solve rd  sharma class 12 Chapter 17  Maxima and Minima excercise 17.3 question 2 subquestion (iii) maths textbook solution

x=1 is point of inflexion

x=-1 is the point of local minimum &$x=\frac{1}{5}$ is the point of local maximum

Hint:

Using chain rule of derivative

Given:

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

Explanation:

Differentiating f with respect to x

\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[-(\mathrm{x}-1)^{3}(\mathrm{x}+1)^{2}\right] \\ &=-\left[(\mathrm{x}-1)^{3} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}+1)^{2}+(\mathrm{x}+1)^{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-1)^{3}\right] \\ &=-\left[(\mathrm{x}-1)^{3} \cdot 2(\mathrm{x}+1)+3(\mathrm{x}+1)^{2}(\mathrm{x}-1)^{2}\right] \\ &=-(\mathrm{x}-1)^{2}(\mathrm{x}+1)[2 \mathrm{x}-2+3 \mathrm{x}+3] \\ &=-(\mathrm{x}-1)^{2}(\mathrm{x}+1)(5 \mathrm{x}+1) \\ &=-\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right)\left(5 \mathrm{x}^{2}+\mathrm{x}+5 \mathrm{x}+1\right) \\ &=-\left(\mathrm{x}^{2}-2 \mathrm{x}+1\right)\left(5 \mathrm{x}^{2}+6 \mathrm{x}+1\right) \\ &=-\left(5 \mathrm{x}^{4}+6 \mathrm{x}^{3}+\mathrm{x}^{2}-10 \mathrm{x}^{3}-12 \mathrm{x}^{2}-2 \mathrm{x}+5 \mathrm{x}^{2}+6 \mathrm{x}+1\right) \\ &=-\left(5 \mathrm{x}^{4}-4 \mathrm{x}^{3}-6 \mathrm{x}^{2}+4 \mathrm{x}+1\right) \end{aligned}

\begin{aligned} &\text { Put } f^{\prime}(x)=0 \\ &-1(x-1)^{2}(x+1)(5 x+1)=0 \\ &(x-1)^{2}(x+1)(5 x+1)=6 \\ &\Rightarrow(x-1)^{2}=0 \\ &x-1=0 \\ &x=1 \\ &x+1=0, \quad 5 x+1=0 \\ &x=-1, \quad x=-\frac{1}{5} \end{aligned}

Thus $x=1$ and $x=-1$ and $x=-\frac{1}{5}$ are the possible points of local minima and maxima

Differentiating f’(x) with respect to x

\begin{aligned} &f^{\prime \prime}(x)=\frac{d}{d x}\left[-\left(5 x^{4}-4 x^{3}-6 x^{2}+4 x+1\right)\right] \\ &=-\left(20 x^{3}-12 x^{2}-12 x+4\right) \\ &=-20 x^{3}+12 x^{2}+12 x-4 \\ &\text { when } x=1 \\ &\begin{aligned} f^{\prime \prime}(1) &=-20(1)^{3}+12(1)^{2}+12(1)-4 \\ &=-20+12+12-4=0 \end{aligned} \end{aligned}

Thus this test is fail as x=1 is point of inflexion

when $x=-1$

\begin{aligned} f^{\prime \prime}(-1) &=-20(-1)^{3}+12(-1)^{2}+12(-1)-4 \\ &=20+12-12-4=16>0 \end{aligned}

So , x = -1 is a point of local minimum

When $x=-\frac{1}{5}$

\begin{aligned} f\left(-\frac{1}{5}\right)=&-20\left(-\frac{1}{5}\right)^{3}+12\left(-\frac{1}{5}\right)^{2}+12\left(-\frac{1}{5}\right)-4 \\ &=-336 / 125<0 \end{aligned}

So, x=-1/5 is the point of local maximum

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