#### Provide solution for rd sharma class 12 chapter 17 Maxima and Minima exercise 17.3 question 2 sub-question (i)

Ex_17.3_2-1

$x=2$ is the local minima and the local minimum value of $f(x)$ at $x=2$  is 0.

$x=\frac{4}{3}$  is the local maxima and the local maximum value of $f(x)$at $x=\frac{4}{3}$   is $x=\frac{4}{27}$.

Hint:

Using chain rule

Given:

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

Explanation:

Differentiating f with respect to x

\begin{aligned} &\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}}{\mathrm{dx}}\left[(\mathrm{x}-1)(\mathrm{x}-2)^{2}\right) \\ &=(\mathrm{x}-1) \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-2)^{2}+(\mathrm{x}-2)^{2} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}-1) \\ &=2(\mathrm{x}-1)(\mathrm{x}-2)+(\mathrm{x}-2)^{2} \quad \text { [Using chain rule] } \\ &=2\left(\mathrm{x}^{2}-2 \mathrm{x}-\mathrm{x}+2\right)+\left(\mathrm{x}^{2}-4 \mathrm{x}+4\right) \\ &=2 \mathrm{x}^{2}-4 \mathrm{x}-2 \mathrm{x}+4+\mathrm{x}^{2}-4 \mathrm{x}+4 \\ &=3 \mathrm{x}^{2}-10 \mathrm{x}+8 \end{aligned}

\begin{aligned} &\text { Put } f^{\prime}(x)=0 \\ &3 x^{2}-10 x+8=0 \\ &x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \\ &=\frac{-10 \pm \sqrt{(10)^{2}-4 \cdot 3 \cdot 8}}{2 \cdot(3)} \\ &=\frac{+10 \pm \sqrt{100-96}}{6} \\ &=\frac{-10 \pm 2}{6} \end{aligned}

\begin{aligned} &\mathrm{x}=\frac{+10+2}{6}\\ &=2\\ &\text { And }\\ &x=\frac{+10-2}{6}\\ &=\frac{4}{3} \end{aligned}

Differentiating $f(x)$ with respect to $x$

\begin{aligned} \mathrm{f}^{\prime \prime}(\mathrm{x}) &=\frac{\mathrm{d}}{\mathrm{dx}}\left(3 \mathrm{x}^{2}-10 \mathrm{x}+\mathrm{i}\right) \\ &=6 \mathrm{x}-10 \end{aligned}

When put x = 2 in f’’(x)

\begin{aligned} \mathrm{f}^{\prime \prime}(2) &=6(2)-10 \\ &=2>0 \end{aligned}

So x=2 is the local Minima

The Local Minimum value of f(x) at x =2 is 0

Put  $x=\frac{4}{3} \text { in } \mathrm{f}^{\prime \prime}(\mathrm{x})$

\begin{aligned} \mathrm{f}^{\prime \prime}(2) &=6(2)-10 \\ &=2>0 \end{aligned}

The local maximum value of $f(x)$at $x=\frac{4}{3}$   is $x=\frac{4}{27}$.