Explain Solution R.D.Sharma Class 12 Chapter 17  Maxima and Minima  Exercise 17.3 Question 2 Sub Question 1 Maths Textbook Solution.

x=2 is the local minima and the local minimum value of $\mathrm{f}(\mathrm{x}) \text { at } \mathrm{x}=2 \text { is } 0$ .

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

Using chain rule

Given:

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

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} \end{aligned}            [Using chain rule]

\begin{aligned} &=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}

Put$\mathrm{f}^{\prime}(\mathrm{x})=0$

\begin{aligned} &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} x &=\frac{+10+2}{6} \\ &=2 \end{aligned}

And

\begin{aligned} 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 $x=2 \text { in } f^{\prime \prime}(x)$

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

So $x=2$ is local minima

The local minimum value of $\mathrm{f}(\mathrm{x}) \text { at } \mathrm{x}=2 \text { is } 0$

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

\begin{aligned} &\mathrm{f}^{\prime \prime}\left(\frac{4}{3}\right)=6\left(\frac{4}{3}\right)-10 \\ &=8-10 \\ &=-2<0 . \text { So } \mathrm{x}=\frac{4}{3} \end{aligned}is local maxima

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