#### Explain solution RD Sharma class 12 chapter Maxima and Minima exercise 17.4 question 1 sub question (iv) maths

Absolute Maximum $=14 \sqrt{2} \text { at } x=9$

Absolute Minimum $=\frac{-2}{3 \sqrt{3}} \text { at } x=4 / 3$

Hint:

Check the value of f(x) at critical and end points

Given:

$f(x)=(x-2) \sqrt{x-1} \text { in }(1,9)$

Explanation:

$f^{\prime}(x)=(x-2) \times \frac{1}{2 \sqrt{x-1}}+\sqrt{x-1}$

\begin{aligned} &=\frac{x-2+(\sqrt{x-1})(\sqrt{x-1})}{2 \sqrt{x-1}} \\\\ &=\frac{x-2+2(x-1)}{2 \sqrt{x-1}} \\\\ &=\frac{3 x-4}{2 \sqrt{x-1}} \end{aligned}

\begin{aligned} &f^{\prime}(x)=0 \\\\ &\frac{3 x-4}{2 \sqrt{x-1}}=0 \\\\ &3 x-4=0 \\\\ &x=\frac{4}{3} \end{aligned}

Now, check the value of $f(x) \text { at } x=1,4 / 3,9$

\begin{aligned} &f(1)=(1-2) \sqrt{1-1}=0 \\\\ &f\left(\frac{4}{3}\right)=\left(\frac{4}{3}-2\right) \sqrt{\frac{4}{3}-1} \end{aligned}

\begin{aligned} &=\frac{-2}{3} \sqrt{\frac{1}{3}}=\frac{-2}{3 \sqrt{3}} \\\\ &f(9)=(9-2) \sqrt{9-1} \\\\ &=7 \sqrt{8}=14 \sqrt{2} \end{aligned}

Hence,

Absolute maximum $=14 \sqrt{2} \text { at } x=9$

Absolute minimum $=\frac{-2}{3 \sqrt{3}} \text { at } x=4 / 3$