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Maxima and Minima — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSMaxima and Minima5 Marks
Question
Find the absolute maximum and the absolute minimum value of the following functions in the given intervals:
$\text{f}(\text{x})=(\text{x}-2)\sqrt{\text{x}-1}\ \text{in}\ [1,9]$
✓

Answer

Given, $\text{f}(\text{x})=(\text{x}-2)\sqrt{\text{x}-1}\ $
$\Rightarrow\text{f}'(\text{x})=\sqrt{\text{x}-1}+\frac{(\text{x}-2)}{2\sqrt{\text{x}-1}} $
For a local maximum or a local minimum, We must have f'(x) = 0
$\Rightarrow\sqrt{\text{x}-1}+\frac{(\text{x}-2)}{2\sqrt{\text{x}-1}}=0 $
$ \Rightarrow2(\text{x}-1)+(\text{x}-2)=0$
$\Rightarrow2\text{x}-2+\text{x}-2=0$
$\Rightarrow3\text{x}-4=0$
$\Rightarrow3\text{x}=4$
$\Rightarrow\text{x}=\frac{4}{3}$
Thus, the critical points of f are $1,\frac{4}{3}$ and 9.
Now, $ \text{f}(1)=(1-2)\sqrt{1-1}=0$ 
$\text{f}\Big(\frac{4}{3}\Big)=\Big(\frac{4}{3}-2\Big)\sqrt{\frac{4}{3}-1}=\frac{-2}{3}\times\frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}$
$\text{f}(9)=(9-2)\sqrt{9-1}=14\sqrt{2}$
Hence, the absolute maximum value when x = 9 is $14\sqrt{2}$  and the absolute minimum value when $\text{x}=\frac{4}{3}$ is $-\frac{2}{3\sqrt{3}}$.

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