Question

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

Answer 1

\[\text { Given }: f\left( x \right) = \left( x - 2 \right)\sqrt{x - 1}\]

\[ \Rightarrow f'\left( x \right) = \sqrt{x - 1} + \frac{\left( x - 2 \right)}{2\sqrt{x - 1}}\]

\[\text { For a local maximum or a local minimum, we must have }\]

\[f'\left( x \right) = 0\]

\[ \Rightarrow \sqrt{x - 1} + \frac{\left( x - 2 \right)}{2\sqrt{x - 1}} = 0\]

\[ \Rightarrow 2\left( x - 1 \right) + \left( x - 2 \right) = 0\]

\[ \Rightarrow 2x - 2 + x - 2 = 0\]

\[ \Rightarrow 3x - 4 = 0\]

\[ \Rightarrow 3x = 4 \]

\[ \Rightarrow x = \frac{4}{3} \]

\[\text { Thus, the critical points of f are } 1, \frac{4}{3} \text { and } 9 . \]

\[\text { Now }, \]

\[ f\left( 1 \right) = \left( 1 - 2 \right)\sqrt{1 - 1} = 0\]

\[ f\left( \frac{4}{3} \right) = \left( \frac{4}{3} - 2 \right)\sqrt{\frac{4}{3} - 1} = \frac{- 2}{3} \times \frac{1}{\sqrt{3}} = - \frac{2}{3\sqrt{3}}\]

\[ f\left( 9 \right) = \left( 9 - 2 \right)\sqrt{9 - 1} = 14\sqrt{2}\]

\[\text { Hence, the absolute maximum value when x } = 9 \text{ is }14\sqrt{2}\text { and the absolute minimum value when x } = \frac{4}{3} \text{ is }- \frac{2}{3\sqrt{3}} . \]