Question

f(x) = \[x + \sqrt{1 - x}, x \leq 1\] .

Answer 1

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

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

\[\text { For the local maxima or minima, we must have }\]

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

\[ \Rightarrow 1 - \frac{1}{2\sqrt{1 - x}} = 0\]

\[ \Rightarrow \sqrt{1 - x} = \frac{1}{2}\]

\[ \Rightarrow 1 - x = \frac{1}{4} \]

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

\[\text { Thus }, x = \frac{3}{4} \text { is the possible point of local maxima or local minima }. \]

\[\text { Now }, \]

\[f''\left( x \right) = - \frac{\frac{1}{4\sqrt{1 - x}}}{4\left( 1 - x \right)}\]

\[\text { At }x = \frac{3}{4}: \]

\[ f''\left( \frac{3}{4} \right) = - \frac{\frac{1}{4\sqrt{1 - \frac{3}{4}}}}{4\left( 1 - \frac{3}{4} \right)} = - \frac{1}{2} < 0\]

\[\text { So,} x = \frac{3}{4} \text { is the point of local maximum }. \]

\[\text { The local maximum value is given by }\]

\[f\left( \frac{3}{4} \right) = \frac{3}{4} + \sqrt{1 - \frac{3}{4}} = \frac{5}{4}\]