#### Question

Find the maximum and minimum values of y = tan \[x - 2x\] .

#### Solution

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

\[ \Rightarrow f'\left( x \right) = \sec^2 x - 2\]

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

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

\[ \Rightarrow \sec^2 x - 2 = 0\]

\[ \Rightarrow \sec^2 x = 2\]

\[ \Rightarrow \sec x = \pm \sqrt{2}\]

\[ \Rightarrow x = \frac{\pi}{4} \text { and } \frac{3\pi}{4}\]

\[\text { Thus, x }= \frac{\pi}{4} \text { and }x = \frac{3\pi}{4}\text { are the possible points of local maxima or a local minima } . \]

\[\text { Now,} \]

\[f''\left( x \right) = 2 \sec^2 x \tan x\]

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

\[ f''\left( \frac{\pi}{4} \right) = 2 \sec^2 \left( \frac{\pi}{4} \right) \tan \left( \frac{\pi}{4} \right) = 4 > 0\]

\[\text { So, }x = \frac{\pi}{4} \text { is a point of local minimum } . \]

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

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

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

\[ f''\left( \frac{3\pi}{4} \right) = 2 \sec^2 \left( \frac{3\pi}{4} \right) \tan \left( \frac{3\pi}{4} \right) = - 4 < 0\]

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

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

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