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Find the Maximum and Minimum Values of Y = Tan X − 2 X . - Mathematics

Question

Find the maximum and minimum values of y = tan $x - 2x$ .

Solution

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

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

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

$f'\left( x \right) = 0$

$\Rightarrow \sec^2 x - 2 = 0$

$\Rightarrow \sec^2 x = 2$

$\Rightarrow \sec x = \pm \sqrt{2}$

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

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

$\text { Now,}$

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

$\text { At }x = \frac{\pi}{4}:$

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

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

$\text { The local minimum value is given by }$

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

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

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

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

$\text { The local maximum value is given by }$

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

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