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Find the points of local maxima or local minima, if any, of the following functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be :

f(x) = sin 2x - x, \(-\frac{\pi}{2}\)≤ x ≤ \(\frac{\pi}{2}\)

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Best answer

We have, 

f(x) = sin 2x – x 

Differentiate w.r.t x, we get, 

f‘(x) = 2cos 2x – 1, 

For, the point of local maxima and minima, 

f’(x) = 0 

2cos2x – 1 = 0

= cos 2x = \(\frac{1}{2}\) = cos\(\frac{\pi}{3}\)

= 2x = \(\frac{\pi}{3}\),\(-\frac{\pi}{3}\)

= x = \(\frac{\pi}{6}\),\(-\frac{\pi}{6}\)

At x = \(-\frac{\pi}{6}\)

f’(x) changes from –ve to + ve

Since, 

x = \(-\frac{\pi}{6}\) is a point of Maxima

At x = \(\frac{\pi}{6}\) 

f‘(x) changes from –ve to + ve

Since, 

x = \(\frac{\pi}{6}\) is point of local maxima

Hence, local max value f(\(\frac{\pi}{6}\)) = \(-\frac{\sqrt3}{2}\) - \(\frac{\pi}{6}\) 

local min value f(\(\frac{\pi}{6}\)) = \(-\frac{\sqrt3}{2}\) + \(\frac{\pi}{6}\) 

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