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(\(\frac{\pi}{6}\)) = \(-\frac{\sqrt3}{2}\) + \(\frac{\pi}{6}\)