We have,
f(x) = 2 sin x – x
Differentiate w.r.t x, we get,
f‘(x) = 2cos x – 1 = 0
For, the point of local maxima and minima,
f’(x) = 0
cos x = \(\frac{1}{2}\) = cos\(\frac{\pi}{3}\)
= x = \(-\frac{\pi}{3}\),\(\frac{\pi}{3}\)
At x = \(-\frac{\pi}{3}\)
f’(x) changes from –ve to + ve
Since,
x = \(-\frac{\pi}{3}\) is a point of Minima with value = \(-\sqrt 3\) \(-\frac{\pi}{3}\)
At x = \(\frac{\pi}{3}\)
f‘(x) changes from –ve to + ve
Since,
x = \(\frac{\pi}{3}\) is point of local maxima with value = \(\sqrt 3\) \(-\frac{\pi}{3}\)
Hence, local max value f(\(\frac{\pi}{3}\)) = \(\sqrt 3\) \(-\frac{\pi}{3}\)
local min value f(\(-\frac{\pi}{3}\)) = \(-\sqrt 3\) \(-\frac{\pi}{3}\)