We have,
f(x) = sin 2x
Differentiate w.r.t x, we get,
f ‘(x) = 2cos 2x, 0 < x,π
For, the point of local maxima and minima,
f ’(x) = 0
= 2x = \(\frac{\pi}{2}\),\(\frac{3\pi}{2}\)
= x = \(\frac{\pi}{4}\),\(\frac{3\pi}{4}\)
At x = \(\frac{\pi}{4}\)
f ’(x) changes from –ve to + ve
Since,
x = \(\frac{\pi}{4}\) is a point of Maxima
At x = \(\frac{3\pi}{4}\)
f ‘(x) changes from –ve to + ve
Since,
x = is point of Minima.
Hence, local max value f (\(\frac{\pi}{4}\)) = 1
local min value f (\(\frac{3\pi}{4}\)) = – 1