Find the maximum and minimum values of `x + sin2x` on `[0,2pi]`.

Explanation

Let ` f(x) = x + sin 2x`,
`rArr f'(x) = 1 + 2 cos 2x`
For maxima/minima, f'(x) = 0
` rArr 1+2 cos 2x = 0`
` rArr cos 2x = - 1/2`
` rArr cos 2x =- cos. Pi/3`
` rArr cos 2x = cos (pi- pi/3) or cos (pi+pi/3)`
or` cos (3pi-(pi)/3) or (3pi+(pi)/3)`
`rArr 2x = (2pi)/3, (4pi)/3, (8pi)/3, (10 pi)/3`
` rArr x = pi/3, (2pi)/3, (4pi)/3, (5pi)/3 in [0, 2 pi]`
Now, we find the values of f(x) at the points ` x = pi/3, (2pi)/3,(4pi)/3,(5pi)/3` and at the end points of the interval `[0, 2pi]`.
at ` x=0, f(0) = 0 + sin 0 =0`
at ` x =pi/3, f(pi/3) = pi/3+sin.(2pi)/3`
`= pi/3 + sin . pi/3=pi/3+sqrt3/2`
at ` x = (2pi)/2 , f((2pi)/3) = (2pi)/3 + sin .(4pi)/3 = (2pi)/3 - sin . pi/3`
`= (2pi)/3 - sqrt3/2`
at ` x = (4pi)/3, f ((4pi)/3) = (4pi)/3 + sin .(8pi)/3 = (4pi)/3 + sin.(2pi)/3`
`= (4pi)/3+sqrt3/2`
at ` x = (5pi)/3, f ((5pi)/3)= (5pi)/3 + sin .(10pi)/3 = (5pi)/3 - sin.(2pi)/3`
` = (5pi)/3 - sqrt3/2`
at ` x = 2 pi, f(2pi) = 2 pi + sin 4 pi = 2 pi + 0 = 2 pi`
So f has maximum value ` 2pi at x = 2 pi` and minimum value 0 at x = 0 .