Find the maximum and minimum values of `f(x)=x+sin2x` in the interval `[0,\ 2pi]`

Explanation

Let `f(x)=x+sin2x`
`f'(x)=1+2 cos2x`
Now, `f'(x)=0 implies cos2x=-1/2=-cos( pi/3)=cos(pi-pi/3)=cos ((2pi)/3)`
`2x=2pi pm (2pi)/3 , n in Z`
`implies x=npipm pi/3 , n in Z`
`implies x=pi/3,2 (pi/3), 4(pi/3),5(pi/3) in [0,2pi]`
`f(pi/3)=pi/3+sin2(pi/3)=pi/3+sqrt3/2`
`f(2(pi/3))=2(pi/3)+sin4(pi/3)=2(pi/3)-sqrt3/2`
`f(4(pi/3))=4(pi/3)+sin8(pi/3)=4(pi/3)+sqrt3/2`
`f(5(pi/3))=5(pi/3)+sin10(pi/3)=5(pi/3)-sqrt3/2`
`f(0)=0+sin0=0`
`f(2pi)=2pi+sin4pi=2pi+0=2pi`
Hence we can conclude that the absolute maximum value of f(x) in the interval `[0,2pi]`is `2pi`occuring at `x=2pi` and the absolute minimum value of `f(x)` in the interval `[0,2pi]` is 0 occurring at `x=0`