Answer
The maximum is $f(2\pi) =2\pi$ and the minimum is $f( 0 ) =0$
Work Step by Step
Given $$y=x+\sin x, \quad[0,2 \pi]$$
Since
\begin{align*}
\frac{dy}{dt}&=1+\cos x
\end{align*}
To find critical points
\begin{align*}
f'(t)&=0\\
1+\cos x &=0
\end{align*}
Then $x= \pi$. Because $\pi\in [0,2\pi]$, then $f(x)$ has a critical point $x=\pi$ , since
\begin{align*}
f(0)&=0 \\
f(\pi )&= \pi\\
f(2\pi)&=2\pi
\end{align*}
Then the maximum is $f(2\pi) =2\pi$ and the minimum is $f( 0 ) =0$
