4. If the function $f(x)=\frac{1}{3} x^{2}+\frac{2}{3} x-\frac{11}{3}$ has both its domain and range as the closed interval $M$, then $M=$
4. $M=[-4,-1]$ or $M=\left[-4, \frac{1+3 \sqrt{5}}{2}\right]$

Analysis: It is easy to know that the vertex coordinates of the parabola are $(-1,-4)$. We discuss in three cases:
(1) If $M$ is to the right of the axis of symmetry, then the roots of $\frac{1}{3} x^{2}+\frac{2}{3} x-\frac{11}{3}=x$ are $\frac{1 \pm 3 \sqrt{5}}{2}$. Since $\frac{1-3 \sqrt{5}}{2}<-1$, it does not meet the requirements;
(2) If $M$ is to the left of the axis of symmetry, let $M=[a, b]$, then $f(a)=b, f(b)=a$, solving gives $a=-4, b=-1$, i.e., $M=[-4,-1]$;
(3) If the axis of symmetry is within $M$, then $a=-4$, and since $f(-4)=-1$, we have $b=\frac{1+3 \sqrt{5}}{2}$, i.e., $M=\left[-4, \frac{1+3 \sqrt{5}}{2}\right]$.

Combining (1), (2), and (3), we know that $M=[-4,-1]$ or $M=\left[-4, \frac{1+3 \sqrt{5}}{2}\right]$.