2. If $x, y \geqslant 0$, and $x+y=1$, then the range of $\sqrt[3]{7 x+1}+$ $\sqrt[3]{7 y+1}$ is $\qquad$ .
2. $[3, \sqrt[3]{36}]$.

Let $f(x)=\sqrt[3]{7 x+1}(x \geqslant 0)$. Then
$$
\begin{array}{l}
f^{\prime}(x)=\frac{7}{3}(7 x+1)^{-\frac{2}{3}}, \\
f^{\prime \prime}(x)=-\frac{98}{9}(7 x+1)^{-\frac{5}{3}} \leqslant 0 .
\end{array}
$$

By Jensen's inequality, we have
$$
\begin{array}{l}
\frac{f(x)+f(y)}{2} \leqslant f\left(\frac{x+y}{2}\right) \\
=f\left(\frac{1}{2}\right)=\sqrt[3]{\frac{9}{2}} .
\end{array}
$$

Therefore, $\sqrt[3]{7 x+1}+\sqrt[3]{7 y+1} \leqslant \sqrt[3]{36}$, with equality when $x=y=\frac{1}{2}$.
Notice that,
$$
\begin{array}{l}
\sqrt[3]{7 x+1}=\sqrt[3]{1+3 x+3 x+x} \\
\geqslant \sqrt[3]{1+3 x+3 x^{2}+x^{3}}=x+1 .
\end{array}
$$

Similarly, $\sqrt[3]{7 y+1} \geqslant y+1$.
In summary, the range sought is $[3, \sqrt[3]{36}]$.