14. If the function $f(x)$ defined on the set $A$ satisfies: for any $x_{1}, x_{2} \in A$, we have
$$
f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant \frac{1}{2}\left[f\left(x_{1}\right)+f\left(x_{2}\right)\right],
$$

then the function $f(x)$ is called a concave function on $A$.
(1) Determine whether $f(x)=3 x^{2}+x$ is a concave function on $\mathbf{R}$.
(2) If the function $f(x)=m x^{2}+x$ is a concave function on $\mathbf{R}$, find the range of real numbers for $m$.
14. (1) Given $f_{1}(x)=3 x_{1}^{2}+x_{1}$,
$$
\begin{array}{l}
f_{2}(x)=3 x_{2}^{2}+x_{2}, \\
f\left(\frac{x_{1}+x_{2}}{2}\right)=3\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\frac{x_{1}+x_{2}}{2},
\end{array}
$$

then $f\left(\frac{x_{1}+x_{2}}{2}\right)-\frac{1}{2}\left[f\left(x_{1}\right)+f\left(x_{2}\right)\right]$
$$
\begin{aligned}
= & 3\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\frac{x_{1}+x_{2}}{2}- \\
& \frac{1}{2}\left(3 x_{1}^{2}+x_{1}+3 x_{2}^{2}+x_{2}\right) \\
= & -3\left(\frac{x_{1}-x_{2}}{2}\right)^{2} \leqslant 0,
\end{aligned}
$$

i.e., $f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant \frac{1}{2}\left[f\left(x_{1}\right)+f\left(x_{2}\right)\right]$.
Thus, $f(x)=3 x^{2}+x$ is a concave function on $\mathbf{R}$.
(2) Since the function $f(x)=m x^{2}+x$ is a concave function on $\mathbf{R}$, for any $x_{1}, x_{2} \in \mathbf{R}$,
$$
\begin{array}{l}
f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant \frac{1}{2}\left[f\left(x_{1}\right)+f\left(x_{2}\right)\right] \\
\Rightarrow m\left(\frac{x_{1}+x_{2}}{2}\right)^{2}+\frac{x_{1}+x_{2}}{2} \\
\quad \leqslant \frac{1}{2}\left(m x_{1}^{2}+x_{1}+m x_{2}^{2}+x_{2}\right) \\
\Rightarrow m\left(\frac{x_{1}-x_{2}}{2}\right)^{2} \geqslant 0 .
\end{array}
$$

Therefore, $m \geqslant 0$.