14. (25 points) (1) If $0 \leqslant x \leqslant 1$, prove:
$$
x-\frac{3}{8} \leqslant 2 x^{4} \leqslant 2 x \text {; }
$$
(2) If $x, y, z \geqslant 0$, and $x+y+z=1$, find
$$
f(x, y, z)=2 x^{4}+y^{2}+z
$$

the maximum and minimum values.
14. (1) Since $0 \leqslant x \leqslant 1$, we have
$$
\begin{array}{l}
2 x^{4}-2 x=2 x(x-1)\left(x^{2}+x+1\right) \leqslant 0, \\
2 x^{4}-\left(x-\frac{3}{8}\right)=\frac{1}{8}(2 x-1)^{2}\left(4 x^{2}+4 x+3\right) \geqslant 0 .
\end{array}
$$

Therefore, $x-\frac{3}{8} \leqslant 2 x^{4} \leqslant 2 x$.
[Note] The result can also be proven using derivatives.
(2) On one hand,
$$
\begin{array}{l}
f(x, y, z)=2 x^{4}+y^{2}+z \\
\leqslant 2\left(x^{4}+y^{2}+z\right) \leqslant 2(x+y+z)=2,
\end{array}
$$

with equality holding if and only if $x=1, y=z=0$.
Thus, the maximum value of $f(x, y, z)$ is 2.
On the other hand,
$$
\begin{array}{l}
f(x, y, z)=2 x^{4}+y^{2}+z \\
\quad=2 x^{4}+y^{2}+(1-x-y) \\
=\left(2 x^{4}-x\right)+\left(y^{2}-y\right)+1 \\
\geqslant\left(-\frac{3}{8}\right)+\left(-\frac{1}{4}\right)+1=\frac{3}{8},
\end{array}
$$

with equality holding if and only if $x=y=\frac{1}{2}, z=0$.
Thus, the minimum value of $f(x, y, z)$ is $\frac{3}{8}$.