14. If the function $f(x)=-\frac{1}{2} x^{2}+\frac{13}{2}$ has a minimum value of $2a$ and a maximum value of $2b$ on the interval $\left.a, b\right]$, find $[a, b]$.
14. Discuss the interval $[a, b]$ in 3 cases.
(1) If $0 \leqslant a < b$,
$$
f(x) \text{ is monotonically increasing on } [a, b], \text{ so } f(a) = 2a, f(b) = 2b, \text{ i.e., }
$$
$$
2a = -\frac{1}{2}a^2 + \frac{13}{2}, 2b = -\frac{1}{2}b^2 + \frac{13}{2}.
$$

Solving the equation $\frac{1}{2}x^2 + 2x - \frac{13}{2} = 0$, we get $x = 1$ or $x = 3$. Therefore,
$$
[a, b] = [1, 3].
$$
(2) If $a < 0 < b$,
$$
f(x) \text{ is monotonically decreasing on } [a, 0] \text{ and monotonically increasing on } [0, b].
$$

Thus, $f(x)$ attains its minimum value at $x = a$, which is $2a$. That is,
$$
2a = -\frac{1}{2}a^2 + \frac{13}{2}, \text{ solving for } a \text{ gives } a = -2 \pm \sqrt{17}.
$$

Therefore,
$$
[a, b] = \left[-2 - \sqrt{17}, \frac{13}{4}\right].
$$
(3) When $a < b \leqslant 0$, $f(x)$ is monotonically increasing on $[a, b]$, so $f(a) = 2a, f(b) = 2b$, i.e.,
$$
2a = -\frac{1}{2}a^2 + \frac{13}{2}, 2b = -\frac{1}{2}b^2 + \frac{13}{2}.
$$

Since the roots of the equation $\frac{1}{2}x^2 + 2x - \frac{13}{2} = 0$ have opposite signs, there is no interval satisfying $a < b < 0$.
In summary, the required intervals are
$$
[1, 3] \text{ or } \left[-2 - \sqrt{17}, \frac{13}{4}\right].
$$