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 $[a, b]$, find $[a, b]$.
$$
[1,3],\left[-2-\sqrt{17}, \frac{13}{4}\right]
$$
14.【Analysis and Solution】From the given information, we can discuss in 3 cases:
(1) When $0 \leqslant a0, \therefore f(x)$ takes the minimum value $2a$ at point $a$, we have $2a=-\frac{1}{2}a^2+\frac{13}{2}, a=-2-\sqrt{17}, \therefore$ $[a, b]=\left[-2-\sqrt{17}, \frac{13}{4}\right]$.
(3) When $a<b \leqslant 0$, $f(x)$ is monotonically increasing on $[a, b]$, so $f(a)=2a, f(b)=2b$, we have $\left\{\begin{array}{l}2a=-\frac{1}{2}a^2+\frac{13}{2} \\ 2b=-\frac{1}{2}b^2+\frac{13}{2}\end{array}\right.$ but the roots of the equation $\frac{1}{2}x^2+2x-\frac{13}{2}=0$ are not of the same sign, hence there are no values that satisfy the condition in this interval.
$\therefore$ The required intervals are $[1,3],\left[-2-\sqrt{17}, \frac{13}{4}\right]$.