Example 14 (2000 National High School League Question) 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 $[a, b]$.
Solve by discussing the interval $[a, b]$ in three cases.
(1) If $0 \leqslant a < b$, then $f(x)$ is monotonically decreasing on $[a, 0]$ and monotonically increasing on $[0, b]$. Therefore, the minimum value of $f(x)$ is $f(0) = 0$, which contradicts the given condition. Hence, this case is not possible.
(2) If $a < 0 < b$, then $f(x)$ is monotonically decreasing on $[a, 0]$ and monotonically increasing on $[0, b]$. The minimum value of $f(x)$ is $f(0) = 0$, which is not possible since $f(x) > 0$ for $x \neq 0$. Therefore, $f(x)$ must take its minimum value at $x = a$, which is $2a$, i.e., $2a = -\frac{1}{2}a^2 + \frac{13}{2}$. Solving this, we get $a = -2 - \sqrt{17}$. Thus, we have $[a, b] = \left[-2 - \sqrt{17}, \frac{13}{4}\right]$.
(3) When $a < b \leqslant 0$, $f(x)$ is monotonically increasing on $[a, b]$, so $f(a) = 2a$ and $f(b) = 2b$, i.e., $2a = -\frac{1}{2}a^2 + \frac{13}{2}$ and $2b = -\frac{1}{2}b^2 + \frac{13}{2}$. Since the roots of the equation $\frac{1}{2}x^2 + 2x - \frac{13}{2} = 0$ are of opposite signs, there is no interval satisfying $a < b < 0$.
In summary, the required intervals are $[1, 3]$ or $\left[-2 - \sqrt{17}, \frac{13}{4}\right]$.