Find the minimum value of the function
\[f(x) = \sqrt{-x^2 + 4x + 21} - \sqrt{-x^2 + 3x + 10}.\]
We can write the function as
\[f(x) = \sqrt{(7 - x)(3 + x)} - \sqrt{(5 - x)(2 + x)}.\]This shows that the function is defined only for $-2 \le x \le 5.$  Also, $(7 - x)(3 + x) - (5 - x)(2 + x) = x + 11 > 0$ on this interval, which means that $f(x)$ is always positive.

Then
\begin{align*}
[f(x)]^2 &= (7 - x)(3 + x) - 2 \sqrt{(7 - x)(3 + x)} \sqrt{(5 - x)(2 + x)} + (5 - x)(2 + x) \\
&= -2x^2 + 7x + 31 - 2 \sqrt{(7 - x)(2 + x)(5 - x)(3 + x)} \\
&= 2 + (7 - x)(2 + x) - 2 \sqrt{(7 - x)(2 + x)} \sqrt{(5 - x)(3 + x)} + (5 - x)(3 + x) \\
&= 2 + \left[ \sqrt{(7 - x)(2 + x)} - \sqrt{(5 - x)(3 + x)} \right]^2 \ge 2.
\end{align*}Therefore, $f(x) \ge \sqrt{2}.$

Equality occurs when $(7 - x)(2 + x) = (5 - x)(3 + x),$ or $x = \frac{1}{3}.$  We conclude that the minimum value is $\boxed{\sqrt{2}}.$