jmc

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 $$\begin{array}{rl}[f(x){]}^2& =(7-x)(3+x)-2\sqrt{(7-x)(3+x)}\sqrt{(5-x)(2+x)}+(5-x)(2+x)\\ & =-2x^2+7x+31-2\sqrt{(7-x)(2+x)(5-x)(3+x)}\\ & =2+(7-x)(2+x)-2\sqrt{(7-x)(2+x)}\sqrt{(5-x)(3+x)}+(5-x)(3+x)\\ & =2+{\left[\sqrt{(7-x)(2+x)}-\sqrt{(5-x)(3+x)}\right]}^2\ge 2.\end{array}$$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 $\boxed{\sqrt{2}}.$