jmc

We can write $$f(x)=\sqrt{(x-5{)}^2+3^2}-\sqrt{x^2+4}.$$Let $P=(x,0),$ $A=(5,3),$ and $B=(0,2).$ Then $f(x)=PA-PB.$



By the Triangle Inequality, $PA\le AB+PB,$ so $$f(x)=PA-PB\le AB=\sqrt{26}.$$Equality occurs when $x=-10$ (which makes $P,$ $B,$ and $A$ collinear).

We can write $$\begin{array}{rl}f(x)& =\sqrt{x^2-10x+34}-\sqrt{x^2+4}\\ & =\frac{(\sqrt{x^2-10x+34}-\sqrt{x^2+4})(\sqrt{x^2-10x+34}+\sqrt{x^2+4})}{\sqrt{x^2-10x+34}+\sqrt{x^2+4}}\\ & =\frac{(x^2-10x+34)-(x^2+4)}{\sqrt{x^2-10x+34}+\sqrt{x^2+4}}\\ & =\frac{-10x+30}{\sqrt{x^2-10x+34}+\sqrt{x^2+4}}.\end{array}$$If $x\le 3,$ then $f(x)\ge 0,$ so assume $x>3,$ so $$f(x)=-10\cdot \frac{x-3}{\sqrt{x^2-10x+34}+\sqrt{x^2+4}}.$$If $3<x\le 5,$ then $$\frac{x-3}{\sqrt{x^2-10x+34}+\sqrt{x^2+4}}=\frac{x-3}{\sqrt{(x-5{)}^2+9}+\sqrt{x^2+4}}\le \frac{2}{3+4}=\frac{2}{7}<\frac{1}{2},$$so $f(x)>-5.$

If $x>5,$ then $$\begin{array}{rl}\frac{x-3}{\sqrt{x^2-10x+34}+\sqrt{x^2+4}}& =\frac{x-3}{\sqrt{(x-5{)}^2+9}+\sqrt{x^2+4}}\\ & <\frac{x-3}{x-5+x}\\ & =\frac{x-3}{2x-5}\\ & <\frac{x-3}{2x-6}=\frac{1}{2},\end{array}$$so $f(x)>-5.$

Furthermore, as $x$ becomes very large, $$\frac{x-3}{\sqrt{x^2-10x+34}+\sqrt{x^2+4}}=\frac{1-\frac{3}{x}}{\sqrt{1-\frac{10}{x}+\frac{34}{x^2}}+\sqrt{1+\frac{4}{x^2}}}$$approaches $\frac{1}{1+1}=\frac{1}{2},$ so $f(x)$ approaches $-5.$

Therefore, the range of $f(x)$ is $\boxed{(-5,\sqrt{26}]}.$