Example 11 For real number $x$, find the maximum value of the function $f(x)=$ $\sqrt{8 x-x^{2}}-\sqrt{14 x-x^{2}-48}$.
Solution 1: The domain of $f(x)$ is $[6,8]$, $u(x)=$ $\sqrt{8 x-x^{2}}=\sqrt{16-(x-4)^{2}}$, when $x=6$, $u_{\max }=\sqrt{12} ; v(x)=-\sqrt{14 x-x^{2}-48}=$ $-\sqrt{1-(x-7)^{2}}$, when $x=6$, $v_{\operatorname{mex}}=0$, thus when $x=6$, $f(x)$ has a maximum value of $\sqrt{12}=2 \sqrt{3}$.

Solution 2: The domain of $f(x)$ is $[6,8]$, let $u(x)=$ $\sqrt{8 x-x^{2}}, v(x)=\sqrt{14 x-x^{2}-48}, u^{2}-v^{2}=48$ $-6 x: \because x \in[6,8], \therefore 0 \leqslant 48-6 x \leqslant 12$, $\because 0 \leqslant u^{2}-v^{2} \leqslant 12 \quad$ (1). $\because y=u-v$, $\therefore u=y+v$ substitute into (1) to get: $y^{2}+2 v y \leqslant 12$, it is easy to know $y \geqslant 0, v=\sqrt{1-(x-7)^{2}} \geqslant 0 \quad$ (2) $\therefore y^{2} \leqslant y^{2}+2 v y \leqslant 12, \therefore y \leqslant 2 \sqrt{3}$, when $x=6$, (1) and (2) simultaneously take the equality. Therefore, $f(x)$ has a maximum value of $\sqrt{12}$ $=2 \sqrt{3}$.

Solution 3: The domain of $f(x)$ is $[6,8], f(x)=$ $\sqrt{8-x}(\sqrt{x}-\sqrt{x-6})=\frac{6 \sqrt{8-x}}{\sqrt{x}+\sqrt{x-6}}$,
$\because \sqrt{8-x}, \frac{1}{\sqrt{x}+\sqrt{x-6}}$ are decreasing functions on $[6,8]$, thus when $x=6$ $f(x)$ has a maximum value of $\sqrt{12}=2 \sqrt{3}$.