Variation 4 Given $a, b, c, d \in \mathbf{R}^{+}$, determine the monotonicity of the function $f(x)=a \sqrt{c-x}+b \sqrt{x+d}$.
Solve: From $c-x \geqslant 0, x+d \geqslant 0$, we get the domain of the original function as $-d \leqslant x \leqslant c$.

Differentiating the function $f(x)$, we get
$$f^{\prime}(x)=-\frac{a}{2 \sqrt{c-x}}+\frac{b}{2 \sqrt{x+d}}$$

From $f^{\prime}(x)=0$, we get $x=\frac{b^{2} c-a^{2} d}{a^{2}+b^{2}}$.
When $x \in\left[-d, \frac{b^{2} c-a^{2} d}{a^{2}+b^{2}}\right)$, $f^{\prime}(x)>0$,
at this time, the function $f(x)$ is an increasing function;
When $x \in\left(\frac{b^{2} c-a^{2} d}{a^{2}+b^{2}}, c\right]$, $f^{\prime}(x)<0$, at this time, the function $f(x)$ is a decreasing function.

Therefore, the function $f(x)=a \sqrt{c-x}+b \sqrt{x+d}$ is increasing on $\left[-d, \frac{b^{2} c-a^{2} d}{a^{2}+b^{2}}\right]$ and decreasing on $\left[\frac{b^{2} c-a^{2} d}{a^{2}+b^{2}}, c\right]$.