Example 7 (2000 National College Entrance Examination for Two Provinces and One Municipality) Let the function $f(x)=\sqrt{x^{2}+1}-a x$, where $a>0$, find the range of real number $a$ so that $f(x)$ is a monotonic function on $(0,+\infty)$.

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Solve $f^{\prime}(x)=\left(\sqrt{x^{2}+1}\right)^{\prime}-(a x)^{\prime}=\frac{1}{2} \frac{2 x}{\sqrt{x^{2}+1}}-a=\frac{x}{\sqrt{x^{2}+1}}-a$.
Since $x>0$,
then $00$, i.e., $x=-\frac{a}{\sqrt{1-a^{2}}}$ (discard), we get the critical point $x=\frac{a}{\sqrt{1-a^{2}}}$. Thus,
when $x \in\left(0, \frac{a}{\sqrt{1-a^{2}}}\right)$, $f^{\prime}(x)0$.
Therefore, $f(x)$ is a decreasing function on $\left(0, \frac{a}{\sqrt{1-a^{2}}}\right)$ and an increasing function on $\left(\frac{a}{\sqrt{1-a^{2}}},+\infty\right)$,
i.e., when $0<a<1$, $f(x)$ is not a monotonic function on $(0,+\infty)$.
Combining (1) and (2), we know that when $a \geqslant 1$, $f(x)$ is a decreasing function on $(0,+\infty)$; when $0<a<1$, $f(x)$ is not a monotonic function on $(0,+\infty)$.