22. (10 points) Given $t>0$,
$$
\begin{array}{l}
f(t)=\sqrt{t}+\frac{1}{\sqrt{t}}+\sqrt{t+\frac{1}{t}+1}, \\
g(t)=\sqrt{t}+\frac{1}{\sqrt{t}}-\sqrt{t+\frac{1}{t}+1} .
\end{array}
$$

Find the minimum value of $f(t)$ and the maximum value of $g(t)$.
Three, 22. From the inequality $x+\frac{1}{x} \geqslant 2(x>0)$, we can get $f(t) \geqslant 2+\sqrt{3}$ (with equality if and only if $t=1$).
$\therefore$ The minimum value of $f(t)$ is $2+\sqrt{3}$.
Also, $\because g(t)=\frac{1}{f(t)}$,
$\therefore$ The maximum value of $g(t)$ is $\frac{1}{2+\sqrt{3}}$, which is $2-\sqrt{3}$.