14. Given the functions $f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{x+\frac{1}{x}+1}(x>0), g(x)=\sqrt{x}+$ $\frac{1}{\sqrt{x}}-\sqrt{x+\frac{1}{x}+1}(x>0)$, find the minimum value of $f(x)$ and the maximum value of $g(x)$.
14. Let $t=\sqrt{x}+\frac{1}{\sqrt{x}}$, then $t \geqslant 2, f(x)$ becomes $y=t+\sqrt{t^{2}-1}$, which is an increasing function on $[2,+\infty]$. Therefore, when $t=2$, i.e., $x=1$, $f_{\min }=2+\sqrt{3}$. $g(x)$ becomes $y=t-\sqrt{t^{2}-1}=\frac{1}{t+\sqrt{t^{2}-1}}$, which is a decreasing function on $[2,+\infty)$. Therefore, when $t=2$, i.e., $x=1$, $g_{\text {max }}=2-\sqrt{3}$.