14. Given the functions $f(x)=1-|1-2 x|$ and $g(x)=(x-1)^{2}$ defined on the interval $[0,1]$, and $F(x)=\left\{\begin{array}{l}g(x), \text { if } f(x) \geqslant g(x) ; \\ f(x) \text {, if } f(x)<g(x) .\end{array}\right.$ Find: (1) The analytical expression of $F(x)$; (2) The maximum value of $F(x)$; (3) Solve the equation $F(x)=\frac{1}{3}$.
14. (1) \( f(x)=\left\{\begin{array}{l}2 x, 0 \leqslant x \leqslant \frac{1}{2} \\ 2-2 x, \frac{1}{2}<x \leqslant 1 .\end{array}\right. \) Draw the graphs of \( f(x) \) and \( g(x) \) in the same coordinate system. From the graph, we can see that \( F(x)=\left\{\begin{array}{l}2 x, 0 \leqslant x \leqslant 2-\sqrt{3} \\ (x-1)^{2}, 2-\sqrt{3}<x \leqslant 1 .\end{array}\right. \)
(2) \( F_{\max }=F(2-\sqrt{3})=4-2 \sqrt{3} \).
(3) If \( 2 x=\frac{1}{3} \), then \( x=\frac{1}{6} \in[0,2-\sqrt{3}] \); If \( (x-1)^{2}=\frac{1}{3} \), then we get \( x=1 \pm \frac{\sqrt{3}}{3} \), but \( 1+\frac{\sqrt{3}}{3} \notin(2-\sqrt{3}, 1] \), so we discard it. Therefore, \( x=\frac{1}{6} \) or \( x=1-\frac{\sqrt{3}}{3} \).