14. Let \( f(x) = \frac{a^x}{1 + a^x} \) where \( a > 0, a \neq 1 \), and \([m]\) denotes the greatest integer not exceeding the real number \( m \). Find the range of the function \(\left[f(x) - \frac{1}{2}\right] + \left[f(-x) - \frac{1}{2}\right]\).
14. Let $g(x)=f(x)-\frac{1}{2}$, then $g(-x)=f(-x)-\frac{1}{2}=\frac{a^{-x}}{1+a^{-x}}-\frac{1}{2}=\frac{1}{a^{x}+1}-\frac{1}{2}=$ $\frac{\left(1+a^{x}\right)-a^{x}}{1+a^{x}}-\frac{1}{2}=\frac{1}{2}-\frac{a^{x}}{1+a^{x}}=-g(x)$, so $\left[f(x)-\frac{1}{2}\right]+\left[f(-x)-\frac{1}{2}\right]=$ $[g(x)]+[-g(x)]$. Since $a^{x}>0$, we have $0<\frac{a^{x}}{1+a^{x}}<1$, so $-\frac{1}{2}<g(x)=f(x)-\frac{1}{2}<\frac{1}{2}$. When $-\frac{1}{2}<g(x)<0$, $[g(x)]+[-g(x)]=-1+0=-1$; when $g(x)=0$, $[g(x)]+$ $[-g(x)]=0$; when $0<g(x)<\frac{1}{2}$, $[g(x)]+[-g(x)]=0+(-1)=-1$.