f(x)=1/(2^x-1)+1/2 吗? 还是 f(x)=1/2^(x-1)+1/2 ?是F（X）= [1/( 2X—1)]+1/2，这样看清楚了吗^表示次方f(x)=1/(2^x-1)+1/2f(-x)=1/(2^(-x)-1)+1/2=2^x/(1-2^x)+1/2=(2*2^x+(1-2^x))/(1-2^x)/2=-(2^x+1)/(2^x-1)/2=-(2^x-1+2)/(2^x-1)/2=-(1+2/(2^x-1))/2=-(1/2+1/(2^x-1))=-f(x)所以 f(x)是奇函数 f'(x)=2^xln2/(2^x-1)^2>0所以 f(x)单调递增