f(x)=a-2/2^x+1设x1＜x2f(x1)-f(x2)=a-2/2^x1+1-(a-2/2^x2+1)=-2/2^x1+2/2^x2=2(2^x1-2^x2)/(2^x1*2^x2)∵x1＜x2∴2^x1-2^x2＜0∵2^x1*2^x2＞0∴f(x1)-f(x2)＜0∴f(x1)＜(x2)∴f(x)是增函数