1,因为f(x)=(a*2^x -1)/1+2^x (a∈R)是R上的奇函数,所以
f(x)+f(-x)=0,( a*2^x -1)/(1+2^x)+[a*2^(-x) -1] / [1+2^(-x)]=0,
(a*2^x -1)/1+2^x+(a-2^x )/(2^x+1)=0,(a-1)(2^x+1)/(2^x+1)=0,
解得:a=1.函数f(x)为:f(x)=(2^x-1)/(1+2^x).
2,令实数x1、x2,x1
1,因为f(x)=(a*2^x -1)/1+2^x (a∈R)是R上的奇函数,所以
f(x)+f(-x)=0,( a*2^x -1)/(1+2^x)+[a*2^(-x) -1] / [1+2^(-x)]=0,
(a*2^x -1)/1+2^x+(a-2^x )/(2^x+1)=0,(a-1)(2^x+1)/(2^x+1)=0,
解得:a=1.函数f(x)为:f(x)=(2^x-1)/(1+2^x).
2,令实数x1、x2,x1