Question

已知函数f(x)=(10^x-10^-x)/(10^x+10^-x)判断f(x)的值域

Answer 1

解f(x)=(10^x-10^-x)/(10^x+10^-x)

=(（10^x）^2-1)/((10^x)^2+1)

令t=10^x t＞0

则原函数变为y=（t^2-1)/(t^2+1)

=（t^2+1-2)/(t^2+1)

=1-2/(t^2+1)

由t＞0

故t^2＞0

即t^2+1＞1

即0＜1/(t^2+1)＜1

即0＜2/(t^2+1)＜2

即-2＜-2/(t^2+1)＜0

即-1＜1-2/(t^2+1)＜1

故-1＜y＜1

故原函数的值域为（-1,1）.