Question

函数f(x)=(3^x-1)/(3^x+1)的值域是什么

Answer 1

令a=3^x
则a>0
y=f(x)=(a-1)/(a+1)
=(a+1-2)/(a+1)
=(a+1)/(a+1)-2/(a+1)
=1-2/(a+1)

a+1>1
0<1/(a+1)<1
-2<-2/(a+1)<0
1-2<1-2/(a+1)<1+0
所以值域(-1,1)

Answer 2

令t=3^x,则t>0
原函数化为f(t)=(t-1)(t+1)
f'(t)=2t=0 --> t=0 -->f(0)=-1,f(+∞)=+∞

故值域为(0,+∞)