求函数y=2^x-1/2^x+1的值域

解:
令t=2^x>0
则
y=(t-1)/(t+1)
=1-2/(t+1)
因为t>0
-2<-2/(t+1)<0
-1<1-2/(t+1)<1
-1<t<1
值域是:(-1,1)

y=2^x-1/2^x+1=1-2/(2^x+1)

2^x>0
2^x+1>1
0<2/(2^x+1)<2
-2<-2/(2^x+1)<0
-1<y<1
值域为(-1,1)

解:
令t=2^x>0
则
y=(t-1)/(t+1)
=1-2/(t+1)
因为t>0
-2<-2/(t+1)<0
-1<1-2/(t+1)<1
-1<y<1
值域是:(-1,1)

令2^x=t(t>0)
y=(t-1)/(t+1)
=(t-1)/(t+1)
=1-2/(t+1)
又t>0
-2<-2/(t+1)<0
-1<1-2/(t+1)<1
-1<y<1
值域是:(-1,1)