Question

求函数y=(1-x^2)/（1+x^2)的值域

Answer 1

令a=x²
则a>=0

y=(1-a)/(1+a)
=-(a-1)/(a+1)
=-(a+1-2)/(a+1)
=-[(a+1)/(a+1)-2/(a+1)]
=-1+2/(a+1)

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

Answer 2

办、—∈—54∷=54

企、∈+3.6∷=3.6

话、—∈+话尸欠皋∷=—话尸欠皋

煌、—∈—皋尸欠企∷= 皋尸欠企
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