Question

求y=x²/(x²+1)的值域

解：y=1/[(x²+1)/x²]
=1/(1+1/x²)
然后怎么做？

Answer 1

解：y=1/[(x²+1)/x²]
=1/(1+1/x²)
因为1/x²>0
所以1/(1+1/x²)<1
所以0≤y<1
另解：（有界函数法）
x^2=yx^2+y
x^2=y/(1-y)≥0
y(y-1)≤0(y≠1)
0≤y＜1

Answer 2

解：将函数变形
x^2=yx^2+y
x^2-yx^2=y
x^2=y/(1-y)≥0
y(y-1)≤0(y≠1)
0≤y＜1