Question

【急!】已知函数f(x)=x^2/x^2+1,求函数f(x)的值域.

Answer 1

f(x)=x²/(x²+1)=1-1/(x²+1)

因为x²+1≥1

所以0＜1/(x²+1)≤1

-1≤-1/(x²+1)＜0

-1+1≤1-1/(x²+1)＜0+1

即0≤1-1/(x²+1)＜1

所以f(x)的值域为[0,1)