If ` f:[0,oo) to [0,1), " and " f(x)=(x)/(1+x)` then check the nature of the function.

Explanation

Given that `f:[0,oo) to [0, oo),f(x)=(x)/(x+1)`
Let `f(x_(1))=f(x_(2))`
`implies (x_(1))/(x_(1)+1)=(x_(2))/(x_(2)+1)`
`implies x_(1)x_(2)+x_(1)=x_(1)x_(2)+x_(2)`
`implies x_(1)=x_(2)`.
Thus f(x) is one-one.
Now let `y=(x)/(1+x)`
`implies y+yx=x`
`implies x=(y)/(1-y)`
As ` x ge 0, (y)/(1-y) ge 0`
`implies (y)/(1-y) le 0`
`implies 0 le y lt 1` or range of f(x) is `[0,1).`
Thus f(x) is onto.