The function f : [0, ∞) → R given by f(x) =x/(x+1), is

We have,

$f(x) =\frac{x}{x+1}=1-\frac{1}{1+x}$

Let $x, y ∈ R$ be such that $f(x) = f(y)$. Then,

$f(x) = f(y)$

$⇒1-\frac{1}{1+x}=1-\frac{1}{1+y}⇒1+x=1+y⇒x=y$

∴ f is one-one.

Clearly, f is not onto as f takes only values less than 1 i.e. Range f = (-∞, 1) ≠ co-domain of f.