Let f:[0,∞)→ℝ defined by f(x)=x/(x+3) , then f is

$f(x_1)=f(x_2)⇒\frac{x_1}{x_1+3}=\frac{x_2}{x_2+3}⇒x_1x_2+3x_1=x_1x_2+3x_2$

$⇒x_1=x_2$ one-one nature

$y=\frac{x}{x+3}⇒yx+3y=x$

so $x=\frac{3y}{y-1}$

so for $y=1$ 

x doesn't exist