If f: [0, ∞)→ [0, 1), and f(x) = x/(1+x) then check the nature of the…

Given that $f: [0, ∞) → [0, ∞), f(x) = \frac{x}{1+x}$

Let $f(x_1) = f(x_2)$

$⇒\frac{x_1}{x_1+1}=\frac{x_2}{x_2+1}$

$⇒x_1x_2+ x_1=x_1x_2 + x_2$

$⇒x_1=x_2$

Thus f(x) is one-one.

Now let $y =\frac{x}{1+x}$

$⇒y+yx = x$

$⇒x=\frac{y}{1-y}$

As $x≥0, \frac{y}{1-y}≥0$

$⇒ 0≤y<1$ or range of f(x) is [0, 1)

Thus f(x) is onto.