If f : ( 0 , ∞ ) → ( 0 , ∞ ) and f x = x 1 + x , then f is

Step-by-step solution

Here, f : 0 ,   ∞ → 0 ,   ∞ , i.e. domain is ( 0 ,   ∞ ) and codomain is 0 ,   ∞ . For one-one, f x = x 1 + x ⇒     f ⁡ ′ x = 1 1 + x 2 > 0 ,   ∀   x ∈ 0 ,   ∞ ∴    f ( x ) is increasing in its domain. Thus, f ( x ) is one-one in its domain. For onto (we find range) f x = x 1 + x   ⇒   y = x 1 + x  ⇒  y + y x = x ⇒    x = y 1 - y