Question

Show that the function f: R → {x ϵ R: -1 < x < 1} defined by f(x) =$\dfrac{{\text{x}}}{{1{\text{ + |x|}}}}$, x ϵ R is a one to one function and onto function.

Answer 1

Hint – To prove that the function is one – one and onto, we pick a positive and negative value from the set of values x can take and see if it satisfies the definitions.
Complete Step-by-Step solution:
f: R
→ {x ϵ R: -1 < x < 1}
X belongs to the set of rational numbers.
f(x) =$\dfrac{{\text{x}}}{{1{\text{ + |x|}}}}$, x ϵ R
Let’s check this relation for ±$\left( {\dfrac{1}{2}} \right)$ϵ$\left( { - 1,1} \right)$
$
   \Rightarrow {\text{f}}\left( {\dfrac{1}{2}} \right) = \dfrac{{\dfrac{1}{2}}}{{1 + \dfrac{1}{2}}} = \dfrac{1}{3} \\
  {\text{and f}}\left( { - \dfrac{1}{2}} \right) = \dfrac{{ - \dfrac{1}{2}}}{{1 - \dfrac{1}{2}}} = - 1 \\
$
Therefore, f(x) is different for each value in x ϵ (-1, 1)
⟹f is one – one.
Also, f(x) exists and gives a real value for all x ϵ (-1, 1)
⟹f is onto.
Hence f is one – one and onto.

Note – In order to solve this type of question the key is to know the definition of a one – one and onto function.
A function f is 1 -to- 1 if no two elements in the domain of f correspond to the same element in the range of f. In other words, each x in the domain has exactly one image in the range.
A onto function is expressing the relationship of a set to its image under a mapping when every element of the image set has an inverse image in the first set.