Examples of Functions

Example:
Find the range of the function f\left( {\text{x}} \right) =  \frac{{{\text{x}} + 1}}{{{\text{x}} - 1}}.
Solution:
We have

f\left( {\text{x}} \right) =  \frac{{{\text{x}} + 1}}{{{\text{x}} - 1}}


Put{\text{x}} = 1

f\left( {\text{1}} \right) = \frac{{{\text{1}}  + 1}}{0} = \infty


Thus, the domain is, \forall \;{\text{x}} \in \mathbb{R} - \left\{ 1  \right\}.
Now for the range, we have

\begin{gathered} f\left( {\text{x}} \right) =  \frac{{{\text{x}} + 1}}{{{\text{x}} - 1}} \\ \therefore {\text{y}} = \frac{{{\text{x}} +  1}}{{{\text{x}} - 1}} \\ \Rightarrow {\text{xy}} - {\text{y}} =  {\text{x}} + 1 \\ \Rightarrow {\text{xy}} - {\text{x}} =  {\text{y}} + 1 \\ {\text{x}} = \frac{{{\text{y}} +  1}}{{{\text{y}} - 1}} \\ \end{gathered}


For {\text{y}} = 1

{\text{x}} = \frac{{{\text{1}} + 1}}{0} = \infty


So, the range of the function fis \left\{ {{\text{y}}:{\text{y}} \ne 1} \right\} =  \left] { - \infty ,1} \right[\;\; \cup \;\;\left] {1,\infty } \right[.
Example:
Let f\left( {\text{x}} \right) =  \frac{{\text{x}}}{{{{\text{x}}^2} - 16}}. Find the domain and range of f.
Solution:
We have

f\left( {\text{x}} \right) =  \frac{{\text{x}}}{{{{\text{x}}^2} - 16}}


For{\text{x}} = 4

f\left( 4 \right) = \frac{4}{{16 -  16}} = \infty


For{\text{x}} =   - 4

f\left( { - 4} \right) = \frac{{ -  4}}{{16 - 16}} = \infty


Thus, the domain is\forall \;{\text{x}} \in \mathbb{R} - \left\{ {4, -  4} \right\}.
Now for the Range, we have

\begin{gathered} f\left( {\text{x}} \right) =  \frac{{\text{x}}}{{{{\text{x}}^2} - 16}} \\ \Rightarrow {\text{y}} =  \frac{{\text{x}}}{{{{\text{x}}^2} - 16}} \\ \Rightarrow {\text{y}}\left( {{{\text{x}}^2}  - 16} \right) = {\text{x}} \\ \Rightarrow {\text{y}}{{\text{x}}^2} -  {\text{x}} - 16{\text{y}} = 0 \\ \Rightarrow {\text{x}} = \frac{{ - \left( {  - 1} \right) \pm \sqrt {{{\left( { - 1} \right)}^2} - 4\left( {\text{y}}  \right)\left( { - 16{\text{y}}} \right)} }}{{2\left( {\text{y}} \right)}} \\ \Rightarrow {\text{x}} = \frac{{1 \pm \sqrt  {1 - 64{{\text{y}}^2}} }}{{2{\text{y}}}} \\ \end{gathered}


For {\text{y}} = 0

{\text{x}}  = \frac{{1 \pm \sqrt {1 + 0} }}{0} = \infty

Thus, the range of the function f = R - \left\{ 0  \right\}