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 $$f$$is $$\left\{ {{\text{y}}:{\text{y}} \ne 1} \right\} = \left] { – \infty ,1} \right[\;\; \cup \;\;\left] {1,\infty } \right[$$.

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 $${\text{y}} = 0$$
\[{\text{x}} = \frac{{1 \pm \sqrt {1 + 0} }}{0} = \infty \]