Obtaining Information about the Graph of a Function

Consider the function \(f ( x) =\dfrac{x+1}{x+2}\).

1. What is the domain of \(f\)?
2. Is the point \(\left( 1,\dfrac{1}{2}\!\right) \) on the graph of \(f\)?
3. If \(x=2\), what is \(f( x) \)? What is the corresponding point on the graph of \(f\)?
4. If \(f( x) =2\), what is \(x\)? What is the corresponding point on the graph of \(f\)?
5. What are the \(x\)-intercepts of the graph of \(f\) (if any)? What point(s) on the graph of \(f\) correspond(s) to the \(x\)-intercept(s)?

Solution (a) The domain of \(f\) consists of all real numbers except \(-2;\) that is, the set \(\{ x|x\neq -2\} \).

(b) \[ \begin{eqnarray*} &&{\rm When}\ x=1, {\rm then}\ f ( 1 )\underset{\underset{\color{#0066A7}{x=1}}{\color{#0066A7}{\uparrow}}} {=} \dfrac{1+1}{1+2}=\dfrac{2}{3}. \hbox{The point } \left( 1, \dfrac{2}{3}\!\right) \hbox{ is on the graph of } f; \hbox{ the}\\ \end{eqnarray*} \] point \(\left( 1,\dfrac{1}{2}\!\right)\) is not on the graph of \(f\).

(c) \[ \begin{eqnarray*} &&\hbox{If } x=2, \hbox{then } f ( 2) \underset{\underset{\color{#0066A7}{x=2}}{\color{#0066A7}{\uparrow}}} {=} \dfrac{2+1}{2+2}=\dfrac{3}{4}. \hbox{The point } \left( 2, \dfrac{3}{4}\!\right) \hbox{ is on the graph of } f.\\ \end{eqnarray*} \]

(d) If \(f( x) =2\), then \(\dfrac{x+1}{x+2}=2.\) Solving for \(x\), we find \[ \begin{eqnarray*} x+1 &=&2( x+2) =2x+4 \nonumber \\ x &=&-3 \end{eqnarray*} \]

The point \(( -3,2) \) is on the graph of \(f\).

(e) The \(x\)-intercepts of the graph of \(f\) occur when \(y=0\). That is, they are the solutions of the equation \(f( x) =0\). The \(x\) -intercepts are also called the real zeros or roots of the function \(f\).

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The real zeros of the function \(f( x) =\dfrac{x+1}{x+2}\) satisfy the equation \(x+1=0\) or \(x=-1\). The only \(x\)-intercept is \(-1\), so the point \(( -1,0) \) is on the graph of \(f.\)