\chapter{Functions and graphs}
\section{Properties of functions}
\subsection{Finding domain, range and the sign}
Recall from Unit 1.2 that a point \((x,y)\) is on the graph of a function \(f\)
when \(y\) is the output assigned to the input \(x\).
This means that \(x\) is in the domain of \(f\). So to find the domain of \(f\) from its graph,
we need to find all \(x\)-coordinates of points of the graph.
Similarly, \((x,y)\) is on the graph when \(y\) is in the range of \(f\). So to find the range
of \(f\) from its graph, we need
to find all \(y\)-coordinates of points of the graph.
Finding the sign of \(f\) means finding where \(f(x)\) is positive, negative or zero.
To do this, we use the fact that if \((x,y)\) is on the graph,
then \(y=f(x)\). So to find where \(f\) is positive we need to look for all points that have a positive \(y\)-coordinate.
This means, all points above the \(x\)-axis.
In the same way, to find where \(f(x)\) is negative, we need to find all points below the \(x\)-axis.
To find where \(f(x)\) is zero, we need to look for the points where the graph
intersects the \(x\)-axis (if any exist).
These points are the \(x\)\textit{-intercepts} of the
function.
Similarly, the \(y\)-\textit{intercept} is the point where the graph intersects the
\(y\)-axis (if it exists).
Note that a function can have at most one \(y\)-intercept
(because of the vertical line test), but it can have several \(x\)-intercepts.
\begin{example}
The graph of \(f(x)\) is shown in the picture.
\[ \img{U1_8F1.jpg}{}{15em}{}\]
We see that the \(x\)-coordinates of all points of the graph extend to the left
up to \(x=-4\) (included, because of the solid dot),
and to the right up to \(x=6\) (excluded, because of the open circle).
So the domain of \(f\) in interval notation is
\[D=[-4,6)\]
In the same way, the \(y\)-coordinates extend down to \(y=-2\) (excluded) and up
to \(y=3\) (included). So the range of \(f\) is
\[R=(-2,3]\]
We also see that the graph is above the \(x\)-axis when the \(x\)-coordinate is between
\(-4\) (included) and \(2\) (excluded). So \(f(x)\) is positive on the interval
\([-4,2)\).
Similarly, \(f\) is negative on the interval \((2,6)\), and it is zero for \(x=2\).
So the \(x\)-intercept is \((2,0)\).
Be careful not to confuse a point with an interval. Unfortunately, the notation commonly used
(and that we also use in this
course) is the same. So in the previous sentence \((2,6)\) is an \textbf{interval}, representing all the numbers between \(2\)
and \(6\) (with end-points excluded), while \((2,0)\) is the \textbf{single point} that has \(x\)-coordinate \(2\) and \(y\)-coordinate
\(0\). When you see the symbol \((a,b)\), there is no way to know if it is an interval or a point, other than reading what the context is.
Finally, the
\(y\)-intercept is \((0,1)\), because that is where the graph intersects the \(y\)-axis.
\end{example}
\subsection{Increasing, decreasing or constant}
We say that a function \(f\) is \textit{increasing} if its graph rises as we move
from left to right, and \textit{decreasing}
if its graph falls. See the pictures for examples.
\[
\begin{array}{ccc}
\img{U1_8F4.jpg}{}{10em}{} & \img{U1_8F5.jpg}{}{10em}{} & \img{U1_8F6.jpg}{}{10em}{}\\
\mbox{Increasing} & \mbox{Increasing} & \mbox{Increasing}\\[2ex]
\img{U1_8F7.jpg}{}{10em}{} & \img{U1_8F8.jpg}{}{10em}{} & \img{U1_8F9.jpg}{}{10em}{}\\
\mbox{Decreasing} & \mbox{Decreasing} & \mbox{Decreasing}
\end{array}
\]
Be careful not to misunderstand the meaning of an arrow on a graph. The arrow only means
that the graph extends indefinitely. It does not tell us anything about increasing or
decreasing. For example, the following graph shows an increasing function with domain \((-\infty,3]\)
\[
\begin{array}{c}
\img{U1_8F10.jpg}{}{12em}{}\\
\mbox{Increasing}
\end{array}
\]
while the next graph shows a decreasing function with the domain \((-\infty,4)\)
\[
\begin{array}{c}
\img{U1_8F11.jpg}{}{12em}{}\\
\mbox{Decreasing}
\end{array}
\]
A function \(f\) is \textit{constant} if its graph is a horizontal straight line:
\[
\begin{array}{c}
\img{U1_8F12.jpg}{}{12em}{}\\
\mbox{A constant function}
\end{array}
\]
A function may be increasing on a part of its domain, and decreasing or constant on
other parts.
\begin{example}
Consider the function in the picture below:
\[
\begin{array}{c}
\img{U1_8F13.jpg}{}{12em}{}
\end{array}
\]
The function is increasing on \((0,2)\),
constant on [2,4] and decreasing on \((4,6)\).
\end{example}
Problems
\problem
The graph of \(f\) is shown in the picture.
\[ \img{U1_8F2.jpg}{}{12em}{}\]
Find the following:
\[
\begin{array}{ll}
\mbox{(a) Domain of }f & \mbox{(b) Range of }f\\
\mbox{(c) Intervals where \(f(x)\) is positive} & \mbox{(d) Intervals where \(f(x)\) is negative} \\
\mbox{(e) \(x\)-intercept(s)} & \mbox{(f) \(y\)-intercept} \\
\mbox{(g) zeros of \(f\)} & \mbox{(h) Points where \(f(x)\) is \(-2\)}\\
\mbox{(i) Points where \(f(x)\) is \(4\)}
\end{array}
\]
\begin{sol}
\begin{enumerate}
\item
The arrow on the left side means that the graph extends indefinitely. So the \(x\)-coordinates
(and the domain) will extend to
\(-\infty\). We note that there is a gap in the graph from \(x=1\) to
\(x=2\). Also, \(x=1\) is included, \(x=2\) is excluded,
and \(x=6\) is included. So the domain is
\[\fbox{\(D=(-\infty,1]\cup(2,6]\)}\]
\item
The \(y\)-coordinates will extend down to \(-\infty\), because of the arrow at the left
end of the graph.
They extend up to \(y=4\) (excluded). So the range is
\[\fbox{\(R=(-\infty,4)\)}\]
\item
The graph is above the \(x\)-axis from \(x=-2\) (excluded) to \(x=1\) (included),
then again from \(x=2\) (excluded) to
\(x=4\) (excluded). So \(f(x)\) is positive on
\[\fbox{\((-2,1]\cup(2,4)\)}\]
\item
The graph is below the \(x\)-axis from \(-\infty\) to \(x=-2\) (excluded),
then again from \(x=4\) (excluded)
to \(x=6\) (included). So \(f(x)\) is negative on
\[\fbox{\((-\infty,-2)\cup(4,6]\)}\]
\item
The graph intersects the \(x\)-axis at \(x=-2\) and \(x=4\). So there are two \(x\)
intercepts: \(\fbox{\((-2,0)\) and \((4,0)\)}\).
\item
The graph intersects the \(y\) axis at \(y=2\). So the \(y\)-intercept is \(\fbox{\((0,2)\)}\).
\item
\(f(x)\) is zero when the graph intersects the \(x\)-axis, at the \(x\)-intercepts.
So \(f(x)\) is zero when \(\fbox{\(x=-2\) or \(x=4\)}\).
\item
There are two points where the \(y\) coordinate is \(-2\): \(\fbox{\((-3,-2)\) and \((5,-2)\)}\).
\item
There is no point where \(f(x)\) is 4 (the point \((2,4)\) is not on the graph,
because of the open circle).
\end{enumerate}
\end{sol}
\mproblem
The graph of \(f(x)\) is shown in the picture.
\[ \img{U1_8F3.jpg}{}{12em}{}\]
Find the following:
\[
\begin{array}{ll}
\mbox{(a) Domain of \(f\)} & \mbox{(b) Range of \(f\)} \\
\mbox{(c) Intervals where \(f(x)\) is positive} &\mbox{(d) Intervals where \(f(x)\) is negative} \\
\mbox{(e) \(x\)-intercept(s)} & \mbox{(f) \(y\)-intercept} \\
\mbox{(g) zeros of \(f\)} & \mbox{(h) Points where \(f(x)\) is \(-2\)} \\
\mbox{(i) Points where \(f(x)\) is \(2\)}
\end{array}
\]
\problem
The graph of \(f\) is shown in the picture.
\[\img{U1_8F14.jpg}{}{15em}{}\]
Answer the following questions:
\begin{enumerate}
\item
Where is \(f\) increasing?
\item
Where is \(f\) decreasing?
\item
Where is \(f\) constant?
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
The graph extends indefinitely to the left (because of the arrow), and it falls
(as we move from left to right)
up to \(x=-1\). So \(f\) is decreasing on \((-\infty,-1]\).
\item
The graph rises (as we move from left to right) from \(x=3\) (included) to \(x=5\) (excluded).
So \(f\) is increasing on \([3,5)\).
\item
The graph is horizontal from \(x=-1\) to \(x=1\). So \(f\) is constant on \([-1,1]\).
\end{enumerate}
\end{sol}
\mproblem
The graph of \(f\) is shown in the picture.
\[\img{U1_8F15.jpg}{}{15em}{}\]
Answer the following questions:
\begin{enumerate}
\item
Where is \(f\) increasing?
\item
Where is \(f\) decreasing?
\item
Where is \(f\) constant?
\end{enumerate}