\chapter{Functions and graphs}
\section{Implicit definition of domain}
\subsection{The domain of a function defined by a formula}
When a function is defined by a formula, and there is no other given information,
the domain is understood to be all the values of the input variable for which the formula
can be used. To check if a formula can be used for the problems in this unit, remember that
real numbers are either positive, or negative or zero.
\begin{example}
For the function
\[f(x)=3x^2-7x+1\]
the formula can be used for any real number (positive, negative, or zero). This means that the
domain is the set of all real numbers. In interval notation, we write:
\[\fbox{\(D=(-\infty,\infty)\)}\]
\end{example}
Make sure to always give your answers in interval notation for the domain of a function.
\begin{example}
For the function
\[f(x)=\frac{1}{x}\]
the input value \(x=0\) is not in the domain, because \(1/0\) is undefined.
But every other number (positive or negative)
will be in the domain. So the domain is the set of all real numbers
that are either less than zero (negative), or greater than zero (positive).
In interval notation, this is:
\[\fbox{\(D=(-\infty,0)\cup (0,\infty)\)}\]
\end{example}
If a formula contains a square root, we must remember that the expression under the square
root sign cannot be negative.
\begin{example}
For the function
\[f(x)=\sqrt{x}\]
the input value \(x=-2\) (or any other negative number)
is not in the domain, because \(\sqrt{-2}\) is not a real number.
But \(x=0\) is in the domain, because \(\sqrt{0}=0\) is well-defined, and of course
every positive number is also in the domain.
So the domain is
\[\fbox{\(D=[0,\infty)\)}\]
\end{example}
In just the same way, the domain of \(\sqrt[4]{x}\), or \(\sqrt[6]{x}\), or any even root function, will be \([0,\infty)\),
because the even root of a negative number does not exits.
But there is no problem with \textbf{odd} roots of negative numbers. So the domain of \(\sqrt[3]{x}\),
or \(\sqrt[5]{x}\), or any other odd root function will be \((-\infty, \infty)\).
Some functions may have more than one denominator containing the variable. Then we need to make sure that
none of the denominators become zero.
\begin{example}
Let
\[f(x) =\frac{1}{2-\frac{1}{x}}.\]
We need to exclude the value \(x=0\), because otherwise the fraction \( \displaystyle \frac{1}{x}\) has a zero in the
denominator. But we also need to make sure that the denominator \(\displaystyle 2-\frac{1}{x}\) is not zero.
So we need to solve the equation
\[2-\frac{1}{x}=0,\]
and the solution is \(\displaystyle x=\frac{1}{2}\) (check that!). So we need to exclude \(x=0\) and \(\displaystyle x= \frac{1}{2}\),
and the domain is
\[D=(-\infty,0)\cup \left(0,\frac{1}{2}\right)\cup \left(\frac{1}{2},\infty\right).\]
\end{example}
We can summarize how to find the domain for the functions discussed in this section as
follows:
\begin{itemize}
\item If the input variable is not in any denominator or inside any square root, and it
is only raised to whole, positive exponents, then the domain is the set of all real numbers,
\(D=(-\infty,\infty)\).
\item
If the input variable appears in a denominator,
we cannot use any value that will make the denominator
zero. In this case, we set
\[\mbox{Denominator }=0,\]
solve the equation, and exclude
the solutions from the domain.
\item If the input variable is inside a square root,
we cannot use any value that
will result in the square root of a negative number. In this case we set
\[\mbox{Expression inside square root }\geq 0,\]
and solve the inequality. The solution of the inequality will be the domain.
\item If the input variable is inside a square root, and the square root is the denominator
of a fraction,
we cannot use any value that will result in a zero denominator or the square root of a
negative number. In this case we set
\[\mbox{Expression inside square root }> 0,\]
and we solve the inequality. As before, the domain of the function is the solution of the
inequality.
\end{itemize}
Problems
\problem
Find the domain of the functions, and write the answer in interval notation:
\begin{enumerate}
\item \( f(t)=5t^4-3t^3+4t-\sqrt{3}\)
\item \( \displaystyle{g(x)=\frac{x-5}{2x+8}} \)
\item\( \displaystyle{h(x)= \frac{x+1}{x^2+5x+6}}\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item \(f(t)=5t^4-3t^3+4t-\sqrt{3}\)
In the given formula, the variable \(t\) has only positive whole exponents ( \(t,t^3,t^4\)),
and it is not in any denominators or inside a square root. So the formula
can be used for any value of \(t\), and the domain is the set of all real numbers,
or, in interval notation,
\[\fbox{\(D=(-\infty,\infty)\)}.\]
\item \(\displaystyle{g(x)= \frac{x-5}{2x+8}}\)
The formula cannot be used when the denominator is zero. To find the values that must be
excluded, we set the denominator equal
to zero and we solve the equation:
\[
\begin{array}{rcll}
2x+8&=& 0 & \mbox{Set the denominator equal to 0}\\[2ex]
2x+8 \color{red}{-8} &=& 0 \color{red}{-8} & \mbox{Subtract \(8\) from both sides}\\[2ex]
2x &=& -8 & \mbox{Simplify}\\[2ex]
\displaystyle{\frac{2x}{\color{red}{2}}}& = & \displaystyle{\frac{-8}{\color{red}{2}}}
& \mbox{Divide by \(2\)}\\[2ex]
x & = & -4 & \mbox{Simplify}
\end{array}
\]
So \(x=-4\) will make the denominator zero and it must be excluded. Any other \(x\)-value will be OK.
So the domain is:
\[\fbox{\(D=(-\infty,-4)\cup(-4,\infty)\)}\]
\item \(\displaystyle{h(x)= \frac{x+1}{x^2+5x+6}}\)
As before, we must exclude the values that make the denominator zero.
So we set the denominator equal to zero and solve the
equation (this time it's a quadratic equation):
\[
\begin{array}{rcll}
x^2+5x+6&=& 0 & \mbox{Set the denominator equal to 0}\\[2ex]
(x+3)(x+2) &=& 0 & \mbox{Solve it by factoring}\\[2ex]
x+3=0 \mbox{ \ } \mbox{ or } \mbox{ \ } x+2=0 & & & \mbox{Set each factor equal to 0}\\[2ex]
x=-3 \mbox{ \ } \mbox{ or } \mbox{ \ } x=-2 & & &
\end{array}
\]
Any value of \(x\) other than \(-2\) or \(-3\) can be used. So the domain is
\[\fbox{\(D=(-\infty,-3)\cup(-3,-2)\cup(-2,\infty)\)}.\]
\end{enumerate}
\end{sol}
\mproblem
Find the domain of the functions, and write the answer in interval notation:
\begin{enumerate}
\item
\(\displaystyle{ f(x)= \frac{2x}{x-3}}\)
\item
\(g(x)=4x^3+x\sqrt{2}-1\)
\item
\(\displaystyle{h(x)=\frac{x+1}{x^2-x-6}}\)
\end{enumerate}
\problem
Find the domain of the functions, and write the answer in interval notation:
\begin{enumerate}
\item
\(f(x)=\sqrt{3x+9}\)
\item
\(\displaystyle h(x)= \frac{x}{\sqrt{2x+1}}\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item \(f(x)=\sqrt{3x+9}\)
The expression under the square root sign cannot be negative. So we set it \(\geq 0\) and solve:
\[
\begin{array}{rcll}
3x+9 & \geq & 0\\
\ \\
3x+9\color{red}{-9} & \geq & 0\color{red}{-9} &\mbox{Subtract 9 from both sides} \\
\ \\
3x & \geq & -9 & \mbox{Simplify}\\
\ \\
\displaystyle \frac{3x}{\color{red}{3}} & \geq & \displaystyle \frac{-9}{\color{red}{3}} &\mbox{Divide both sides by 3}\\
\ \\
x & \geq & -3
\end{array}\]
The domain consists of all numbers greater than or equal to \(-3\), or
\[\fbox{\(D=[-3,\infty)\)}\]
Note the bracket [ on the left side, because the endpoint \(-3\) is included.
\item
The formula \(\displaystyle h(x)= \frac{x}{\sqrt{2x+1}}\) contains both a square root and
a denominator. This time we need the expression inside the square
root to be \textbf{strictly} greater than zero (we cannot allow zero in the denominator).
So we need to solve the inequality \(2x+1>0\):
\[
\begin{array}{rcll}
2x+1 & > & 0 & \mbox{Set the expresion inside square root in the denominator \(>0\)}\\
\ \\
2x+1\color{red}{-1} & > & 0 \color{red}{-1} & \mbox{Subtract \(1\) from both sides}\\
\ \\
2x & > & -1 & \mbox{Simplify} \\
\ \\
\displaystyle{\frac{2x}{\color{red}{2}}} & > & \displaystyle{\frac{-1}{\color{red}{2}}} & \mbox{Divide both sides by 2}\\
\ \\
x & > & \displaystyle{-\frac{1}{2}} & \mbox{Simplify}
\end{array}
\]
So the domain of the function is
\[\fbox{\(D=\left(-\frac{1}{2},\infty\right)\)}\]
\end{enumerate}
\end{sol}
\mproblem
Find the domain of the functions, and write the answer in interval notation
\begin{enumerate}
\item \(f(x)=\sqrt{2x-6}\)
\item
\(h(t)=\displaystyle \frac{2t}{\sqrt{4t-2}}\)
\end{enumerate}