5.3: Absolute Value Functions

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The most popular piecewise defined function is the absolute value function.

$|x|=\left\{ \begin{array} {ll} x & x\ge 0 \\ -x & x<0 \end{array} \right. $

Clearly the domain of this function is $(-\infty,\infty)$, and we can see its graph below:

$clipboard_e237ea6b9cc62001d4f1533b21a0f5bd2.png$

Solving equations with absolute values require us to solve two separate equations (replacing the absolute value term with $+$ or $-$ itself) and then check our answers.

Example $\PageIndex{1}$

Solve the following equations involving absolute values.

(a) $\big|x+2\big|=3$

(b) $\big|t+1\big|=2t-3$

Solution

(a) Since the inside could be positive or negative, we have two equations to solve:

$x+2=3$ and $-(x+2)=3$

This gives us two possible solutions:

$x=1$ and $x=-5$

We must check our answers to see if they indeed satisfy the original equation:

$\big|1+2\big|\stackrel{?}{=}3 \implies 3\stackrel{?}{=}3$ True

$\big|-5+2\big|\stackrel{?}{=}3 \implies 3\stackrel{?}{=}3$ True

Since they are both true, we have two solutions to the equation: $x=1,-5$

(b) Since the inside could be positive or negative, we have two equations to solve:

$ t+1 =2t-3$ and $ -(t+1) =2t-3$

This gives us two possible solutions:

$ t=4$ and $ t=\dfrac{2}{3}$

We must check our answers to see if they indeed satisfy the original equation:

$\big|4+1\big|\stackrel{?}{=}2\cdot 4-3 \implies 5\stackrel{?}{=}5$ True

$\big|\dfrac{2}{3}+1\big|\stackrel{?}{=}2\cdot \dfrac{2}{3}-3 \implies \dfrac{5}{3}\stackrel{?}{=}-\dfrac{5}{3}$ False

Therefore we have only one solution to the equation: $t=4$

Solving inequalities can be a little more work, but we can create something similar to a sign chart to get our answers. First, solve the inequality as if it were an equation, then plot the solutions to the equation on a number line, and finally check each interval to see if the original inequality holds.

Example $\PageIndex{2}$

Solve each inequality involving absolute values below and put your answer into interval notation.

(a) $\displaystyle{ |1-\frac{1}{2}u| \ge 2} $

(b) $\displaystyle{ |y|>2y-3} $

Solution

(a) First, we pretend the $\ge $ symbol is an $=$ instead and solve for $u$:

$\displaystyle{ |1-\frac{1}{2}u| =2} $

This turns into two equations $\textcolor{blue}{1-\frac{1}{2}u=2}$ and $\textcolor{brown}{-(1-\frac{1}{2}u)=2}$

We solve them each separately to get $\textcolor{blue}{u=-2}$ and $ \textcolor{brown}{u=6}$

Then we check that our answers satisfy the absolute value equation (which they do in this instance)

We will make a number line for a sign chart, and split it up at these two solutions:

$clipboard_eef8fa8a9afb0b6b89e90b6ce0cdb3ada.png$

In each interval between these roots, we pick any number and see if it satisfies the original inequality

In the left most interval $(-\infty,-2)$, we will pick $u=-4$

$|1-\frac{1}{2}(-4)| \stackrel{?}{\ge} 2 \implies 3\stackrel{?}{\ge} 2$ which is True

This means everything in $(-\infty,-2)$ will make the inequality true.

In the middle interval $(-2,6)$, we will pick $u=0$

$|1-\frac{1}{2}(0)| \stackrel{?}{\ge} 2 \implies 1\stackrel{?}{\ge} 2$ which is False

This means nothing in $(-2,6)$ will make the inequality true.

In the right most interval $(6,\infty)$, we will pick $u=8$

$|1-\frac{1}{2}(8)| \stackrel{?}{\ge} 2 \implies 3\stackrel{?}{\ge} 2$ which is True

This means everything in $(-2,6)$ will make the inequality true.

Lastly, since our inequality was not strict, we know our solutions to the equation are also included in our final answer:

$(-\infty,-2]\cup [6,\infty)$

(b) Again we start by pretending the $ < $ symbol is an $=$ and solving:

$\displaystyle{ |y|=2y-3} $

This turns into two equations $\textcolor{purple}{y=2y-3}$ and $\textcolor{orange}{-y=2y-3}$

We solve them each separately to get $\textcolor{purple}{y=3}$ and $ \textcolor{orange}{y=1}$

Then we check that our answers satisfy the absolute value equation (which $y=1$ does not)

This time when we make a number line for a sign chart, we split it at only the legitimate solution $y=3$:

$clipboard_e33da1f3c5b7bc4c20c6a96d9f4cf3fc7.png$

In each interval between these roots, we pick any number and see if it satisfies the original inequality

In the left interval $(-\infty,3)$, we will pick $y=0$

$|0| \stackrel{?}{>} 2\cdot 0 -3 \implies 0\stackrel{?}{>} -3$ which is True

This means everything in $(-\infty,3)$ will make the inequality true.

In the right most interval $(3,\infty)$, we will pick $y=10$

$|10| \stackrel{?}{>} 2\cdot 10 -3 \implies 10\stackrel{?}{>} 17$ which is False

This means nothing in $(3,\infty)$ will make the inequality true.

Lastly, since our inequality was strict, we know our solutions to the equation are not included in our final answer:

$(-\infty,3)$

This is similar to the sign charts we used for polynomials and rational functions. Those sign charts worked because the only way for the functions to switch from positive to negative was by having it equal to zero. These truth charts work because the only way the inequality can change between being true and being false is at a point where the equation (with the $=$ in it) would hold.

The good news is that the absolute value function is continuous everywhere. Even better, we can say that $\displaystyle{\lim_{x\to a} |f(x)|=\left|\lim_{x\to a}f(x)\right|}$ whenever the latter exists.

When in doubt, just write it as a piecewise function and you'll be fine!

Example $\PageIndex{3}$

Evaluate each limit below involving absolute values

(a) $\displaystyle{ \lim_{t\to -3} \frac{t^2+1}{1+\big|t+5\big|}}$

(b) $\displaystyle{ \lim_{w\to 2^-} \frac{2-\big|2w-6\big|}{w-2}}$

Solution

(a) We start by writing the absolute value as a piecewise function $\textcolor{orange}{\big|t+5\big|=\left\{ \begin{array} {ll} t+5 & t\ge -5 \\ -t-5 & t<-5\end{array} \right.}$

Then use it to write the entire function as piecewise: $\displaystyle{\frac{t^2+1}{1+\textcolor{orange}{|t+5|}}}=\left\{ \begin{array} {ll} \displaystyle{\frac{t^2+1}{1+\textcolor{orange}{(t+5)}}} & t>-5 \\ \displaystyle{\frac{t^2+1}{1+\textcolor{orange}{(-t-5)}}} & t<-5\end{array} \right.$

For all values of $t$ near $-3$, we will be using the top piece:

$\displaystyle{ \lim_{t\to -3} \frac{t^2+1}{1+\big|t+5\big|}=\lim_{t\to -3} \frac{t^2+1}{1+(t+5)}=\lim_{t\to -3} \frac{t^2+1}{t+6}= \frac{(-3)^2+1}{(-3)+6}= \frac{10}{3}}$

(b) We start by writing the absolute value as a piecewise function $\textcolor{purple}{\big|2w-6\big|=\left\{ \begin{array} {ll} 2w-6 & w\ge 3 \\ 6-2w & w<3 \end{array} \right.}$

Then use it to write the entire function as piecewise: $\displaystyle{\frac{2-\textcolor{purple}{|2w-6|}}{w-2}}=\left\{ \begin{array} {ll} \displaystyle{\frac{2-\textcolor{purple}{(2w-6)}}{w-2}} & w\ge 3 \\ \displaystyle{\frac{2-\textcolor{purple}{(6-2w)}}{w-2}} & w<3\end{array} \right.$

For all values of $w$ left of $2$, we will be using the bottom piece:

$\displaystyle{ \lim_{w\to 2^-} \frac{2-|2w-6|}{w-2}=\lim_{w\to 2^-} \frac{2-(6-2w)}{w-2}=\lim_{w\to 2^-} \frac{2w-4}{w-2}=\lim_{w\to 2^-} \frac{2(w-2)}{w-2}=\lim_{w\to 2^-} 2=2}$

Limits of absolute values can always be done in this way, and with practice you won't need to write out the piecewise definition every time. (But it never hurts to do so!)

We can use also limits to find any discontinuities for functions involving the absolute value.

Example $\PageIndex{4}$

Find all discontinuities for each function below.

(a) $\displaystyle{ f(x)=\frac{|x|}{x}}$

(b) $\displaystyle{ g(x)=\frac{x^2-1}{|x|-1}}$

Solution

(a) Since $\displaystyle{\lim_{x\to a} |x|=|a|}$ and $\displaystyle{\lim_{x\to a} x=a}$ for all $a$
then $\displaystyle{ \lim_{x\to a} f(x)=\lim_{x\to a}\frac{|x|}{x}=\frac{|a|}{a}=f(a)}$ for $a\ne 0$

$\displaystyle{ \lim_{x\to 0^+} f(x)=\lim_{x\to 0^+} \frac{|x|}{x}=\lim_{x\to 0^+} \frac{x}{x}=\lim_{x\to 0^+} 1 = 1}$
$\displaystyle{ \lim_{x\to 0^-} f(x)=\lim_{x\to 0^-} \frac{|x|}{x}=\lim_{x\to 0^-} \frac{(-x)}{x}=\lim_{x\to 0^+} -1 = -1}$
Lastly, $f(0)=DNE$

Thus we have a jump discontinuity at $x=0$ which we can see in the graph of $y=f(x)$ below

$clipboard_e1e61dcb483cdcd893cfb78a9dcab3742.png$

(b) Again $\displaystyle{\lim_{x\to a} g(x)=\lim_{x\to a} \frac{x^2-1}{|x|-1}=\frac{a^2-1}{|a|-1}=g(a)}$ for $|a|\ne 1$

$\displaystyle{ \lim_{x\to 1} \frac{x^2-1}{|x|-1}=\lim_{x\to 1} \frac{x^2-1}{x-1}=\lim_{x\to 1} \frac{(x+1)(x-1)}{x-1} = \lim_{x\to 1} x+1=2}$
But $g(1)=DNE$ so we have a removable discontinuity at $x=1$

$\displaystyle{ \lim_{x\to -1} \frac{x^2-1}{|x|-1}=\lim_{x\to -1} \frac{x^2-1}{-x-1}=\lim_{x\to 1} \frac{(x+1)(x-1)}{-(x+1)} = \lim_{x\to -1} -(x-1)=2}$
But $g(-1)=DNE$ so we have a removable discontinuity at $x=-1$ which we can see in the graph of $y=g(x)$ below

$clipboard_e3ca61ad6d922190814e1701948b040ef.png$

This page titled 5.3: Absolute Value Functions was last modified on Thu, 13 Mar 2025 20:04:56 GMT and is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Kenn Huber.

Contents

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Example \(\PageIndex{1}\)

Solution

Example \(\PageIndex{2}\)

Solution

Example \(\PageIndex{3}\)

Solution

Example \(\PageIndex{4}\)

Solution