| Solving Absolute Value Equations | Applications of Absolute Value Functions |
| Solving Absolute Value Inequalities | More Practice |
| Graphs of Absolute Value Functions |
As we saw earlier in the Negative Numbers and Absolute Value section, an absolute value (designated by | |) means take the positive value of whatever is between the two bars. The absolute value is always positive, so you can think of it as the distance from 0. As an example, $ \left| 3 \right|=3$ and $ \left| {-3} \right|=3$. It’s as simple as that!
(Note that we also address absolute values here in the Piecewise Functions section and here in the Rational Functions, Equations, and Inequalities section.)
Solving Absolute Value Equations
Solving absolute equations isn’t too difficult; just have to separate the equation into two different equations (once we isolate the absolute value), since we don’t if what’s inside the absolute value is positive or negative. Then, make the expression on the right-hand side (without the variables) both positive and negative and solve each equation; typically, we will get two answers. We must check our answers, since we may get extraneous solutions (solutions that don’t work)..
There are a few cases with absolute value equations or inequalities where we don’t have to even solve! One is when we have isolated the absolute value, and it is set equal to a negative number, such as $ \left| {x-5} \right|=-4$, or $ \left| {x-5} \right|\le -4$, for example. Since an absolute value can never be negative, we have no solution for this case. The other is when the absolute value is greater than a negative number, such as $ \left| {x-5} \right|>-4$ for example. In this case our answer is all real numbers, since an absolute value is always positive.
Note that we can always solve absolute value equations and inequalities graphically, as shown below.
Here are some problems:
| Solving Absolute Value Problems | Notes |
| $ \begin{array}{c}\color{#800000}{{\left| x \right|-3=20}}\\\left| x \right|=23\\\swarrow \,\,\searrow \\x=23\,\,\,\,\,\,\,\,\,\,x=-23\\x=-23,\,\,23\end{array}$ | First simplify, then separate the absolute value problem into two separate equations, since we don’t know if what the value of $ x$ is positive or negative.
Try the answers in the original equation to make sure they work! |
| $ \displaystyle \begin{array}{c}\color{#800000}{{3\left| {x-2} \right|+5=14}}\\3\left| {x-2} \right|=9\\\left| {x-2} \right|=3\\\swarrow \,\,\,\,\,\,\,\,\,\,\searrow \\x-2\,=\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,\,x-2=-3\\\underline{{\,\,\,\,\,\,\,+2=+2}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline{{\,+2=+2}}\\\,\,\,\,\,x\,\,\,=\,\,\,\,5\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,=-1\end{array}$ $ x\,=\,-1,\,\,5$ | Here’s one a little bit more complicated. Simplify first to separate the absolute value from the rest of the numbers.
Check the answers; the work! |
| $ \displaystyle \begin{array}{c}\color{#800000}{{\left| {{{x}^{2}}-2x} \right|+1=9}}\\\left| {{{x}^{2}}-2x} \right|=8\\\swarrow \,\,\,\,\,\,\,\,\,\,\,\,\searrow \\{{x}^{2}}-2x\,=\,\,8\,\,\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}-2x\,\,=\,-8\\{{x}^{2}}-2x\,\,-8=0\,\,\,\,\,\,\,\,\,\,\,\,{{x}^{2}}-2x\,\,+8=0\\\left( {x-4} \right)\left( {x+2} \right)=0\,\,\,\,\,\,\,\,\,\,x=\,\,\frac{{-\left( {-2} \right)\pm \sqrt{{{{{\left( {-2} \right)}}^{2}}-4\left( 1 \right)\left( 8 \right)}}}}{{2\left( 1 \right)}}\\x=\,\,4,\,\,-2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x=1\pm \sqrt{7}i\end{array}$ | This one’s a little trickier, since it’s not linear. After simplifying, separate the absolute value equations into two; 8 positive, and 8 negative.
For the second case, use the quadratic formula to get the roots: $ \displaystyle \frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$. Note that we get some complex roots (since we had to take the square root of a negative number).
Check all the answers; they all work! |
| $ \displaystyle \frac{2}{{\left| {x-2} \right|}}=5$ $ \displaystyle \,\frac{2}{{\left| {x-2} \right|}}=\frac{5}{1}$ $ \begin{array}{c}2\times 1=5\left| {x-2} \right|\\\,\left| {x-2} \right|=\frac{2}{5}\\\swarrow \,\,\,\,\searrow \end{array}$ $ \displaystyle x-2=\frac{2}{5}\,\,\,\,\,\,\,\,\,\,x-2=-\frac{2}{5}$ $ \displaystyle x=\,2\frac{2}{5},\,\,\,1\frac{3}{5}$ | Here’s another nonlinear equation where the absolute value is originally in the denominator (so it’s a rational function). Treat the absolute value like a variable, and get it out from the denominator by cross multiplying.
Then, continue to solve, and divide up the equations to get the two answers. (Note that if we ended up with “2” as one of the answers, we’d have to eliminate it, since we can’t have a denominator that equals 0).
Check the answers – they work! |
| \begin{array}{c}\color{#800000} {{\,\left| {x-2} \right|=-\left| {-2} \right|}}\,\\\,\left| {x-2} \right|=-2\\\emptyset ,\text{ or no solution}\end{array} | Since the absolute value can never be a negative number, stop right away, and not separate the equations. |
Here’s one more that’s a bit tricky, since we have two expressions with absolute value in it. In this case, we have to separate in four cases, just to be sure we cover all the possibilities. We then must check for extraneous solutions, possible solutions that don’t work.
| Absolute Value Problem | Notes | ||||
| $ \left| {3x-2} \right|+\left| {x+2} \right|=6$
| Separate into four cases, since we don’t know whether $ 3x-2$ and $ x+2$ are positive or negative. Since they are absolute values in the equations, they could be either, but still come out positive.
For example, when the expression $ 3x-2$ is negative, the absolute value of that expression is the negation of it, or $ -3x+2$, to make it positive in the equation. Play around with some numbers and you’ll see this!
When we get all the possible answers, check for extraneous solutions, since we’re dealing with absolute value. We found two answers that worked: $ \displaystyle x=\frac{3}{2}$ and $ x=-1$. You can also put the equation in your graphing calculator to check your answers! |
Here’s another way to approach the absolute value problem above, using number lines:
| Absolute Value Problem | Notes | ||||
| $ \left| {3x-2} \right|+\left| {x+2} \right|=6$ $ \left| {3x-2} \right|=6-\left| {x+2} \right|$
| First, separate the equation so there’s an absolute value on each side (not necessary, but it’s a little easier this way). Then, set each absolute value to 0 to get boundary points, where the absolute values turn from negative to positive, or positive to negative. You can use piecewise functions to define the absolute values expressions for each interval.
Now draw number lines for each absolute value, and then for the whole equation above. We see for the last number line that for $ <-2$, we’ll use $ 2-3x$ and $ -x-2$, between $ -2$ and $ \displaystyle \frac{2}{3}$, we’ll use $ 2-3x$ and $ x+2$, and $ \displaystyle >\frac{2}{3}$, we’ll use $ 3x-2$ and $ x+2$.
After solving for $ x$ in the original equation, we have to check to make sure each value we get for $ x$ falls into the correct interval of the number line. For example, $ \displaystyle -\frac{3}{2}$ isn’t $ <-2$, so we have to “throw it away”. $ -1$ is between $ -2$ and $ \displaystyle \frac{2}{3}$, so it works, and $ \displaystyle \frac{3}{2}$ is $ \displaystyle >\frac{2}{3}$, so it works. We have $ \displaystyle x=\frac{3}{2}$ and $ x=-1$. √ | ||||
Solving Absolute Value Inequalities
Note that we learned about Linear Inequalities here.
When dealing with absolute values and inequalities (just like with absolute value equations), we have to separate the inequality into two different ones, if there are any variables inside the absolute value bars.
First, get the absolute value all by itself on the left (remember to reverse the inequality sign when multiplying or dividing by a negative number). Now, separate the equations. We get the first equation by just taking away the absolute value sign away on the left. The easiest way to get the second equation is to take the absolute value sign away on the left, and do two things on the right: reverse the inequality sign, and change the sign of everything on the right (even if we have variables over there).
We also have to think about whether or not to use “or” or “and” between the two new equations. The way I remember this is that with a $ >\,\text{or}\,\,\ge $ sign, you can remember “gore”: greater than uses “or”. With a $ <\,\text{or}\,\,\le $ sign, think “land”: less than uses “and”.
GORE: Greater Than uses OR
LAND: Less Than uses AND
Note that statement with “or” is a disjunction, which means that it works if only one (or both) parts are true. A statement with “and” is a conjunction, which means it only works if both parts are true.
And again, if we get something like $ \left| {x+3} \right|<0$ (or a negative number), there is no solution, and something like $ \left| {x+3} \right|\ge 0$ (or a negative number), there are infinite solutions (all real numbers).
Also, remember to use open brackets for inequalities that aren’t inclusive ($ <$ and $ <$) and closed brackets for inequalities that are inclusive and include the boundary point ($ \le $ and $ \ge $).
Here are some examples:
| Absolute Value Inequality Problems | Number Line Graph |
| $ \displaystyle \left| x \right|>2$ $ \displaystyle \begin{array}{c}x>2\text{ }\,\text{or }\,x<-2\\x<-2\text{ }\,\text{or }\,x>2\\\left( {-\infty ,-2} \right)\cup \left( {2,\infty } \right)\end{array}$ |  |
| $ \displaystyle -2\left| {x-1} \right|\ge -6$ $ \displaystyle \begin{array}{c}\,\left| {x-1} \right|\le 3\\x-1\le 3\,\text{ and }\,x-1\ge -3\\\,x\le 4\,\text{ and }\,x\ge -2\\\,-2\le x\le 4:\,\,\,\,\left[ {-2,4} \right]\,\,\end{array}$ Note: Some solve “and” inequalities this way, which yields the same results: $ \displaystyle -2\left| {x-1} \right|\ge -6$ $ \displaystyle \begin{array}{c}\,\left| {x-1} \right|\le 3\\-3\le x-1\le 3\\-3+1\le x\le 3+1\\\,\,\,-2\le x\le 4:\,\,\,\,\,\left[ {-2,4} \right]\,\,\end{array}$ | Don’t forget to reverse the inequality sign when dividing by $ -2$.
|
| $ \displaystyle 4\left| x \right|\,+2>6-8x$ $ \displaystyle \begin{array}{c}\left| x \right|>1-2x\\x>1-2x\,\text{ or }\,x<-1+2x\\3x>1\text{ }\,\text{or }\,1<x\end{array}$ $ \displaystyle x>\frac{1}{3}\text{ }\,\text{or }x>1$ $ \displaystyle x>\frac{1}{3}\text{:}\,\text{ }\,\left( {\frac{1}{3},\infty } \right)\,$ | Divide both sides by 4 without worrying about changing the inequality, since 4 is a positive number. Since we have an “or” situation, we can ignore the $ \displaystyle x>1$, since, when $ x>1$, it will always be greater than $ \displaystyle \frac{1}{3}$.
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| $ \displaystyle \left| {3x} \right|\le -3$ $ \displaystyle \emptyset \text{ (no solution)}$ (We don’t even need to solve, since we can’t have a negative absolute value). |  |
| $ \displaystyle \left| {3x} \right|>-3$ $ \displaystyle \mathbb{R}\,\,\,(\text{all real numbers})$ (We don’t even need to solve, since an absolute value will always be greater than a negative.) |  |
| $ x\left| {x-3} \right|>0$ Draw a sign chart to solve (like we saw here in the Quadratic Inequalities section), since we have a quadratic, and the right-hand side is 0. For each interval, put a test value in the inequality (as is, with the absolute value) to see if it’s positive or negative. To get the boundary points, or critical values, set both $ x$ and $ x-3$ to $ 0$. We want the positive intervals (but can’t include $ x=0$ and $ x=3$ because of the $ >$ sign). Since we have an “or” situation, the solution is $ \left( {0,3} \right)\cup \left( {3,\infty } \right)$. You can confirm on a graphing calculator! |  |
There are examples of rational functions with absolute values here in the Rational Functions, Equations, and Inequalities section.
Graphs of Absolute Value Functions
Note that you can put absolute values in your Graphing Calculator (and even graph them!) by hitting MATH, scroll right to NUM, and then hitting 1 (abs) or ENTER.
Absolute Value functions typically look like a V (upside down if the absolute value is negative), where the point at the V is called the vertex. For the absolute value parent function, the vertex is at $ \left( {0,0} \right)$.
We looked at absolute value parent functions and their transformations in the Absolute Value Transformations section, and absolute value functions as piecewise equations here in the Piecewise Functions section.
Note that the general form for the absolute value function is $ f\left( x \right)=a\left| {x-h} \right|+k$, where $ \left( {h,k} \right)$ is the vertex. If $ a$ is positive, the function points down (like a V); if $ a$ is negative, the function points up (like an upside-down V). Here’s a graph of the parent function, and also a transformation:
| Absolute Value Parent Function | Absolute Value Transformation Example | ||||||||||||||||||||||||
| $ \displaystyle y=\left| x \right|$
Absolute Value (Even Function) Domain: $ \left( {-\infty ,\infty } \right)$ Range: $ \left[ {0,\infty } \right)$
End Behavior: $ \begin{array}{l}x\to -\infty \text{, }\,y\to \infty \\x\to \infty \text{, }\,\,\,y\to \infty \end{array}$ | $ \displaystyle \begin{array}{c}y=-3\left| {2x+4} \right|+1\\y=-3\left| {2\left( {x+2} \right)} \right|+1\end{array}$
Without using a t-chart, we can see that the vertex is at $ \left( {-2,1} \right)$ and the graph is upside-down because of the negative sign. It’s also stretched vertically by a factor of 3 and horizontal by a factor of $ \displaystyle \frac{1}{2}$ (or stretched vertically by a factor of 6); thus, other points down can be drawn by going back and forth 1 and down 6. | ||||||||||||||||||||||||
You can solve absolute value equations and equalities with graphing; here are some examples of solving inequalities:
| Absolute Value Inequality: Solve Graphically | Absolute Value Inequality: Solve Graphically |
| $ -\left| {x-2} \right|+3\ge -2$
To graph the left-hand side, the vertex is at $ \left( {2,3} \right)$. Then flip the graph because of the negative sign. The right-hand part of the graph is a horizontal line at $ y=-2$.
| $ \displaystyle \frac{1}{2}\left| x \right|<\frac{1}{2}x-2$
To graph the left-hand side, the vertex is at $ \left( {0,0} \right)$. The graph is then compressed with a factor of $ \displaystyle \frac{1}{2}$. The right-hand part of the graph is a line at $ \displaystyle y=\frac{1}{2}x-2$.
There is no solution to this absolute value inequality (there is no place the first graph will be lower than the second).. |
The solution to this absolute value inequality (where the first graph is higher) is $ \left[ {-3,7} \right]$. Note that we used hard brackets instead of soft brackets, because of the $ \ge $.
Applications of Absolute Value Functions
Absolute Value Functions are in many applications, especially in those involving V-shaped paths and margin of errors, or tolerances. Here are some examples absolute value “word” problems that you may see:
| Absolute Value Word Problem | Solution |
| Two students are bounce-passing a ball between them. The first student bounces the ball from 6 feet high and it bounces 5 feet away from her. The second student is 4 feet away from where the ball bounced.
Create an absolute value equation to represent the situation.
How high the did the ball bounce for the second student to catch it? | First, draw a picture on the coordinate system. For simplicity’s sake, arbitrarily put the first point on the $ y$-axis:
Find the equation of this absolute value function: since the vertex is at $ \left( {5,0} \right)$, $ f\left( x \right)=a\left| {x-5} \right|+0=a\left| {x-5} \right|$. To find $ a$ (which should be positive), use the non-vertex point $ \left( {0,6} \right)$ for $ \left( {x,f\left( x \right)} \right)$: $ \displaystyle 6=a\left| {0-5} \right|;\,\,\,\,\,6=5a;\,\,\,\,a=\frac{6}{5}$. Thus, the equation of the path of the ball is $ \displaystyle f\left( x \right)=\frac{6}{5}\left| {x-5} \right|$. Since the second student is 4 feet away from the vertex, use $ 4+5=9$ for $ x$ to get the height of the ball at that point: $ \displaystyle f\left( x \right)=\frac{6}{5}\left| {9-5} \right|=\frac{6}{5}\times 4=\frac{{24}}{5}=4\frac{4}{5}$ feet. That looks right! |
| Suppose that a coordinate grid is placed over a putt-putt golf course, where Amy is playing golf. The golf ball starts at $ \left( {3,2} \right)$, the hole is at $ \left( {5,2} \right)$, and Amy wants to bank the ball (bounce the ball off of a wall) at $ \left( {4,6} \right)$.
Write an equation for the path of the ball. | First, draw a picture on a coordinate system:
Find the equation of this absolute value function: since the vertex is at $ \left( {4,6} \right)$, $ f\left( x \right)=a\left| {x-4} \right|+6$. To find $ a$ (which should be negative), either put one of the non-vertex points in for $ x$ and $ y$, or notice that the graph is upside down and has a vertical stretch of 4. Use point $ \left( {3,2} \right)$: $ 2=a\left| {3-4} \right|+6;\,\,\,\,\,2=a+6;\,\,\,\,a=-4$. Thus, the equation of the path of the ball is $ f\left( x \right)=-4\left| {x-4} \right|+6$, for $ 3\le x\le 5$ (interval $ \left[ {3,5} \right]$). |
Here are examples that are absolute value inequality applications. Use this rule of thumb: the absolute value of a difference is usually on the left-hand side, the amount that differs or varies is usually on the right-hand side, with a $ <$ or $ \le $ sign in between.
| Absolute Value Inequality Problem | Solution |
| $ x$ differs from 2.5 by less than .8.
Express this situation as an absolute value inequality and solve for $ x$.
| Since $ x$ differs from 2.5 by less than an amount, we could add or subtract 2.5 from $ x$ and that amount would have to be less than .8. By doing this, we can see that $ x$ would have to be between $ 1.7\,\,\,(2.5-.8)$ and $ 3.3\,\,\,(2.5+.8)$. Put this into an absolute value equation since we don’t care if the difference of $ x$ and 2.5 is positive or negative: it still needs to be $ <$ .8: $ \begin{array}{c}\left| {\,x-2.5} \right|<.8\\\,x-2.5<.8\,\,\text{ or }\,\,x-2.5>-.8\\\,x<3.3\,\,\text{ or }\,\,x>1.7\\\,\left( {1.7,\,\,3.3} \right)\end{array}$ Thus, $ x$ would need to be between 1.7 and 3.3. This makes sense since anything outside of these values would be more than .8 units from 2.5. |
| The thermostat at Lindsay’s house is set at 72°, but the temperature may vary by as much as 3°.
Write an absolute value inequality to model this situation, and solve for the range of possible temperatures, $ t$. | Think about this with real numbers before setting up the absolute value inequality. Since the temperature can vary at most 3° either way from 72°, Lindsay’s house could range anywhere from $ 72-3$ degrees to $ 72+3$ degrees, or from 69° to 75° ($ 72\pm 3$). Now, write this using an absolute value. With “at most” 3°, we have $ \le $ 3, and the difference between this temperature and 72 has to be in this range. However, this difference could be positive or negative $ (69-72=-3;\,\,\,75-72=3)$. Therefore, we can write this temperature range as an absolute value and solve: .$ \displaystyle \begin{array}{c}\left| {\,t-72} \right|\le 3\\t-72\le 3\,\,\,\,\,\text{and}\,\,\,\,\,\,\,t-72\ge -3\,\,\,\\\,t\le 75\,\,\,\,\,\,\,\,\,\,\,\,t\ge 69\,\\69\le t\le 75\end{array}$ This is tricky, but the way to do these problems is to always make an absolute value of the difference of the variable and the starting point (such as 72°) and make this $ \le $ the amount that varies (the 3°). |
| A bird is approaching Erin, a photographer, and she films it. She starts her video when the bird is 100 feet horizontally from her, and continues filming until the bird is at least 50 feet past her. The bird is flying at a rate of 30 feet per second.
Write (don’t solve) an equation to find the range of times after Erin starts filming that the bird within 50 feet (horizontally) from her. |  This one’s a little tricky since we have to start measuring the distance (and thus the time) from point 0 feet, which is 100 feet away from Erin. Erin is at the 100-ft mark, so the distance (rate times time) relative to Erin before the bird gets to Erin is $ 100-30t$, and after the bird gets to Erin is $ 30t-100$ (try real numbers!).
These distances must be within 50 feet, so the absolute value inequality to find these times is $ \left| {100-30t} \right|\le 50$, or $ \left| {30t-100} \right|\le 50$. |
| At one restaurant, fresh lobsters are rejected if they weigh less than 1 pound, or more than 2.2 pounds.
Write an absolute value inequality to represent when lobsters are kept (not rejected) in the restaurant. |  Here’s the trick for these types of problems: if we start out with the middle (average) of 1 and 2.2, we can go lower or higher by a certain amount and this would represent when lobsters are kept. Then we can turn the problem into an absolute value problem.
The average of the lowest and highest values is $ \displaystyle \frac{{1+2.2}}{2}=1.6$. Therefore, if we start out with 1.6, we can either go down to 1 (by subtracting .6) or up to 2.2 (but adding .6) to get lobsters that aren’t rejected. Thus, we can use the absolute value inequality $ \left| {x-1.6} \right|<.6$ to describe the weight of lobsters that aren’t rejected. Try it; it works! |
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