FG4 Absolute value functions

August 11, 2022 - 2:41pm — AJ (e67821)

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The absolute value of a number x gives a measure of its size or magnitude regardless of whether it is positive or negative. If a number is plotted on a number line then its absolute value can be considered to be the distance from zero.

Introduction

The absolute value of a number gives a measure of its size or magnitude regardless of whether it is positive or negative.

If a number is plotted on a number line then its absolute value can be considered to be the distance from zero.

The absolute value of a number or a pro-numeral is designated by two vertical lines such as $\left|\,\cdot\,\right|$. For example the absolute value of the pro-numeral $x$ is $\left|x\right|.$

Examples

$\left|2\right|=2$
$\left|-2\right|=2$
$\left|-4+3\right|=\left|-1\right|=1$
$\left|-8\right|+\left|-1\right|=8+1=9$
$\left|x\right|=7$ $\Rightarrow$ $x=7$ or $x=-7$

The Absolute Value Function and its Graph

The absolute value function is a hybrid function11 A hybrid function involves two or more cases. Each case depends on the domain of the function. defined as follows:22 In what follows, $\mathbb{R}$ is the set of real numbers.

\[\begin{align*} f:\mathbb{R}\rightarrow\mathbb{R},\textrm{where $f(x)=$ $\left|x\right|$ } & =\begin{cases} -x, & x<0\\ x & x\geq0 \end{cases} \end{align*}\]

with graph

The domain of $f(x)$=$\left|x\right|$ is $\mathbb{R}$ and the range of $f(x)$ is $\mathbb{R}^{+}\cup\left\{ 0\right\}$. That is the set of all positive real numbers and zero.

The graph of $y=\left|x\right|$ may be translated in the same way as the graphs of other functions. Compare the graphs of the following functions with that of $y=\left|x\right|$

The graph of $y=\left|x-2\right|$ is shown below

The graph of $y=\left|x\right|+1$ is shown below

The graph of $y=3-\left|x\right|=-\left|x\right|+3$ is shown below

and is the graph of $y=\left|x\right|$ reflected in the $x$ axis followed by a vertical shift of three units up.

In general, to sketch the graph of $y=\left|f\left(x\right)\right|$, we need to sketch the graph of $y=f\left(x\right)$ first and then reflect in the $x$-axis the portion of the graph which is below the $x$-axis.

Sketch $\left\{ \left(x,y\right):y=\left|x^{2}-1\right|\right\}$

The graph of this function is the graph of $y=x^{2}-1$ with the portion below the $x$-axis reflected in the $x$-axis and is shown below:

Equations and Inequalities Involving $\left|f\left(x\right)\right|$

Because $y=\left|f\left(x\right)\right|$ is a hybrid function, two cases must be considered when solving equations and inequalities.

Examples

Solve $\left|x-2\right|=3$

Solution:

If $\left|x-2\right|=3$ we must consider the two cases: \[\begin{align*} x-2 & =3\\ x & =3+2\\ & =5 \end{align*}\] and \[\begin{align*} x-2 & =-3\\ x & =-3+2\\ & =-1. \end{align*}\]

Hence the answer is $x=-1$ and $x=3.$

Solve $\left|2x+1\right|=\left|x-5\right|$.

Solution:

With an absolute value expression on each side of the equation it is easier to square both sides.
\[\begin{align*} \left|2x+1\right| & =\left|x-5\right|\\ \left(2x+1\right)^{2} & =\left(x-5\right)^{2}\\ 4x^{2}+4x+1 & =x^{2}-10x+25\\ 4x^{2}+4x+1-x^{2}+10x-25 & =0\\ 3x^{2}+14x-24 & =0\\ (3x-4)(x+6) & =0 \end{align*}\] So \[\begin{align*} 3x-4 & =0\\ x & =\frac{4}{3} \end{align*}\] or \[\begin{align*} x+6 & =0\\ x & =-6. \end{align*}\] Hence the answer is $x=4/3$ and $x=-6.$

Find the set of $x\in\mathbb{R}$ such that $\left|\frac{2-x}{3}\right|<4$.

Solution: 55 Care must be taken when multiplying or dividing an inequality by a negative number. In such cases the inequality is reversed. The answer to this type of question is in fact a set as it involves an infinite number of solutions.

We have

\[\begin{align*} \left|\frac{2-x}{3}\right| & <4. \end{align*}\] Multiplying each side by $3:$ \[\begin{align*} \left|2-x\right| & <12 \end{align*}\] and so

\[\begin{align*} -12<2-x & <12. \end{align*}\] Adding $2$ to all sides we get: \[\begin{align*} -14<-x & <10. \end{align*}\] Multiplying by $-1$, and noting the reversal of the inequality signs,

\[\begin{align*} 14>x & >-10\,\textrm{ or }-10<x<14. \end{align*}\] Hence the answer is that $x$ is greater than $-10$ but less than $14$ $.$ More formally, this may be expressed as a set \[\begin{align*} \left\{ x\in\mathbb{R}:-10<x<14\right\} . \end{align*}\]

Find the set of $x\in\mathbb{R}$ such that $\left|\frac{x-2}{3}\right|\geq2$.

Solution:

Multiply both sides by $3$ to get \[\begin{align*} \left|x-2\right| & \geq6. \end{align*}\] Hence \[\begin{align*} x-2 & \geq6\\ x & \geq8 \end{align*}\] or \[\begin{align*} x-2 & \leq-6\\ x & \leq-4. \end{align*}\] Hence the answer is \[\begin{align*} \left\{ x:x\leq-4\right\} & \cup\left\{ x:x\geq8\right\} . \end{align*}\]

Exercise 1

Evaluate:

$\text{a) $\left|-11\right|\quad\text{b) $\left|-9+4\right|\quad\text{c) $-\left|4\right|-\left|-5\right|\quad\text{d) $\left|-12\right|-\left|3\right|\quad\text{e) $\left|-30\right|\div\left|5\right|$ }$ }$ }$ }$ }$

Answers

$\text{a) $11\quad\text{b) $5\quad\text{c) $-9\quad\text{d) $9\quad\text{e) $6$ }$ }$ }$ }$ }$

Exercise 2

Sketch the graph of

$\text{a) $y=\left|x+4\right|\quad\quad\text{b) $y=\left|x-1\right|-3\qquad\text{c) $y=\left|3-x^{2}\right|$ }$ }$ }$

Answers

Exercise 3

Find for $x\in\mathbb{R}$

$\text{a) $\left\{ x:\left|x\right|=6\right\} $ }$

$\text{b) $\left\{ x:\left|x-1\right|<3\right\} $ }$

$\text{c) $\left\{ x:\left|\frac{x-3}{2}\right|\geq1\right\} $ }$

$\text{d) $\left\{ x:\left|\frac{x}{2}\right|=\left|x+2\right|\right\} $ }$

Answers

$\text{a}.\ \left\{ -6,6\right\} \quad\text{b}.\ \left\{ x:-2<x<4\right\} \quad\text{c}.\ \left\{ x:x\leq1\right\} \cup\left\{ x:x\geq5\right\} \quad\text{d}.\ \left\{ -4,-\frac{4}{3}\right\}$

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