FG4 Absolute value functions

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

1. $$\left|2\right|=2$$

2. $$\left|-2\right|=2$$

3. $$\left|-4+3\right|=\left|-1\right|=1$$

4. $$\left|-8\right|+\left|-1\right|=8+1=9$$

5. $$\left|x\right|=7$$ $$\Rightarrow$$ $$x=7$$ or $$x=-7$$

The Absolute Value Function and its Graph

The absolute value function is a hybrid function1 A hybrid function involves two or more cases. Each case depends on the domain of the function. defined as follows:2 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 $$y=\left|x\right|$$

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

and is the graph of $$y=\left|x\right|$$ translated horizontally two units to the right. 3 The graph of $$y=\left|x+2\right|$$ is the graph of $$y=\left|x\right|$$ shifted two units to the left.

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

and is the graph of $$y=\left|x\right|$$ translated vertically one unit up. 4 The graph of $$y=\left|x\right|-1$$ is the graph of $$y=\left|x\right|$$ translated vertically one unit down.

1. 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.

1. 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

1. 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 $$x=3.$$

1. 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 $$x=-6.$$

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

Solution: 5 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*}

1. 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:

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

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

Exercise 2

Sketch the graph of

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

Exercise 3

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

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

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

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

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

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