NZ Level 7 (NZC) Level 2 (NCEA) Domain and Range of Absolute Value Functions I
Lesson

Consider the absolute value function definition given below and the associated graph.

 $\left|3x+2\right|$|3x+2|$=$= $3x+2$3x+2$,$, $x\ge-\frac{2}{3}$x≥−23​ $-\left(3x+2\right)$−(3x+2)$,$, $x<-\frac{2}{3}$x<−23​ The domain of this function is the whole real line - any value of $x$x can be substituted in, large or small, positive or negative. On the other hand, the range of the absolute value function illustrated above is the interval $[0,\infty)$[0,). Any non-negative value is a possible output, but we cannot obtain a negative output from any $x$x-value.

Also note that $|-3x-2|$|3x2| is just another way to define the same function. For a given value of $x$x the function $-3x-2$3x2 (without absolute value signs) has almost the same output as the function $3x+2$3x+2 - it just differs by its sign. The absolute value around each of these functions then makes the outputs exactly the same.

This is true in general: For all values of $a$a and $b$b (each positive or negative),

$|-ax-b|=|(-1)\times(ax+b)|=|-1|\times|ax+b|=|ax+b|$|axb|=|(1)×(ax+b)|=|1|×|ax+b|=|ax+b|.

This trick can be convenient whenever we want to move minus signs around inside the absolute value.

#### Worked example

Solve: Describe the natural domain and the range of the function given by the expression $|-x+2|$|x+2|.

Think: The absolute value function can take any number into it - no values of $x$x can break the function. What kinds of numbers $k$k could be the output? In mathematical terms, for what values of $k$k can $k=|-x+2|$k=|x+2|? Think about what value of $x$x makes $k=0$k=0. We could also use the trick discussed above and discuss the expression $|x-2|$|x2|, since it is equal to $|-x+2|$|x+2|, but it is not necessary.

Do: The domain of the function is the entire real line, $\left(-\infty,\infty\right)$(,).

As for the range, we can tell that $x=2$x=2 produces $k=0$k=0, and any value of $x$x either less than or greater than $2$2 will produce a positive value of $k$k. To confirm this, let's consider each case separately.

If $x>2$x>2 then $-x+2$x+2 will be negative. This means

$k=-(-x+2)=x-2$k=(x+2)=x2,

and by substituting a large enough $x$x we can make $k>0$k>0 as large as we like.

If $x<2$x<2 then $-x+2$x+2 will be positive. This means $k=-x+2$k=x+2, and by substituting a large enough $x$x we can make $k>0$k>0 as large as we like once again.

Taking these cases together we conclude that $k\ge0$k0 is the range. We have confirmed in algebraic symbols what is evident from a graph of the function: the range is the non-negative real numbers. Reflect: Do all absolute value functions have domain $\left(-\infty,\infty\right)$(,) and range $\left[0,\infty\right)$[0,)? When might this not be the case?

#### Practice questions

##### Question 1

Consider the function that has been graphed.

1. What is the domain of the function?

all real $x$x

A

$x<3$x<3

B

$x\ge0$x0

C

$x>0$x>0

D

all real $x$x

A

$x<3$x<3

B

$x\ge0$x0

C

$x>0$x>0

D
2. What is the range of the function? Give your answer as an inequality.

##### Question 2

Consider the graph of the function below.

1. What is the domain of the function?

$\left(0,\infty\right)$(0,)

A

$\left(-\infty,2\right)$(,2)

B

$\left[0,\infty\right]$[0,]

C

$\left(-\infty,\infty\right)$(,)

D

$\left(0,\infty\right)$(0,)

A

$\left(-\infty,2\right)$(,2)

B

$\left[0,\infty\right]$[0,]

C

$\left(-\infty,\infty\right)$(,)

D
2. What is the range of the function? Give your answer using interval notation.

##### Question 3

Consider the function defined as $y=\left|4x+12\right|$y=|4x+12|.

1. Fill in the gaps to completely define the expression $\left|4x+12\right|$|4x+12|.

 $-4x-12$−4x−12 for $x$x$<$<$\editable{}$ $\left|4x+12\right|$|4x+12| $=$= $\editable{}$ for $x=-3$x=−3 $4x+12$4x+12 for $x$x$>$>$\editable{}$
2. What is the domain of the function?

$x\ge12$x12

A

$x\ge-3$x3

B

$x\le-3$x3

C

all real $x$x

D

$x\ge12$x12

A

$x\ge-3$x3

B

$x\le-3$x3

C

all real $x$x

D
3. What is the range of the function? Give your answer as an inequality.

### Outcomes

#### M7-2

Display the graphs of linear and non-linear functions and connect the structure of the functions with their graphs

#### 91257

Apply graphical methods in solving problems