Absolute Value Functions

Lesson

For any function, we understand the *domain *to be the set of values that the independent variable can take. The *range* is the image of the domain values under the effect of the function.

Absolute value functions of the form $|f(x)|$|`f`(`x`)| are defined on some subset of the real numbers, their domain, and this subset is mapped by the function into the non-negative real numbers. Thus, the range of an absolute value function of this form is a subset of the positive numbers.

A simple function of this form can be subjected to further shifts and reflections that may result in negative values appearing in the range. For example, $-|f(x)|$−|`f`(`x`)| is always negative, and $A-|f(x)|$`A`−|`f`(`x`)| is negative for $|f(x)|>A$|`f`(`x`)|>`A`.

The absolute value function applied to linear functions produces graphs made of line segments.

Thus, for example, $y=|x-4|$`y`=|`x`−4| is a combination of $y=4-x$`y`=4−`x` when $x\le4$`x`≤4, and $y=x-4$`y`=`x`−4 when $x>4$`x`>4. The natural domain, in this example, is the whole of the set of real numbers. The function maps the domain *into *the real numbers but, more precisely, the image or range of the function is the set of numbers greater than or equal to zero.

Applying a vertical shift and a reflection to the same function, we might ask for the domain and range of, say, $y=5-|x-4|$`y`=5−|`x`−4|. The domain is the set of real numbers, as before, but in this case, the range includes both positive and negative numbers.

The maximum value the function can take is $5$5, when $|x-4|=0$|`x`−4|=0. That is, when $x=4$`x`=4.

The two parts of the function are, thus, $y=x+1$`y`=`x`+1 when $x\le4$`x`≤4 and $y=9-x$`y`=9−`x` when $x>4$`x`>4.

The function has the value zero when $|x-4|=5$|`x`−4|=5. This occurs when $x=9$`x`=9 and when $x=-1$`x`=−1. Between these domain values, the function is positive and it is negative otherwise. The function can be made as large in the negative direction as we wish by making $|x-4|$|`x`−4| arbitrarily large.

Thus, the range is the set $y\le5$`y`≤5.

The absolute value function applied to quadratic functions produces graphs made of parabolic arcs.

The function given by $p(x)=x^2+2x-3$`p`(`x`)=`x`2+2`x`−3 has zeros at $x=1$`x`=1 and $x=-3$`x`=−3. Between these domain values, the function is negative and it is positive elsewhere. Thus, the critical points of the absolute value function $q(x)=|x^2+2x-3|$`q`(`x`)=|`x`2+2`x`−3| are at $x=1$`x`=1 and $x=-3$`x`=−3.

We can write $q(x)$`q`(`x`) as a combination of $q(x)=-p(x)$`q`(`x`)=−`p`(`x`) for $-3\le x\le1$−3≤`x`≤1 and $q(x)=p(x)$`q`(`x`)=`p`(`x`) elsewhere.

The natural domain is the set of real numbers and the range is the set of non-negative real numbers.

From $q(x)$`q`(`x`), we might construct another function, $r(x)=A-q(x)$`r`(`x`)=`A`−`q`(`x`). As in the linear example, the constant $A$`A` shifts the graph vertically and the negative sign reflects the graph about the line $q(x)=A$`q`(`x`)=`A`.

We might further complicate the construction by applying the absolute value function to just the linear part of $p(x)$`p`(`x`). Thus, we could have $y(x)=x^2+|2x-3|$`y`(`x`)=`x`2+|2`x`−3|.

As before, the domain can be all of the real numbers. The function maps the domain into the non-negative real numbers because both $x^2$`x`2 and $|2x-3|$|2`x`−3| are always non-negative. The function value can be made very large by allowing $x$`x` to be sufficiently large but we also wish to know the minimum value of the function in order to specify the range.

A critical value occurs where $|2x-3|=0$|2`x`−3|=0. That is, at $x=\frac{3}{2}$`x`=32. For $x\le\frac{3}{2}$`x`≤32, the function can be written $y_1(x)=x^2-2x+3$`y`1(`x`)=`x`2−2`x`+3, and for $x>\frac{3}{2}$`x`>32 we have $y_2(x)=x^2+2x-3$`y`2(`x`)=`x`2+2`x`−3.

Now, $y_2$`y`2 has zeros at $x=-3$`x`=−3 and $x=1$`x`=1 and a minimum between these values that must be less than zero. This minimum is not in the domain of definition of $y_2$`y`2. So, the minimum we seek must be the minimum of $y_1$`y`1.

The quadratic $y_1$`y`1 is never negative. By completing the square, we can rewrite it as $y_1(x)=x^2-2x+1+2=(x-1)^2+2$`y`1(`x`)=`x`2−2`x`+1+2=(`x`−1)2+2. This expression is at its smallest when $x-1=0$`x`−1=0. That is, when $x=1$`x`=1, which is within the domain of $y_1$`y`1.

At this point, $y(x)=2$`y`(`x`)=2. Hence, we conclude that the range of $y(x)=x^2+|2x-3|$`y`(`x`)=`x`2+|2`x`−3| is $y\ge2$`y`≥2.

Consider the function that has been graphed.

Loading Graph...

What is the domain of the function?

all real $x$

`x`A$x<3$

`x`<3B$x\ge0$

`x`≥0C$x>0$

`x`>0Dall real $x$

`x`A$x<3$

`x`<3B$x\ge0$

`x`≥0C$x>0$

`x`>0DWhat is the range of the function? Give your answer as an inequality.

Consider the graph of the function below:

Loading Graph...

What is the domain of the function?

$x\ge2$

`x`≥2Aall real $x$

`x`B$x>0$

`x`>0C$x>2$

`x`>2D$x\ge2$

`x`≥2Aall real $x$

`x`B$x>0$

`x`>0C$x>2$

`x`>2DWhat is the range of the function? Give your answer as an inequality.

Consider the function $y=\frac{\left|x-3\right|}{x-3}$`y`=|`x`−3|`x`−3.

Select the single option that states the domain of the function.

$x>3$

`x`>3A$x$

`x`$\in$∈$\mathbb{R}$ℝ, $x\ne3$`x`≠3B$x>0,x<0$

`x`>0,`x`<0C$x$

`x`$\in$∈$\mathbb{R}$ℝD$x>3$

`x`>3A$x$

`x`$\in$∈$\mathbb{R}$ℝ, $x\ne3$`x`≠3B$x>0,x<0$

`x`>0,`x`<0C$x$

`x`$\in$∈$\mathbb{R}$ℝDSimplify the function for $x>3$

`x`>3.Simplify the function for $x<3$

`x`<3.Hence or otherwise, state the range of the function.

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

Apply graphical methods in solving problems