Absolute Value Functions

NZ Level 7 (NZC) Level 2 (NCEA)

Solutions to 1-variable Linear Absolute Value Functions

Lesson

Like other functions, a function that has the absolute value symbols in its definition becomes an equation when its function value is specified. The equation may then be solved in order to find the values of the domain variable that are mapped to the specified function value.

The function defined over the real numbers by $y(x)=2|x-1|+3$`y`(`x`)=2|`x`−1|+3 becomes an equation that can be solved when the value of the function $y$`y` is specified. We might have, say, $y=21$`y`=21. Then we look for solutions to the equation $2|x-1|+3=21$2|`x`−1|+3=21.

To be sure of finding all the solutions to such an equation, we must recognise that the function definition contains the definitions of two different functions that are applicable over separate parts of the domain.

In the example given above, we see that $y(x)=2x+1$`y`(`x`)=2`x`+1 when $x\ge1$`x`≥1, and $y(x)=5-2x$`y`(`x`)=5−2`x` when $x\le1$`x`≤1. Thus, there are two equations that need to be solved.

We solve $2x+1=21$2`x`+1=21 and $5-2x=21$5−2`x`=21 separately and find that the solutions are $x=10$`x`=10 and $x=-8$`x`=−8, respectively.

The existence of these solutions can be seen in the following graphical representation.

An equation with linear components may have $0$0, $1$1, $2$2, $3$3 or infinitely many solutions. Care is needed to be sure that all solutions have been found in each case.

To illustrate some of the complications to could conceivably arise, we consider the equation

$3|x+1|-3|x|-2x=1$3|`x`+1|−3|`x`|−2`x`=1

We might think of rearranging this to $3|x+1|-(x+1)=3|x|+x$3|`x`+1|−(`x`+1)=3|`x`|+`x` and then considering the left and the right sides as separate functions that take the same values for certain values of $x$`x`. That is, the graphs intersect.

The left-hand function is equivalent to $-4-4x$−4−4`x` when $x\le-1$`x`≤−1 and to $2x+2$2`x`+2 when $x\ge-1$`x`≥−1.

The right-hand function is equivalent to $-2x$−2`x` when $x\le0$`x`≤0 and to $4x$4`x` when $x\ge0$`x`≥0.

The graphs of these four separate functions intersect in three points.

To find the three intersection points algebraically, we form three pairs of linear equations - one for each pair of intersecting segments.

We wish to know for what value of $x$`x` it is true that $-4-4x=-2x$−4−4`x`=−2`x`, we wish to know when $2x+2=-2x$2`x`+2=−2`x` and, we wish to know when $2x+2=4x$2`x`+2=4`x`.

These equations have, respectively, the solutions $x=-2$`x`=−2, $x=-\frac{1}{2}$`x`=−12 and $x=1$`x`=1, which should be verified by substitution into the left-hand-side of the original equation.

The equation $3|x+1|-3|x|-2x=1$3|`x`+1|−3|`x`|−2`x`=1 can also be solved without reference to its graph. The equation has three different representations, depending on whether $x\le-1$`x`≤−1, $-1\le x\le0$−1≤`x`≤0 or $0\le x$0≤`x`.

These are respectively:

$3(-x-1)-3(-x)-2x=1$3(−`x`−1)−3(−`x`)−2`x`=1, which after simplification is $-3-2x=1$−3−2`x`=1;

$3x+3+3x-2x=1$3`x`+3+3`x`−2`x`=1, which is the same as $4x+3=1$4`x`+3=1, and

$3x+3-3x-2x=1$3`x`+3−3`x`−2`x`=1, which simplifies to $3-2x=1$3−2`x`=1.

The solutions are as before.

Solve the equation $9\left|x\right|=4$9|`x`|=4.

Write both solutions as equations on the same line separated by a comma.

Consider the equation $\left|5x\right|=15$|5`x`|=15.

On the same set of axes, graph the functions $y=\left|5x\right|$

`y`=|5`x`| and $y=15$`y`=15.Loading Graph...Hence determine the solutions to the equation $\left|5x\right|=15$|5

`x`|=15. Write the solutions on the same line, separated by a comma.

Solve $\left|4x-8\right|+1=13$|4`x`−8|+1=13.

Write both solutions as equations on the same line separated by a comma.

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