1.6 Absolute value functions  (Page 4/7)

With both approaches, we will need to know first where the corresponding equality is true. In this case we first will find where $\left|x-5\right|=4.$ We do this because the absolute value is a function with no breaks, so the only way the function values can switch from being less than 4 to being greater than 4 is by passing through where the values equal 4. Solve $\left|x-5\right|=4.$

After determining that the absolute value is equal to 4 at $x=1$ and $x=9,$ we know the graph can change only from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals:

To determine when the function is less than 4, we could choose a value in each interval and see if the output is less than or greater than 4, as shown in [link] .

Because $1\le x\le 9$ is the only interval in which the output at the test value is less than 4, we can conclude that the solution to $\left|x-5\right|\le 4$ is $1\le x\le 9,$ or $\left[1,9\right].$

To use a graph, we can sketch the function $f(x)=\left|x-5\right|.$ To help us see where the outputs are 4, the line $g(x)=4$ could also be sketched as in [link] .

We can see the following:

• The output values of the absolute value are equal to 4 at $x=1$ and $x=9.$
• The graph of $f$ is below the graph of $g$ on $1<x<9.$ This means the output values of $f(x)$ are less than the output values of $g(x).$
• The absolute value is less than or equal to 4 between these two points, when $1\le x\le 9.$ In interval notation, this would be the interval $\left[1,9\right].$

We are trying to determine where $f(x)<0,$ which is when $-\frac{1}{2}|4x-5|+3<0.$ We begin by isolating the absolute value.

Next we solve for the equality $\left|4x-5\right|=6.$

Now, we can examine the graph of $f$ to observe where the output is negative. We will observe where the branches are below the x -axis. Notice that it is not even important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at $x=-\frac{1}{4}$ and $x=\frac{11}{4}$ and that the graph has been reflected vertically. See [link] .

We observe that the graph of the function is below the x -axis left of $x=-\frac{1}{4}$ and right of $x=\frac{11}{4}.$ This means the function values are negative to the left of the first horizontal intercept at $x=-\frac{1}{4},$ and negative to the right of the second intercept at $x=\frac{11}{4}.$ This gives us the solution to the inequality.

In interval notation, this would be $\left(-\infty ,-0.25\right)\cup \left(2.75,\infty \right).$