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### Absolute Value

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- Evaluate \(\left| 5 \right|\) and \(\left| { - 123} \right|\)

Show SolutionTo do these evaluations we need to remember the definition of absolute value.

\[\left| p \right| = \left\{ {\begin{array}{*{20}{c}}{\,\,\,p\hspace{0.25in}{\rm{if }}\,p \ge 0}\\{ - p\hspace{0.25in}{\rm{if }}\,p < 0}\end{array}} \right.\]With this definition the evaluations are easy.

\begin{align*}\left| 5 \right| & = 5\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} & \hspace{0.25in} & {\rm{because }}\,5 \ge 0\\ \left| { -123} \right| & = - \left( { - 123} \right) = 123 & \hspace{0.25in} & {\rm{because }} -123 < 0\end{align*}Remember that absolute value takes any nonzero number and makes sure that it’s positive.

- Eliminate the absolute value bars from \(\left| {3 - 8x} \right|\)

Show SolutionThis one is a little different from the first example. We first need to address a very common mistake with these.

\[\left| {3 - 8x} \right| \ne 3 + 8x\]Absolute value doesn’t just change all minus signs to plus signs. Remember that absolute value takes a number and makes sure that it’s positive or zero. To convince yourself of this try plugging in a number, say \(x = - 10\)

\[83 = \left| {83} \right| = \left| {3 + 80} \right| = \left| {3 - 8\left( { - 10} \right)} \right| \ne 3 + 8\left( { - 10} \right) = 3 - 80 = - 77\]There are two things wrong with this. First, is the fact that the two numbers aren’t even close to being the same so clearly it can’t be correct. Also note that if absolute value is supposed to make nonzero numbers positive how can it be that we got a -77 of out of it? Either one of these should show you that this isn’t correct, but together they show real problems with doing this, so don’t do it!

That doesn’t mean that we can’t eliminate the absolute value bars however. We just need to figure out what values of \(x\) will give positive numbers and what values of \(x\) will give negative numbers. Once we know this we can eliminate the absolute value bars. First notice the following (you do remember how to Solve Inequalities right?)

\begin{align*}3 - 8x & \ge 0\hspace{0.25in} \Rightarrow \hspace{0.25in}3 \ge 8x\hspace{0.25in}\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,x \le \frac{3}{8}\\ 3 - 8x & < 0\hspace{0.25in} \Rightarrow \hspace{0.25in}3 < 8x\hspace{0.25in}\,\,\,\,\, \Rightarrow \hspace{0.25in}\,\,\,\,x > \frac{3}{8}\end{align*}So, if \(x \le \frac{3}{8}\) then \(3 - 8x \ge 0\) and if \(x > \frac{3}{8}\) then \(3 - 8x < 0\). With this information we can now eliminate the absolute value bars.

\[\left| {3 - 8x} \right| = \left\{ {\begin{array}{*{20}{c}}{\,\,\,\,\,\,\,3 - 8x\hspace{0.25in}{\rm{if }}\,x \le \frac{3}{8}}\\{ - \left( {3 - 8x} \right)\hspace{0.25in}{\rm{if }}\,x > \frac{3}{8}}\end{array}} \right.\]Or,

\[\left| {3 - 8x} \right| = \left\{ {\begin{array}{*{20}{c}}{\,\,\,3 - 8x\hspace{0.25in}\hspace{0.25in}{\rm{if }}\,x \le \frac{3}{8}}\\{ - 3 + 8x\hspace{0.25in}\hspace{0.25in}{\rm{if }}\,x > \frac{3}{8}}\end{array}} \right.\]So, we can still eliminate the absolute value bars but we end up with two different formulas and the formula that we will use will depend upon what value of \(x\) that we’ve got.

On occasion you will be asked to do this kind of thing in a calculus class so it’s important that you can do this when the time comes around.

- List as many of the properties of absolute value as you can.

Show SolutionHere are a couple of basic properties of absolute value.

\[\left| p \right| \ge 0\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\left| { - p} \right| = \left| p \right|\]These should make some sense. The first is simply restating the results of the definition of absolute value. In other words, absolute value makes sure the result is positive or zero (if \(p\) = 0). The second is also a result of the definition. Since taking absolute value results in a positive quantity or zero it won’t matter if there is a minus sign in there or not.

We can use absolute value with products and quotients as follows

\[\left| {ab} \right| = \left| a \right|\,\,\left| b \right|\hspace{0.25in}\hspace{0.25in}\left| {\frac{a}{b}} \right| = \frac{{\left| a \right|}}{{\left| b \right|}}\]Notice that I didn’t include sums (or differences) here. That is because in general

\[\left| {a + b} \right| \ne \left| a \right| + \left| b \right|\]To convince yourself of this consider the following example

\[7 = \left| { - 7} \right| = \left| {2 - 9} \right| = \left| {2 + \left( { - 9} \right)} \right| \ne \left| 2 \right| + \left| { - 9} \right| = 2 + 9 = 11\]Clearly the two aren’t equal. This does lead to something that is often called the triangle inequality. The triangle inequality is

\[\left| {a + b} \right| \le \left| a \right| + \left| b \right|\]The triangle inequality isn’t used all that often in a Calculus course, but it’s a nice property of absolute value so I thought I’d include it.