Absolute Value

The majority of students learn about absolute value long before high school. That is, they learn a lot of wrong things about absolute value.

  • They learn that “the absolute value of a number is the number without its sign” or some such nonsense. All numbers, except zero have a sign!  This sort of works with numbers, but becomes a problem when variables appear. True or false | x | = x? True or false | –x | = x? Most kids will say they are both true; in fact, as you know, they are both false.
  • They also learn that “the absolute value of a number is its distance from zero on the number line.” True and works for numbers, but what about variables?
  • They learn that “the absolute value of a number is the larger of the number and its opposite.” True again. How do you use it with variables?
  • They learn \left| x \right|=\sqrt{{{x}^{2}}} which is correct, useful for order-of-operation practice, and useful in other ways later, But they still compute \sqrt{{{\left( -3 \right)}^{2}}}=-3 and  \sqrt{{{x}^{2}}}=x since the square and square “cancel each other out.”

So here is a good vertical team topic. Get to those teachers in elementary and middle school and be sure they are not doing any of the above. They should start with the correct definition in words:

  • The absolute value of a negative number is its opposite.
  • The absolute value of a positive number (or zero) number is the same number.

This works all the time and will continue to work all the time. Teaching anything else will eventually require unlearning what they are using, and unlearning is far more difficult than learning.

When they start using variables and reading symbols translated into English, then the definition becomes their first piecewise define function:

  • \text{ If }x\ge 0,\text{ then }\ \left| x \right|=x;  and if x<0,\text{ then }\left| x \right|=-x
  • \left| x \right|=\left\{ \begin{matrix} x & \text{ if }x\ge 0 \\ -x & \text{ if }x<0 \\ \end{matrix} \right.

When reading this definition be sure to say “the opposite of the number” not “negative x” which in this case is probably a positive number.

Give variations of the two True-False questions above on every quiz and test until everyone gets it right!

When you see absolute value bars and want to be rid of them the first question to ask is, “Is the argument positive or negative? “Any time there is an absolute value situation, this is the way to proceed.

And yes, this does show up on the AP Calculus exams. Consider \int_{0}^{1}{\left| x-1 \right|dx} which appeared as a multiple-choice question a few years ago. Give it a try before reading on.

On the interval of integration, [0,1], \left( x-1 \right)\le 0 so \left| x-1 \right|=-\left( x-1 \right)

\displaystyle \int_{0}^{1}{\left| x-1 \right|dx}=\int_{0}^{1}{-\left( x-1 \right)}dx=\left. -\tfrac{1}{2}{{x}^{2}}+x \right|_{0}^{1}=-\tfrac{1}{2}+1-0=\tfrac{1}{2}

Now try \displaystyle \int_{0}^{1}{\sqrt{{{x}^{2}}-2x+1}\,dx}, or did we do this one already?


1 thought on “Absolutely

  1. When finding the integral of an absolute value where the limits span the function you end up with two expressions to integrate. While the analytical method has merit because it reinforces the definition of absolute value I usually have my students do these geometrically I.e find the x intercept ( assuming a simple abs function that bounces off the xaxis) the find the area of the two right triangles.


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