Absolute Value

The answers to the True-False quiz at the end of the last post are all false. This brings us to absolute value, another topic I want to concentrate on for my upcoming Algebra 1 class. Absolute value becomes a concern in calculus too which I will discuss as the last example below.

There a several “definitions” of absolute value that I’ve seen over the years which I mostly do not like

  • The number without a sign – awful: all numbers except zero have a sign
  • The distance from zero on the number line – true, but not too useful especially with variables
  • The larger of a number and its opposite – true, but not to useful with variables

So I propose to give them an algorithm: If the number is positive, then the absolute value is the same number; if the number is negative, then its absolute value is its opposite. Of course, this is really the definition.

So I’ll soon express this in symbols

\left| a \right|=\left\{ \begin{matrix} a & \text{if }a\ge 0 \\ -a & \text{if }a<0 \\ \end{matrix} \right.

Now interestingly this is probably the first piecewise defined function an Algebra 1 student may see, or at least the first one that’s not artificial.  So this is a good place to start talking about piecewise defined function and the importance of talking about the domain. And of course, we’ll have to take a look at the graph.

Sometimes we will have to start solving equations and inequalities with absolute values. So here is the next thing I understand but do not like and will try to avoid. Solve the equation: \left| x \right|=3 , Answer including work: x=\pm 3. But I think a longer way around is also better:

If x<0 then \left| x \right|=-x=3 so x=-3 or if x>0, then \left| x \right|=x=3 Solution: x=3 or x=-3.

Longer? Sure. I hope that by making the students write that a few times that when they get to solving  \left| x \right|>3 that it will be natural to say

If  x<0 then \left| x \right|=-x>3  so x<-3 or if x>0, then \left| x \right|=x>3 Solution:  x<-3 or x>3

The last case may take a little more discussion. Solve \left| x \right|<3. Starting the same way

If x<0 then \left| x \right|=-x<3 so x>-3 which really means -3<x<0

if x>0, then \left| x \right|=x<3 which really means 0\le x<3 . Then the union of these two sets looks like an intersection. The solution is -3<x<3

Quite often the equation and the two types of inequalities are treated as separate problems: with = you go with \pm  on the other side, with > you have a union pointing away from the origin and with < you have somehow an intersection.  Who needs to remember all that when this idea works all the time?

Example:  Solve \left| 4x-10 \right|<8

If 4x-10<0, then x<\tfrac{5}{2} and \left| 4x-10 \right|=-\left( 4x-10 \right)=-4x+10<8, so -4x<-2 and x>\tfrac{1}{2} or more precisely  \tfrac{1}{2}  If latex 4x-10\ge 0$, then x\ge \tfrac{5}{2}  and \left| 4x-10 \right|=4x-10<8, so 4x<18 and x<\tfrac{9}{2} or more precisely \tfrac{5}{2}<x<\tfrac{9}{2}. The union again becomes an intersection and the answer is \tfrac{1}{2}<x<\tfrac{9}{2}

Finally an example from calculus. On the 2008 AB exam, question 5 asked student to find the particular solution of a differential equation with the initial condition f\left( 2 \right)=0. After separating the variables, integrating, including the “+C” and substituting the initial condition students arrived at this equation which they now need to solve for y:

 \displaystyle \left| y-1 \right|={{e}^{\tfrac{1}{2}-\tfrac{1}{x}}}

How can you lose the absolute value sign? Simple, near the initial condition where y = 0,  \left( y-1 \right)<0 so replace \left| y-1 \right| with -\left( y-1 \right) and then go ahead and solve for y

 \displaystyle -\left( y-1 \right)={{e}^{\tfrac{1}{2}-\tfrac{1}{x}}}

\displaystyle y=1-{{e}^{\tfrac{1}{2}-\tfrac{1}{x}}}

I don’t think I’ll try this one in Algebra 1, but maybe it will come in handy when they get to calculus.


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