# Absolute Value Instructor: Dr. Jo Steig

That is, the absolute value of a number, a, is equal to the distance that ‘a’ is from zero. The two statements (above) are needed to define absolute value in order to insure that the output of an absolute value function is NEVER less than zero. Try the following few calculations on your calculator:

- | – 5 |
- | 55.6 |
- | e – 7 | (here, e represents the irrational number that is defined as the base of the natural log)

**INSIDE EXPRESSION IS A CONSTANT**

If the inside expression is a constant, then writing the expression without absolute values is easy. We just need to insure that out output is nonnegative. That is,

- | – 5 | = 5
- | 55.6 | = 55.6
- | e – 7| = 7 – e, or approximately 4.28

**INSIDE EXPRESSION IS A VARIABLE**

If the expression inside the absolute value includes a variable, there is not much we can do with it as long as the absolute values are there. Therefore, to get any work done we must first write it without absolute values. This is similar to receiving a computer disk on which the information has been condensed to save space. Before you can access the information you must first run a decompression program. Our decompression program for the absolute value is the definition given above.

The following three examples illustrate

(1.) If x 2 then | x – 2 | = – (x – 2) **Justification:** | x – 2 | is either x – 2 or – (x – 2) If x 2, then x – 2 is 0 (try a few numbers and see for yourself!)

But | x – 2 | cannot be negative.

So | x – 2 | = – (x – 2) = 2 – x

(2.) If x 8 then | 5 – x | = – (5 – x )

**Justification:** The reasoning here is similar to the last example. If x 8, then 5 – x 0

So | 5 – x | = – (5 – x) = x – 5

(3.) If x 4 then | x – 4 | = x – 4

**Justification:** This is true because as long as x is larger than 4, x – 4 will be positive.

**BUT, WHAT IF YOU DON’T KNOW ANYTHING ABOUT THE VARIABLES?**

When the inside expression includes variables and you do not know anything about the values of the variables, then writing the expression without absolute values takes a bit more care and our reliance on the definition above becomes more obvious. The end result will be a piecewise defined function that is similar to the original definition that was given (above) for absolute value .

The following example shows that process in detail.

**Example:** Write | 4 – 5x | without using absolute value

**Solution:** The expression inside the absolute value is (4 – 5x). In the original definition for absolute value, replace each occurance of ‘a’ with (4 – 5x). That is,

When we simplify the inequalities in the last statement, (4 – 5x) 0 becomes x 4/5 and (4 – 5x) 0 becomes x 4/5. The final statement is then

**Note:** While this last statement looks more complicated than the beginning statement of | 4 – 5x |,it is in a form that can be added, graphed, integrated (calculus) or differentiated (calculus). At this point we are just trying to develop the skill of writing an absolute value expression without the absolute value symbol. You will investigate addition, integration, and differentiation in a later course.

## Sample Problems

1. Write |x + 2| without absolute values if x 3

2. Write |x + 2| without absolute values if x -2

3. Write |x + 2| without absolute values if x -2

4. Write |5 – x| without absolute values if x 5

5. Write |5 – x| without absolute values if x 5

6. Write |x + 2| + |5 + x| without absolute values if x -2

7. Write |x + 2| + |5 – x| without absolute values if x 8

8. Write |x + 2| + |5 – x| without absolute values if x -3

9. Write | x + 2 | without using absolute value

10. Write |5 – x | without using absolute value

11. Write | 2x – 3 | without using absolute value

12. Write | 3x + 6 | without using absolute value

### Answers to Sample Problems

1. If x 3 then | x + 2 | = x + 2

2. If x – 2 then | x + 2 | = x + 2

3. If x – 2 then | x + 2 | = – (x + 2) = -x – 2

4. If x 5 then | 5 – x | = – ( 5 – x ) = x – 5

5. If x 5 then | 5 – x | = ( 5 – x )

6.If x – 2 then | x + 2 | + | 5 + x | = (x + 2) + (5 + x) = 2x + 7

7. If x 8 then | x + 2 | + | 5 – x | = (x + 2) – (5 – x) = x + 2 – 5 + x = 2x – 3

8. If x – 3 then | x + 2 | + | 5 – x | = – (x + 2) + (5 – x) = -x – 2 + 5 – x = 3 – 2x