# on the TI-89

## Entering |x|:

The absolute value function, |x|, is written as "abs(x)" on the TI-89. You can type the function name from the keyboard, or enter it from the CATALOG or the Math menu as shown below:

 To select "abs(" from the catalog, first press . If "abs(" is not on the screen, press "A" (the "=" key - don't press since the calculator expects a letter). Use the blue arrows to highlight "abs(" as shown, then press . To select "abs(" from the math menu, first press <2nd>"5". The "Number" submenu will be highlighted, so press the right blue arrow and select "2: abs(".

## Inequalities Involving Absolute Value:

The TI-89 won't solve inequalities involving absolute value directly - you need to use properties 5 and 6 on page A7 of the Stewart text to convert the absolute value statement into a "double" inequality. These properties are, more or less:

5. |x| < a if and only if -a < x < a

6. |x| > a if and only if x > a or x < -a

Note:

• The phrase "if and only if" (often abbreviated "iff" or " ") means that both the statement and its converse are true. In other words, "A iff B" means that both "If A (is true) then B (is true)." and "If B then A" are true.
• Statement 5 says "If the absolute value of x is less than a, then x is between -a and a." AND "If x is between -a and a, then the absolute value of x is less than a.".
• Statement 6 says "If the absolute value of x is bigger than a, then either x is bigger than a or x is smaller than -a." AND "If x is bigger than a or x is less than -a, then the absolute value of x is bigger than a.".
• The symbol "x" represents any symbol OR GROUP OF SYMBOLS - the statements aren't true for just "x". Thus, property 5 guarantees that if |4x - 3| < 5, then -5 < 4x - 3 < 5, and also that if -3 < 8p3 + p - 6 < 3, then |8p3 + p - 6| < 3
• The properties above apply whether you are using "" or "<", "" or ">".

Here are some examples:

 This is problem #38 on p. A8 of the Larsen text, which says, in part "solve the inequality ". According to the second property of absolute value shown above, this is equivalent to "". Of course, you still need to interpret the calculator's answer. This is problem #40 on p. A8 of the Larsen text, which says, in part "solve the inequality ". According to the first property of absolute value shown above, this is equivalent to "". Of course, you still need to interpret the calculator's answer.

last update November 26, 2007 by JL Stanbrough