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