Table of Contents MathBits.com Terms of Use TI-84+ Web Site  
Absolute Value Inequalities

I N E Q U A L I T I E S:

Example 1:     Graph:  y < | x + 2 | 

Inequalities. such as this, are graphed as "relations"
(as they are not f1(x) "functions").

  • Open Add Graph. With cursor to the right of
    f1(x) =, hit to erase the "=" sign.
    In the drop down box,
    choose #2 y < for this example.
    Type your inequality.

    To get the absolute value symbol, use template:
    :
    OR type the letters "abs" directly from the alphabet keypad, followed by parentheses and expression.

     Not seeing a dotted graph?
    If your line graph and the shading are the same color, you may not see the graph as a dotted entry since the colors are fusing together.
    Change Color:
    #1 Action, #5 Set Conditions

    Change Line Thickness:
     #1 Action, #4 Attributes





For more directions
on Color and Thickness,
see Graphing Inequalities.



Example 2:    Solve:  | x | < 3

Unfortunately, the TI-Nspire CX II non-CAS will not graph and shade inequalities that do not contain both an "x" and a "y" value (such as Example 1).

The best this calculator can do is show us where the left and right sides of this inequality intersect. We can then "look" at the graph to see for which x-values the sections are "above" or "below" each other (depending upon the inequality).
  • Graph y = | x | and graph y = 3

  • Find the intersection points.
    , #6 Analyze Graph , #4 Intersection

  • Examine the graphs:
    The graphs intersect at (-3,3) and (3,3).
    The x-values of | x | are "    below" (less than) the horizontal red line y = 3, BETWEEN these two intersection points.

  • Answer:  interval notation (-3,3)
    Can also be written as -3 < x < 3.
    Can also be written as x > -3 and x < 3.






Example 3:      Solve: | 2x - 1 | > 0.5x + 2 

Follow the same steps as Example 2.

In this problem, we are looking for x-values for which the absolute value will be "above" or "on" (greater than or equal to) the red line.

For the x-values that are "above or on" to the right and to the left of the two intersection locations, including the intersection points themselves, will make the absolute value greater than the linear equation.

• Answer:  (-∞, -0.4] U [2,∞)
Can also be written as x < -0.4 or x > 2


 

 
Example 4:   Solve:   v1
From the algebraic solution, we know that this is a "two part" type of problem (a compound inequality).

Algebraic solution:
b2        b4
Answer:  The values that make the compound inequality true are f3.
The answers fall in -7 < x < -3 or they fall in 1 < x < 5.

 

Calculator solution: 
• Graph the three equations involved in this solution.
f1(x) = | x + 1 |
f2(x) = 2
f3(x) = 6

• Find the four intersection points.

• For the left hand portion: find the x-values for which the absolute value is "above" (greater than) 2.
The red section: It is above 2 for x > 1 or for x < -3

• For the right hand portion: find the x-values for which the absolute value is "below" (less than) 6.
The black section: It is below 6 for -7 < x < 5

• Find where these two solutions overlap one another
(where the x-values for the black arrows and the red arrows overlap). The green arrows show the overlap.

 

Answer: -7 < x < -3 or 1 < x < 5.


divider
Table of Contents | MathBits.com | Terms of Use