2.5 Piecewise-defined Functions
A function defined symbolically (using a formula), but
using a different formula for different parts of its domain.
Example:
f(x) = 2
-2
if x < 0
if x  0
So, f(-1) = 2, f(2) = -2, f(0) = ??
f(100) = ??
Here’s the graph of f:
y



   





x




 for x = 0, the graph “jumps”
 called a point of discontinuity
 at a point of discontinuity:
o one “end” will usually have a solid dot,
o indicating the value at that point
o the other “end” must have a open dot
o two open dots indicate that the function is
undefined at the point of discontinuity
2.5-1
Absolute value function
f(x) = | x |
The graph:
y



   





x




Solving absolute value equations
 | x | = 5 is satisfied by BOTH x = 5 AND x = -5
 let’s use this idea:
Example: solve | 2x + 3 | + 2 = 5
Step 1: (must always start with |xxx| by itself)
| 2x + 3 | = 3
2x + 3 = 3
2x + 3 = -3
2x = 0
2x = -6
x=0
x = -3
Do they check?
Note: | x | = -5 clearly has no solution
2.5-2
Solving absolute value inequalities




the method shown here is the Alternative Method, p. 150
there is another method shown on p. 148
use either method, but this one is easier and better
in any event, don’t make up your own!
Two simple inequalities to think about:
1. | x | < 5
The solution consists of all points whose
absolute value is less than 5. Let’s graph
them and state the solution:
2. | x | > 5
The solution consists of all points whose
absolute value is greater than 5. Let’s graph
them and state the solution:
2.5-3
THE METHOD
Example:
|x-2|>5
What to do?
Here's what. In the following,  stands for any expression:
What you see: What you write: What you do:
||<a
-a <  < a
solve 3-part inequality
||>a
 < -a or  > a solve 2 inequalities
 careful!whole left-hand side must be enclosed in "|"
signs
 | x - 5 | + 2 > 7 does not fit the pattern
 must rewrite it first as | x - 5 | > 5, then apply the
method
 2-inequality case: always write the or; answer will use 
Example: |x – 2| - 5 > 0
Example: |x – 2| - 5 ≤ 0
2.5-4