Learning about Piecewise Functions
What is a piecewise function?
A piecewise function is just how it sounds: it is simply a function with multiple pieces. These pieces can then be graphed or use to solve real word circumstances.
A piecewise function is just how it sounds: it is simply a function with multiple pieces. These pieces can then be graphed or use to solve real word circumstances.
Solving a Piecewise Function
1. Put the equation in standard form (mx+b)
2. Find the domain that matches your equation
3. Replace x in the corresponding equation
4. Solve
Example: Evaluate for x= 4 and x=0
f(X) ={ 3x if 0 < x
{ 5x+6 if 0 > x
Since the first value for x fits within the domain of the first part of the piecewise function then we say f(x)= 3(4) and that is 12 so one answer is 12.
We do the same for the second and it falls under the domain of the second piece so we plug it into the corresponding equation f(x)= 5(0)+6 which is 6 so the answer to the second part is 6.
Our answers are:
12 and 6
2. Find the domain that matches your equation
3. Replace x in the corresponding equation
4. Solve
Example: Evaluate for x= 4 and x=0
f(X) ={ 3x if 0 < x
{ 5x+6 if 0 > x
Since the first value for x fits within the domain of the first part of the piecewise function then we say f(x)= 3(4) and that is 12 so one answer is 12.
We do the same for the second and it falls under the domain of the second piece so we plug it into the corresponding equation f(x)= 5(0)+6 which is 6 so the answer to the second part is 6.
Our answers are:
12 and 6
Graphing Piecewise functions
1. Find the value where two (or both) equations interact and set that as an x value for both.
2. One equation should not be able to use that point so draw it as a hollow circle. (This allows these functions to pass the vertical line test.)
3. Draw the other point as a dot on its point. ^
4. From that point draw to the end of it's domain (if the domain is not equal to the point and would met another on a vertical line test draw another open circle and repeat)
5. Repeat as needed
6. Done
Example: f(x) ={2x+1 if x < 0
{6x if x > 0
Both questions interact with 0 for the domains so 2(0)+1 is 1 and 6(0) is 0 for the next one.
I have graphed them both on the bottom of the page and extended them to encompass the entire domain.
2. One equation should not be able to use that point so draw it as a hollow circle. (This allows these functions to pass the vertical line test.)
3. Draw the other point as a dot on its point. ^
4. From that point draw to the end of it's domain (if the domain is not equal to the point and would met another on a vertical line test draw another open circle and repeat)
5. Repeat as needed
6. Done
Example: f(x) ={2x+1 if x < 0
{6x if x > 0
Both questions interact with 0 for the domains so 2(0)+1 is 1 and 6(0) is 0 for the next one.
I have graphed them both on the bottom of the page and extended them to encompass the entire domain.
Finding the Equation of a Piecewise Function
The objective of this procedure is, given a piecewise function that is graphed, to determine the equation of that piecewise function.
1.) Determine the equation of each function
(For Linear Equations)
 Find two points from the same function
 Using the slope formula (see right), determine the slope of the function
 Use the slope to set up an algebraic equation (y=mx+b)
 Using one of the points from earlier, plug in those values for x and y
 Solve for "b"
 Express the equation in the form "y=mx+b"
(For Other Functions)
 Use transformations and parent functions to determine what the equation is
Repeat for all functions in the graph
2.) Determine Domain Restrictions
 Look at where each function begins and/or ends
 An open circle means a "less than" or a "greater than" sign
 A closed circle means a "less than or equal to", a "greater than or equal to", or an "equal to" sign
 For each function there should be at least one domain restriction, and no more than two
3.) State each function with its corresponding domain restriction(s) in the same form as all piecewise functions,
y={ (equation 1) if (domain restriction 1)
{ (equation 2) if (domain restriction 2)
and so on.
This is the equation of the piecewise function.
1.) Determine the equation of each function
(For Linear Equations)
 Find two points from the same function
 Using the slope formula (see right), determine the slope of the function
 Use the slope to set up an algebraic equation (y=mx+b)
 Using one of the points from earlier, plug in those values for x and y
 Solve for "b"
 Express the equation in the form "y=mx+b"
(For Other Functions)
 Use transformations and parent functions to determine what the equation is
Repeat for all functions in the graph
2.) Determine Domain Restrictions
 Look at where each function begins and/or ends
 An open circle means a "less than" or a "greater than" sign
 A closed circle means a "less than or equal to", a "greater than or equal to", or an "equal to" sign
 For each function there should be at least one domain restriction, and no more than two
3.) State each function with its corresponding domain restriction(s) in the same form as all piecewise functions,
y={ (equation 1) if (domain restriction 1)
{ (equation 2) if (domain restriction 2)
and so on.
This is the equation of the piecewise function.
Creating Equation Sample
1.) Both of the functions are linear
For the first function, I find the points (0,3) and (1.5,0)
The slope of the first equation is (03)/(1.50) = 2
3=2(0)+b ; b=3
y=2x+3 is the first equation
For the second function, I see that it is constant, at five.
This means that the second function is y=5
2.)The first Equation begins at 3 and ends at 1.
The second equation begins at 1 and ends at 5.
For the first equation, the domain restriction is: 3<x<1
For the second equation, the domain restriction is: 1<x<5, as all dots are open
3.) Our final answer is:
y= { 2x+3 if 3<x<1
{ 5 if 1<x<5
For the first function, I find the points (0,3) and (1.5,0)
The slope of the first equation is (03)/(1.50) = 2
3=2(0)+b ; b=3
y=2x+3 is the first equation
For the second function, I see that it is constant, at five.
This means that the second function is y=5
2.)The first Equation begins at 3 and ends at 1.
The second equation begins at 1 and ends at 5.
For the first equation, the domain restriction is: 3<x<1
For the second equation, the domain restriction is: 1<x<5, as all dots are open
3.) Our final answer is:
y= { 2x+3 if 3<x<1
{ 5 if 1<x<5
Real World Applications
The objective of this type of problem combines elements from all three of the other types of problems, and is to, given a word scenario, develop a table, an equation, a graph, and to use those to answer questions about the scenario.
(1.) Identify your variables, typically x and y
(2.) Create a table with the yvalues for certain xvalues
(3.) Use the table to create the equations that define the word problem (Hint: The domain restrictions are usually mentioned in the word problem itself.)
(4.) Follow the steps for the "graphing piecewise functions" above, to graph the equation (note, this is sometimes not required) , including the domain restrictions
(5.) Use the graph (or solving piecewise functions) to answer any questions about the data, often solving a piecewise function for a certain value (See above)
(2.) Create a table with the yvalues for certain xvalues
(3.) Use the table to create the equations that define the word problem (Hint: The domain restrictions are usually mentioned in the word problem itself.)
(4.) Follow the steps for the "graphing piecewise functions" above, to graph the equation (note, this is sometimes not required) , including the domain restrictions
(5.) Use the graph (or solving piecewise functions) to answer any questions about the data, often solving a piecewise function for a certain value (See above)
Real World Application Example
A store charges $15 per tshirt for orders of 50 or fewer Tshirts, $13.50 per tshirt for orders of 75 or fewer but more than 50 tshirts, and $12.50 per tshirt for orders of more than 75 tshirts. Create a table, equation and graph to model this situation.
Zero tshirts
$0

50 tshirts
$750

75 tshirts
$1,012.50

80 tshirts
$1,000

15x , if 0<=x<=50
C(x)= {13.50x, if 51<=x<=75
12.50x, if 76<=x
C(x)= {13.50x, if 51<=x<=75
12.50x, if 76<=x
Q. How much would 55 tshirts cost?
A. It fits into the second domain restriction, so $242.50
Q. How many tshirts would $600 buy?
A. 40 tshirts.
A. It fits into the second domain restriction, so $242.50
Q. How many tshirts would $600 buy?
A. 40 tshirts.