College Algebra Tutorial Operations with Functions

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College Algebra
Tutorial 30B:
Operations with Functions

Learning Objectives

After completing this tutorial, you should be able to:

Find the sum of functions.
Find the difference of functions.
Find the product of functions.
Find the quotient of functions.
Find the composition of functions.

Introduction

In this tutorial we will be working with functions. Note as you are going through this lesson that a lot of the things we are doing we have done before with expressions. Like adding, subtracting, multiplying and dividing. What is new here is we are specifically looking at these same operations with functions this time. I think we are ready to forge ahead.

Tutorial

The following show us how to perform the different operations on functions.

Use the functions and to illustrate the operations:

Sum of f + g

(f + g)(x) = f(x) + g(x)

This is a very straight forward process. When you want the sum of your functions you simply add the two functions together.

Example 1: If and then find (f + g)(x)

*Add the 2 functions

*Combine like terms

Difference of f - g

(f - g)(x) = f(x) - g(x)

Another straight forward idea, when you want the difference of your functions you simply take the first function minus the second function.

Example 2: If and then find (f - g)(x) and (f - g)(5)

*Take the difference of the 2 functions
*Subtract EVERY term of the 2nd ( )

*Plug 5 in for x in the diff. of the 2 functions found above

Since the difference function had already been found, we didn't have to take the difference of the two functions again. We could just merely plug in 5 into the already found difference function.

Product of f g

(f g)(x) = f(x)g(x)

Along the same idea as adding and subtracting, when you want to find the product of your functions you multiply the functions together.

Example 3: If and then find (fg)(x)

*Take the product of the 2 functions

*FOIL method to multiply

Quotient of f/ g

(f /g)(x) = f(x)/g(x)

Well, we don't want to leave division of functions out of the loop. It stands to reason that when you want to find the quotient of your functions you divide the functions.

Example 4: If and then find (f/g)(x) and (f/g)(1)

*Write as a quotient of the 2 functions

*Use the quotient found above to plug 1 in for x

Composite Function

Be careful, when you have a composite function, one function is inside of the other. It is not the same as taking the product of those functions.

Example 5: If and then find

*g is inside of f

*Substitute in x + 2 for g
*Plug x + 2 in for x in function f

Example 6: Let , , and . Find an equation defining each function and state the domain.

a) b) c) d) e) .

*Add the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator CANNOT equal zero:

*The den. x CANNOT equal zero

Putting these two sets together we get the domain:

*Subtract the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator CANNOT equal zero:

*The den. x CANNOT equal zero

Putting these two sets together we get the domain:

*Multiply the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator CANNOT equal zero:

*The den. x CANNOT equal zero

Putting these two sets together we get the domain:

*Divide the 2 functions

Domain:

The radicand CANNOT be negative. In other words it has to be greater than or equal to zero:

*Set radicand of x + 1 greater than or equal to 0 and solve

The denominator h(x) CANNOT equal zero:

*The den. h(x) = x + 3 CANNOT equal zero

Putting these two sets together we get the domain:

*h is inside of f is inside of g
*Substitute in x + 3 for h
*Substitute in sqroot(x + 3 + 1) for f(x + 3)

*Plug sqroot(x + 4) in for x in function g

Domain:

The radicand CANNOT be negative AND the denominator CANNOT equal zero. In other words it has to be greater than zero:

*Set radicand of x + 4 greater than 0 and solve

The domain would be:

Example 7: Let f = {(-3, 2), (-2, 4), (-1, 6), (0, 8)}, and g = {(-2, 5), (0, 7), (2, 9)} and h = {(-3, 0), (-2, 1)}. Find the following functions and state the domain.

a) b) c) d) e)

When you are looking for the domain of the sum of two functions that are given as sets, you are looking for the intersection of their domains.

Since the x values that f and g have in common are -2 and 0, then the domain would be {-2, 0}.

x = -2:

(-2, 9)

*Add together the corresponding y values to x = -2

x = 0:

(0, 15)

*Add together the corresponding y values to x = 0

Putting it together in ordered pairs we get:

When you are looking for the domain of the difference of two functions that are given as sets, you are looking for the intersection of their domains.

Since the x values that f and g have in common are -2 and 0, then the domain would be {-2, 0}.

x = -2:

(-2, -1)

*Subtract the corresponding y values to x = -2

x = 0:

(0, 1)

*Subtract the corresponding y values to x = 0

Putting it together in ordered pairs we get:

When you are looking for the domain of the product of two functions that are given as sets, you are looking for the intersection of their domains.

Since the x values that f and g have in common are -2 and 0, then the domain would be {-2, 0}.

x = -2:

(-2, 20)

*Multiply the corresponding y values to x = -2

x = 0:

(0, 56)

*Multiply the corresponding y values to x = 0

Putting it together in ordered pairs we get:

When you are looking for the domain of the quotient of two functions that are given as sets, you are looking for the intersection of their domains AND values of x that do NOT cause the denominator to equal 0.

The x values that f and h have in common are -3 and -2. However, h(-3) = 0, which would cause the denominator of the quotient to be 0.

So, the domain would be {-2}.

x = -2:

(-2, 4)

*Find the quotient of the corresponding y values to x = -2

Putting it together in an ordered pair we get:

When you are looking for the domain of the composition of two functions that are given as sets, you are looking for values that come from the domain of the inside function AND when you plug those values of x into the inside function, the output is in the domain of the outside function.

The x values of h are -3 and -2. However, g(h(-2)) = g(1), which is undefined. In other words, there are no ordered pairs in g that have a 1 for their x value.

So, the domain would be {-3}.

x = -3:

(-3, 7)

*h(-3) = 0

*Find the y value that corresponds to x = 0

Putting it together in an ordered pair we get:

Example 8: Let and . Write as a composition function using f and g.

When you are writing a composition function keep in mind that one function is inside of the other. You just have to figure out which function is the inside function and which is the outside function.

Note that if you put g inside of f you would get:

*Put g inside of f

Note that if you put f inside of g you would get:

*Put f inside of g

Hey this looks familiar.

Our answer is

Practice Problems

These are practice problems to help bring you to the next level. It will allow you to check and see if you have an understanding of these types of problems. Math works just like anything else, if you want to get good at it, then you need to practice it. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. In fact there is no such thing as too much practice.

To get the most out of these, you should work the problem out on your own and then check your answer by clicking on the link for the answer/discussion for that problem. At the link you will find the answer as well as any steps that went into finding that answer.