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Applying Function Operations Practice Problems

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  • 0:05 Function Operations
  • 0:37 Example #1
  • 1:55 Example #2
  • 3:33 Example #3
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Lesson Transcript
Instructor: Luke Winspur

Luke has taught high school algebra and geometry, college calculus, and has a master's degree in education.

In this lesson, learn how to apply all the different properties of functions to solve complex problems. From function operations and composition to domain and range, get your practice here!

Function Operations

When it comes to functions in algebra, the major things to know are domain and range, function operations, inverse functions and function composition. While it's possible to learn these things separately, it's very easy to combine these topics into a single problem. This provides a good opportunity to review and apply your function knowledge, but it also makes for some pretty confusing problems. The point of this video will be to summarize your function knowledge and give you some practice with these complex questions.

Example #1

First, let's take a look at a twist on function composition. Instead of providing the algebraic function, you might just be given a graph of the function, maybe like this. Just because the information is being presented in a different way doesn't mean we can't answer the same type of questions: maybe find g(g(2)), for example.

Example 1 involves a composition of functions question
Composition of Functions Graph Example 1

This is a composition of functions question because we are substituting one function into another - in this case, one function into itself. Let's start by checking out the inside of this function, g(2), and using the graph to find out this information. In this case, it looks like when we plug x = 2 into the function, it produces an output of 5.

That means we can rewrite our problem to be this - g(5) - and again, go back to the graph to find this value as well. This time, we start by plugging in 5 for the x value, going up to the line, going over and finding that the corresponding output is 7. Therefore, g(g( 2)) = 7!

Example #2

Let's take a look at another example, one that combines multiple function topics. If f(x) = 2x - 7 and g(x) = x - 5, then what is the domain of f(x) / g(x)? For this question, we'll need to know both how to divide functions and also how to determine the domain of a function.

Before we can start thinking about the domain, we'll need to do the division to see what we're dealing with. As long as you don't psych yourself out, function operations like this one aren't too difficult. We simply substitute in what we know each function is into our fraction. That means we end up with f(x) / g(x) = 2x - 7 / x - 5. Now that we have our new function, it's time to start thinking about the domain. We'll need to recall that the domain of a function is most often all real numbers but that there are a few common properties that prevent this from being the case. The main ones are:

Looking at our function here, I don't see any square roots or logs, so we're good there. But we do have a fraction, which means dividing by 0 is a possibility. That would mean that x - 5 would have to be 0 for this to be an issue. Could that happen? Is there a value of x that makes x - 5 = 0? Sure there is, if x = 5! That means that x = 5 needs to be excluded from our domain. If we substituted that into our fraction, we would get an undefined value, which isn't allowed. Therefore, our answer is any real number except 5.

Example #3

The new function, h inverse, as the square root of x - 1
Example 3 Inverse Functions

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