+
3 Tutorials that teach Adding and Subtracting Functions
Take your pick:
Adding and Subtracting Functions

Adding and Subtracting Functions

Rating:
Rating
(0)
Author: Colleen Atakpu
Description:

Adding and Subtracting Functions

(more)
See More

Try Our College Algebra Course. For FREE.

Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to over 2,000 colleges and universities.*

Begin Free Trial
No credit card required

25 Sophia partners guarantee credit transfer.

221 Institutions have accepted or given pre-approval for credit transfer.

* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 20 of Sophia’s online courses. More than 2,000 colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.

Tutorial

Video Transcription

Download PDF

Today we're going to talk about adding and subtracting functions. Remember, a function is just a special type of relationship between two sets of values. So we'll do some examples both adding and subtracting functions.

Let's start by doing some examples adding functions. I've got one function f of x, which is equal to x squared plus 10x and another function, g of x, which is equal to negative 8x plus 3. So first, I want to find f of 3 plus g of negative 5.

So I first to find the value of f of 3, which is the value of my function f of x when x is 3. And I'm going to add to that g of negative 5, which is the value of my function g of x when x is negative 5.

So to find f of 3, I'm going to replace each of my x variables with 3. So f of 3 will become 3 squared plus 10 times 3. 3 squared is going to give me 9. And 10 times 3 is going to give me 30. 9 plus 30 will give me 39.

So I found that f of 3 is 39, or my value of the function f of x is 39 when x is 3. Similarly, to find g of negative 5, I'll substitute negative 5 for my x here. So here I'll have negative 8 times negative 5 plus 3.

Negative 8 times negative 5 is 40, bring down the plus 3. 40 plus 3 will give me 43. So I found that g of negative 5 is 43.

So now finally, to find f of 3 plus g of negative 5, I simply need to combine these two values together. 39 plus 43 is going to give me a value of 82. So I found that f of 3 plus g of negative 5 is equal to 82.

Let's do another example. Here I want to find f of 4 plus g of 4. So similarly, I can find the value of my function f of x, when x is equal to 4 and add to that my value for the function g of x, when x is equal to 4.

However, because my argument, or my input value, is the same for both functions, I can simply add the functions together first, and then evaluate when x is equal to 4. So I'm going to start by adding my two functions together. So I have f of x plus g of x will be equal to x squared plus 10x plus my function g of x, negative 8x plus 3.

I can add these two expressions together by combining like terms. So I have 10x and a negative 8x. I can combine that to give me just 2x. And I bring down my other terms.

So I find that f of x plus g of x is x squared plus 2x plus 3. I can also write f of x plus g of x as f plus g of x. So now that I have found an expression for f plus g of x, or f of x plus g of x, now I can evaluate this at x is equal to 4.

So I'm going to substitute 4 in for my x variable and simplify. 4 squared is 16. 2 times 4 is 8. 16 plus 8 is going to give me 24. And 24 plus 3 will give me 27. So I found that f plus g of 4 is equal to 27.

So let's do some example subtracting functions. I've got a function, f of x, which is equal to 3x squared, and another function, g of x, which is equal to 4x minus 6. So first I want to find the value of f of 1 minus g of 2.

So I can find this by first determining what f of 1 is, and then determining what g of 2 is, and then subtracting those two values. So to find f of 1, I'm going to replace my x variable with 1.

So this is going to become 3 times 1 squared. Simplifying that, 1 squared will give me 1. And multiplying 3 times gives me 3. So I found that the value of f of 1 is 3.

Defining g of 2, I'm going to replace my x variable with 2. So this is going to be called 4 times 2 minus 6. Simplifying 4 times 2 will give me 8. And 8 minus 6 will give me 2. So I found that the value of g of 2 is 2.

So now to find f of 1 minus g of 2, I can combine these two by subtracting 3 minus 2, which gives me 1. So I found that the value of f of 1 minus g of 2 is equal to 1.

Let's do another example. Now I want to find f of 5 minus g of 5. So similarly to this last example, I could find f of 5 and find g of 5, and subtract those two values. But because the argument, or the input value, is the same, I can also determine an expression for f of x minus g of x, and then evaluate that at x is equal to 5.

So to find f of x minus g of x, I have 3x squared and I'm subtracting g of x, which is 4x minus 6. So now because I'm subtracting this expression, I know that that's the same as adding the opposite. So I can write this as 3x squared plus negative 4x.

So the opposite of 4x is negative 4x plus a positive 6. The opposite of negative 6 is positive 6. So now I can look for any like terms, and I see that I don't have any.

So my expression for f of x minus g of x is going to be 3x squared minus 4x plus 6. And I can write that as f minus g of x. So now that I have my expression for f minus g of x, I can find f minus g of 5 by replacing my x variables with 5.

So this becomes 3 times 5 squared minus 4 times 5 plus 6. Simplifying, 5 squared is 25. 3 times 25 is 75. 4 times 5 is 20-- bring down the plus 6. And finally, 75 minus 20 is going to give me 55. Adding 6 to that will give me 61. So I found that f minus g of 5 is 61, which is to same as f of 5 minus g of 5.

So let's go very key points from today. To find the sum or difference of two are more functions, evaluate each function separately and combine the values for each function. For a given value a in the domain of f of x and g of x, f of a plus g of a equals f plus g of a. And of a minus g of a equals f minus g of a.

So I hope that these key points and examples helped you understand a little bit more about adding and subtracting functions. Keep using your notes and keep on practicing, and soon you'll be a pro. Thanks for watching.

Notes on "Adding and Subtracting Functions"

Overview

(00:00 - 00:14) Introduction

(00:15 - 04:45) Adding Functions Examples

(04:46 - 08:55) Subtracting Functions Examples

(08:56 - 09:37) Summary

Key Formulas

None

Key Terms

None