where did the 1500 come from?

It is most likely the average expected loss of crops in kg when harvesting.

Could someone please explain where 6750a came from in Problem One? How was that number found?

The problem gave you: M(C(a))=6750a−1400 This was created by combining the 2 functions C(a) and M(c) by making C(a) as the input to M(c). Here's how that was done... We were also given: C(a)=7500a−1500 M(c) = 0.9c - 50 Insert C(a) as the input into M(c) and here's what M(C(a)) looks like before simplifying: M(C(a))=0.9(7500a−1500)−50 After you simplify, you get M(C(a)) = 6750a−1400 Hope this helps.

Why create and use composite functions when you can just break it down to simpler and smaller equations which still give you the correct answer?

i mean its much slicker just to have one equation inside of another. that way, you dont have to deal with all the separate pieces

im lost how it became 5000a-400...

the function was M(p)=0.2p-200 to start with. In order to make the function relate acres(a) to money(M), we must put the two relevent equations together. To do this, we multiply the function for Money(M) by the function for acres. So we get this: M(p(a))=o.2(25,000a-1000)-200. (we replaced the (p) on both sides of the equation with the function (p(a)).) Then from there all we need to do is just multiply 25000a and -1000 by o.2. When we do this we get M(p(a))=5000a-200-200. Then we just add -200 to -200, and that gives us M(p(a))=5000a-400. Hope this helps! :)

For the example equation on the top, does the lowercase "c" refer to "C(a)"? Does c = C(a)? If so, is it a rule to make it lowercase and why :(

Capital C is the name of the function that calculates kg of corn from planting "a" acres of corn. lowercase "c" is the variable used in the next function: M(c). It represents the input into function M. It happens to also represent kg of corn. This was done to help you see linkage opportunities between the functions. But, they do mean different things. "C" is name of a function, and "c" is the input to a different function.

How do you find the domain of a composite function?

The domain of a composite function f(g(x)) is all x in the domain of g such that g(x) is in the domain of f. Let's break this down. First off, the x has to be in the domain of g; if g(x) were say 1/x, then x = 0 could not be in the composite domain. Second of all, even if g(x) is defined, it has to be in the domain of f. Say f(x) equals 1 / (x - 1). Then if you choose an x such that g(x) = 1, making f(g(x)) = 1 / 0, that x cannot be in the domain of the composite function. Hope that I helped.

My question is why do we have to cram everything into one function when even though it might take longer, we could do the separate functions and solve it like that? I wish there was an article about that too because it would make more sense to me.

A good question. Different strokes for different folks. The received wisdom seems to be that reducing the number of steps in arriving at a solution to a problem is better. Imagine if f(x) = 4x^2 and g(x) = sqrt(x) g(f(x)) = g(4x^2) = sqrt(4x^2) = 2x The composite function g(x) = 2x is many orders of magnitude simpler than doing f(x) first and then g(x), especially if x is very large or, as I saw in in article on Khan Academy, x is not a "nice" number (e.g. x could be pi or 2.3727). However, if f(x) = 2(sqrt x) and g(x) = 9x^3, there's no harm in not composing the two into one composite function. g(f(x)) = g(2*sqrt x) = 9(2*sqrt x)^3, that's a lot of keystrokes on a calculator and easy to make (silly) mistakes.

can i get some help with this its kinda getting confusing?

Could you be more specific on what parts you are not understanding?

How would you find the value of the function if like you had f(g(-1)) how would you put that into in equation to solve?

Since this is currently real world problems, having a negative amount of land is impossible. You would solve it the same way though such as the potato farmer problem by solving P of -1, or substituting it at the end.

I noticed that the first problem with the corn has the function 6750a-1400 where the 1400 came from subtracting 50 from 1350. If I'm not mistaken, subtracting 50 from 1350 is actually 1300

The 1350 is "-1350", you are treating it as "+1350". -1350-50 = -1400 When the 2 numbers have the same sign, we add them and keep the common sign. Or, consider a number line. If you are at -1350 and need to subtrct 50, you move further left to -1400. Hope this helps.

Main content

Intro to composing functions

Google Classroom

Learn why we'd want to compose two functions together by looking at a farming example.

Cam is a farmer. Each year he plants seeds that turn into corn. The function below gives the amount of corn,

C

, in kilograms (kg), that he expects to produce if he plants corn on

a

acres of land.

$C (a) = 7500 a - 1500$ ‍

For example, if Cam plants two acres, he expects to produce

C (2) = 7500 (2) - 1500 = 13, 500

kg

of corn.

What Cam really wants to know is how much money he will make from selling this corn. So he uses the following function to predict the amount of money,

M

, in dollars, that he will earn from selling

c

kilograms of corn.

$M (c) = 0.9 c - 50$ ‍

So if Cam produces

13,500 kg

of corn, he can expect to make

M (13,500) = 0.9 (13,500) - 50 = $ 12,100

Notice that Cam has to use two separate functions to get from acres planted to expected earnings. The first function,

C

, takes acres to corn, while the second function,

M

, takes corn to money.

Wouldn't it be great if Cam could write a function that turned planted acres directly into expected earnings?

Creating a new function

We can indeed find the function that takes acres planted directly to expected earnings! To find this new function, let's think about the most general question: how much money does Cam expect to make if he plants corn seed on

a

acres of land?

Well, if Cam plants corn on

a

acres, he expects to produce

C (a)

kilograms of corn. And if he produces

C (a)

kilograms of corn, he expects to make

M (C (a))

dollars.

So, to find a general rule that converts

a

acres directly into expected earnings, we can find the expression

M (C (a))

But just how do we do this? Well, notice that in the expression

M (C (a))

, the input of function

M

C (a)

. So, to find this expression, we can substitute

C (a)

in for

c

in function

M

$\begin{aligned} M (c) & = 0.9 c - 50 \\ M (C (a)) & = 0.9 (C (a)) - 50 \\ = 0.9 (7500 a - 1500) - 50 Since C (a) = 7500 a - 1500 \\ = 6750 a - 1350 - 50 \\ = 6750 a - 1400 \end{aligned}$ ‍

So the function

M (C (a)) = 6750 a - 1400

converts acres planted directly into expected earnings. Let's use this new function to predict the amount of money that Cam would make from planting corn on two acres.

$M (C (2)) = 6750 (2) - 1400 = $ 12,100$ ‍

Cam can expect to make

$ 12,100

from planting corn on two acres of land, which is consistent with our previous work!

Defining composite functions

We just found what is called a composite function. Instead of substituting acres planted into the corn function, and then substituting the amount of corn produced into the money function, we found a function that takes the acres planted directly to the expected earnings.

We did this by substituting

C (a)

into function

M

, or by finding

M (C (a))

. Let's call this new function

M \circ C

, which is read as "

M

composed with

C

We now know that

(M \circ C) (a) = M (C (a))

. This, in fact, is the formal definition of function composition!

Visualizing the two methods

Here's a visual to help interpret the above definition.

Using both functions

C

and

M

, function

C

—the corn function—takes two to 13,500. Then, function

M

—the money function—takes 13,500 to

$

12,100.

Using the composite function, we see that function

M \circ C

takes two directly to

$

12,100.

The two are equivalent!

Now let's practice some problems.

Problem 1

Using the functions presented in the example, how much can Cam expect to earn if he sells all the corn produced on 1.5 acres?

For reference: $C (a) = 7500 a - 1500$ ‍, $M (c) = 0.9 c - 50$ ‍ and $M (C (a)) = 6750 a - 1400$ ‍

dollars

Since we just built the composite function that takes acres planted directly to expected earnings, let's use that to answer the question.

We know

M (C (a)) = 6750 a - 1400

. We can find how much Cam expects to earn if he plants corn on 1.5 acres by substituting 1.5 in for

a

in the above formula.

$\begin{aligned} M (C (a)) & = 6750 a - 1400 \\ M (C (1.5)) & = 6750 (1.5) - 1400 \\ = 8725 \end{aligned}$ ‍

Cam can expect to earn

$

8725 by selling the corn produced on 1.5 acres of land.

Alternatively, we could also use function

C

to turn acres into corn, and then function

M

to turn corn into money; however, since we already had the composite function, using it was more efficient!

Problem 2

Ben is a potato farmer. The function

P (a) = 25,000 a - 1000

gives the amount of potatoes,

P

, in kilograms, that he expects to produce from planting potatoes on

a

acres of land. The function

M (p) = 0.2 p - 200

gives the amount of money,

M

, in dollars, that Ben expects to make if he produces

p

kilograms of potatoes.

How much money can Ben expect to make if he sells all of the potatoes produced on the 3 acres?

One way to do this is to substitute acres planted into the potato function to find the amount of potatoes produced. Then, we can substitute this number into the money function. Let's do this.

First, let's find

P (3)

$\begin{aligned} P (3) & = 25,000 (3) - 1000 \\ = 74,000 \end{aligned}$ ‍

Now let's find

M (74,000)

$\begin{aligned} M (74,000) & = 0.2 (74,000) - 200 \\ = 14,600 \end{aligned}$ ‍

Alternatively, we could also find the composite function, and then substitute three into this function. This is shown below.

$\begin{aligned} M (P (a)) & = 0.2 (P (a)) - 200 \\ = 0.2 (25,000 a - 1000) - 200 \\ = 5000 a - 200 - 200 \\ = 5000 a - 400 \\ M (P (3)) & = 5000 (3) - 400 \\ = 14,600 \end{aligned}$ ‍

Either way, we see that Ben can expect to make

$

14,600 if he sells all of the potatoes produced on 3 acres of land.

Problem 3

Which of the following expressions gives the amount of money that Ben expects to make if he plants potatoes on $a$ ‍ acres of land?

Want to join the conversation?

Sort by:

ashley santana
Posted 8 years ago. Direct link to ashley santana's post “where did the 1500 come f...”
where did the 1500 come from?
Button navigates to signup pageComment on ashley santana's post “where did the 1500 come f...”
(28 votes)
Answer
- xilverd020
  Posted 7 years ago. Direct link to xilverd020's post “It is most likely the ave...”
  It is most likely the average expected loss of crops in kg when harvesting.
  Comment on xilverd020's post “It is most likely the ave...”
  (45 votes)
Liah C.
Posted 7 years ago. Direct link to Liah C.'s post “Could someone please expl...”
Could someone please explain where 6750a came from in Problem One? How was that number found?
Button navigates to signup pageComment on Liah C.'s post “Could someone please expl...”
(16 votes)
Answer
- Kim Seidel
  Posted 7 years ago. Direct link to Kim Seidel's post “The problem gave you: M(...”
  The problem gave you: M(C(a))=6750a−1400
  This was created by combining the 2 functions C(a) and M(c) by making C(a) as the input to M(c). Here's how that was done...
  We were also given:
  C(a)=7500a−1500
  M(c) = 0.9c - 50
  Insert C(a) as the input into M(c) and here's what M(C(a)) looks like before simplifying:
  M(C(a))=0.9(7500a−1500)−50
  After you simplify, you get M(C(a)) = 6750a−1400
  
  Hope this helps.
  Comment on Kim Seidel's post “The problem gave you: M(...”
  (50 votes)
avs68325
Posted 9 months ago. Direct link to avs68325's post “Why create and use compos...”
Why create and use composite functions when you can just break it down to simpler and smaller equations which still give you the correct answer?
Button navigates to signup pageComment on avs68325's post “Why create and use compos...”
(12 votes)
Answer
- Forever Learner
  Posted 9 months ago. Direct link to Forever Learner's post “i mean its much slicker j...”
  i mean its much slicker just to have one equation inside of another. that way, you dont have to deal with all the separate pieces
  Button navigates to signup page
  (22 votes)
KayleeS415
Posted 10 months ago. Direct link to KayleeS415's post “im lost how it became 500...”
im lost how it became 5000a-400...
Button navigates to signup pageButton navigates to signup page
(8 votes)
Answer
- mail4kissy
  Posted 10 months ago. Direct link to mail4kissy's post “the function was M(p)=0.2...”
  the function was M(p)=0.2p-200 to start with. In order to make the function relate acres(a) to money(M), we must put the two relevent equations together. To do this, we multiply the function for Money(M) by the function for acres. So we get this: M(p(a))=o.2(25,000a-1000)-200. (we replaced the (p) on both sides of the equation with the function (p(a)).) Then from there all we need to do is just multiply 25000a and -1000 by o.2. When we do this we get M(p(a))=5000a-200-200. Then we just add -200 to -200, and that gives us M(p(a))=5000a-400.
  Hope this helps! :)
  Button navigates to signup page
  (25 votes)
Kobe M
Posted 10 months ago. Direct link to Kobe M's post “I noticed that the first ...”
I noticed that the first problem with the corn has the function 6750a-1400 where the 1400 came from subtracting 50 from 1350. If I'm not mistaken, subtracting 50 from 1350 is actually 1300
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Kim Seidel
  Posted 10 months ago. Direct link to Kim Seidel's post “The 1350 is "-1350", you ...”
  The 1350 is "-1350", you are treating it as "+1350".
  -1350-50 = -1400
  When the 2 numbers have the same sign, we add them and keep the common sign. Or, consider a number line. If you are at -1350 and need to subtrct 50, you move further left to -1400.
  
  Hope this helps.
  Button navigates to signup page
  (21 votes)
Claire Jung
Posted 9 months ago. Direct link to Claire Jung's post “For the example equation ...”
For the example equation on the top, does the lowercase "c" refer to "C(a)"? Does c = C(a)? If so, is it a rule to make it lowercase and why :(
Button navigates to signup pageComment on Claire Jung's post “For the example equation ...”
(6 votes)
Answer
- Kim Seidel
  Posted 9 months ago. Direct link to Kim Seidel's post “Capital C is the name of ...”
  Capital C is the name of the function that calculates kg of corn from planting "a" acres of corn.
  
  lowercase "c" is the variable used in the next function: M(c). It represents the input into function M. It happens to also represent kg of corn. This was done to help you see linkage opportunities between the functions. But, they do mean different things. "C" is name of a function, and "c" is the input to a different function.
  Button navigates to signup page
  (10 votes)
Shalaya
Posted 8 years ago. Direct link to Shalaya's post “How would you find the va...”
How would you find the value of the function if like you had f(g(-1)) how would you put that into in equation to solve?
Button navigates to signup pageComment on Shalaya's post “How would you find the va...”
(5 votes)
Answer
- Dawson Stokes
  Posted 8 years ago. Direct link to Dawson Stokes's post “Since this is currently r...”
  Since this is currently real world problems, having a negative amount of land is impossible. You would solve it the same way though such as the potato farmer problem by solving P of -1, or substituting it at the end.
  Comment on Dawson Stokes's post “Since this is currently r...”
  (2 votes)
Silverleaf.yen
Posted 5 years ago. Direct link to Silverleaf.yen's post “How do you find the domai...”
How do you find the domain of a composite function?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Alex
  Posted 5 years ago. Direct link to Alex's post “The domain of a composite...”
  The domain of a composite function f(g(x)) is all x in the domain of g such that g(x) is in the domain of f.
  
  Let's break this down. First off, the x has to be in the domain of g; if g(x) were say 1/x, then x = 0 could not be in the composite domain. Second of all, even if g(x) is defined, it has to be in the domain of f. Say f(x) equals 1 / (x - 1). Then if you choose an x such that g(x) = 1, making f(g(x)) = 1 / 0, that x cannot be in the domain of the composite function. Hope that I helped.
  Button navigates to signup page
  (8 votes)
Lambert,Blaze
Posted 8 months ago. Direct link to Lambert,Blaze's post “My question is why do we ...”
My question is why do we have to cram everything into one function when even though it might take longer, we could do the separate functions and solve it like that? I wish there was an article about that too because it would make more sense to me.
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- Agent Smith
  Posted 8 months ago. Direct link to Agent Smith's post “A good question. Differen...”
  A good question. Different strokes for different folks. The received wisdom seems to be that reducing the number of steps in arriving at a solution to a problem is better.
  Imagine if f(x) = 4x^2 and g(x) = sqrt(x)
  g(f(x)) = g(4x^2) = sqrt(4x^2) = 2x
  The composite function g(x) = 2x is many orders of magnitude simpler than doing f(x) first and then g(x), especially if x is very large or, as I saw in in article on Khan Academy, x is not a "nice" number (e.g. x could be pi or 2.3727).
  
  However, if f(x) = 2(sqrt x) and g(x) = 9x^3, there's no harm in not composing the two into one composite function. g(f(x)) = g(2*sqrt x) = 9(2*sqrt x)^3, that's a lot of keystrokes on a calculator and easy to make (silly) mistakes.
  Comment on Agent Smith's post “A good question. Differen...”
  (7 votes)
Jason Reed
Posted 8 years ago. Direct link to Jason Reed's post “can i get some help with ...”
can i get some help with this its kinda getting confusing?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- xilverd020
  Posted 7 years ago. Direct link to xilverd020's post “Could you be more specifi...”
  Could you be more specific on what parts you are not understanding?
  Button navigates to signup page
  (4 votes)

Algebra (all content)

Course: Algebra (all content) > Unit 7