ARGH! I typed in the exact same answer as the one in the "I need help" paragraph but it won't register?

It won't work because the fraction is not simplified in 30/4. You have to simplify it into 15/2 and then it will work

Is there ever a circumstance where f(g(x)) = x but g(f(x)) isn't equal to x?

Interesting question! This could occur if f(x) fails to be 1-to-1 (and so fails to be invertible). For example, if f(x) = x^2 and g(x) = sqrt(x), then f(g(x)) = [sqrt(x)]^2 = x but g(f(x)) = sqrt(x^2) = |x|. Have a blessed, wonderful day!

Would the functions still be inverses if their result was equal but not x? For example f(g(x)) = x - 24 and g(f(x)) = x - 24. Would these functions still be inverses of each other?

No. If you use "x" as the input to either f(g(x)) or g(f(x)), you must get "x" as the final out. To be inverse functions, each must completely reverse the operations performed by the other function. In your case, your final result is different than "x", so some operations is not being reversed.

Can anyone SHOW me why f(g(x)) = g(f(x)) = x?

This might be a bit late... So, the reason why functions are inverses if f(g(x)) = x and g(f(x)) = x is because of the fundamental definition of inverse functions: that is, the input and output values are switched. So, let's take f(g(x)) = x. What is it saying? It is saying that if you INPUT the output of function g into function f, you get the input of function g as the OUTPUT. To make it more clear: x is the input of g, and g(x) is the output. However, inputting the output of g into f causes f to output x, which is the input of g. Now, for g(f(x)) = x, it is essentially the same thing. f(x) = output of f and x = input of f. Now, inputting f(x) - the output of f, into g gets you the output x - the input of f. See, it's just inverse functions in play. Hope this helped!

I NEED HELP! I'm working on a problem and have the g(f(x))=2(x+1/2)-1 as a problem I'm just trying to figure it out with the fraction. i stink writing stuff out so if you can understand please help me;(

Your working is wrong bokeum. It will be 2x+2/2 which wil become 2x/2 +2/2 to simplify it. It will become x+1-1 the (-1) from the original question. so it will equal x.

How are ever gonna use this in the real world?

you might not but if you go into a career that needs math then probably

in 2, how did they get from =4(4/1 x−10)+10 to =x-40+10

Multiply everything in brackets by 4. 4*x/4 = x 4*-10 = -40 4(x/4 - 10) + 10 = x-40+10. You stated the problem in the wrong form. It should be 4(x/4 - 10) + 10 instead of 4(4/x - 10) +10

Is the third question correct? because im getting x-4 for h(f(x)) no matter how many times i try

h(f(x)) = (3/2)( f(x) +8) [*write function*] h(f(x)) = (3/2)( ((2/3)x-8) +8) [*put f(x) in*] h(f(x)) = (3/2)( ((2/3)x) [-8, +8 *cancels out*] h(f(x)) = x [*here (3/2)*(2/3)=1*] f(h(x)) = (2/3)(h(x))-8 [write function] f(h(x)) = (2/3)((3/2)(x+8))-8 [*put h(x) in*] f(h(x)) = (2/3)((3/2)(x)(8))-8 [*expand, noting that the 3/2 applies to the x _and_ the +8*] f(h(x)) = (2/3)(((3/2)x)+12)-8 [*the (3/2)8 simplifies to 12* _BUT_ *is still enclosed with the 2/3 bracket*] f(h(x) = x+8-8 [*here (2/3)*(3/2) is 1, (2/3)12 is 8*] f(h(x) = x [*simplify*] h(f(x)) = x f(h(x)) = x The two functions are invertible. The main point I think you have got caught on is the conversion of +8 to 12 due to the 3/2, but then exiting that out of *all* of the brackets so that it is then not affected by the 2/3, which would result in x+4, instead of x. Hope this helps :D

Main content

Algebra (all content)

Course: Algebra (all content) > Unit 7

Lesson 21: Verifying that functions are inverses (Algebra 2 level)

Verifying inverse functions by composition

Google Classroom

Learn how to verify whether two functions are inverses by composing them. For example, are f(x)=5x-7 and g(x)=x/5+7 inverse functions?

This article includes a lot of function composition. If you need a review on this subject, we recommend that you go here before reading this article.

Inverse functions, in the most general sense, are functions that "reverse" each other. For example, if a function takes

a

b

, then the inverse must take

b

a

Let's take functions

f

and

g

for example:

f (x) = \frac{x + 1}{3}

and

g (x) = 3 x - 1

Notice how

f (5) = 2

and

g (2) = 5

[Please show me the calculations]

Here we see that when we apply

f

followed by

g

, we get the original input back. Written as a composition, this is

g (f (5)) = 5

But for two functions to be inverses, we have to show that this happens for all possible inputs regardless of the order in which

f

and

g

are applied. This gives rise to the inverse composition rule.

The inverse composition rule

These are the conditions for two functions

f

and

g

to be inverses:

$f (g (x)) = x$ ‍ for all $x$ ‍ in the domain of $g$ ‍
$g (f (x)) = x$ ‍ for all $x$ ‍ in the domain of $f$ ‍

This is because if

f

and

g

are inverses, composing

f

and

g

(in either order) creates the function that for every input returns that input. We call this function “the identity function".

Example 1: Functions $f$ ‍ and $g$ ‍ are inverses

Let's use the inverse composition rule to verify that

f

and

g

above are indeed inverse functions.

Recall that

f (x) = \frac{x + 1}{3}

and

g (x) = 3 x - 1

Let's find

f (g (x))

and

g (f (x))

$f (g (x))$ ‍	$g (f (x))$ ‍
$\begin{aligned} f (g (x)) & = \frac{g (x) + 1}{3} \\ = \frac{3 x - 1 + 1}{3} \\ = \frac{3 x}{3} \\ = x \end{aligned}$ ‍	$\begin{aligned} g (f (x)) & = 3 (f (x)) - 1 \\ = 3 (\frac{x + 1}{3}) - 1 \\ = x + 1 - 1 \\ = x \end{aligned}$ ‍

So we see that functions

f

and

g

are inverses because

f (g (x)) = x

and

g (f (x)) = x

Example 2: Functions $f$ ‍ and $g$ ‍ are not inverses

f (g (x))

g (f (x))

is not equal to

x

, then

f

and

g

cannot be inverses.

Let's try this for

f (x) = 5 x - 7

and

g (x) = \frac{x}{5} + 7

$f (g (x))$ ‍	$g (f (x))$ ‍
$\begin{aligned} f (g (x)) & = 5 (g (x)) - 7 \\ = 5 (\frac{x}{5} + 7) - 7 \\ = x + 35 - 7 \\ = x + 28 \end{aligned}$ ‍	$\begin{aligned} g (f (x)) & = \frac{f (x)}{5} + 7 \\ = \frac{5 x - 7}{5} + 7 \\ = x - \frac{7}{5} + 7 \\ = x + \frac{28}{5} \end{aligned}$ ‍

So functions

f

and

g

are not inverses because

f (g (x)) \neq x

and

g (f (x)) \neq x

(Note here, that we could have concluded that

f

and

g

were not inverses after showing that

f (g (x)) = x + 28

Check your understanding

In general, to check if

f

and

g

are inverse functions, we can compose them. If the result is

x

, the functions are inverses. Otherwise, they are not.

1) $f (x) = 2 x + 7$ ‍ and $h (x) = \frac{x - 7}{2}$ ‍

Write simplified expressions for $f (h (x))$ ‍ and $h (f (x))$ ‍ in terms of $x$ ‍.

f (h (x)) =

h (f (x)) =

Are functions $f$ ‍ and $h$ ‍ inverses?

[I need help!]

2) $f (x) = 4 x + 10$ ‍ and $g (x) = \frac{1}{4} x - 10$ ‍

Write simplified expressions for $f (g (x))$ ‍ and $g (f (x))$ ‍ in terms of $x$ ‍.

f (g (x)) =

g (f (x)) =

Are functions $f$ ‍ and $g$ ‍ inverses?

[I need help!]

3) $f (x) = \frac{2}{3} x - 8$ ‍ and $h (x) = \frac{3}{2} (x + 8)$ ‍

Write simplified expressions for $f (h (x))$ ‍ and $h (f (x))$ ‍ in terms of $x$ ‍.

f (h (x)) =

h (f (x)) =

Are functions $f$ ‍ and $h$ ‍ inverses?

[I need help!]

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YanSu
Posted 8 years ago. Direct link to YanSu's post “ARGH! I typed in the exac...”
ARGH! I typed in the exact same answer as the one in the "I need help" paragraph but it won't register?
Button navigates to signup pageComment on YanSu's post “ARGH! I typed in the exac...”
(20 votes)
Answer
- brand3636
  Posted 8 years ago. Direct link to brand3636's post “It won't work because the...”
  It won't work because the fraction is not simplified in 30/4. You have to simplify it into 15/2 and then it will work
  Comment on brand3636's post “It won't work because the...”
  (38 votes)
alexander erlichherzog
Posted 6 years ago. Direct link to alexander erlichherzog's post “Is there ever a circumsta...”
Is there ever a circumstance where f(g(x)) = x but g(f(x)) isn't equal to x?
Button navigates to signup pageButton navigates to signup page
(20 votes)
Answer
- Ian Pulizzotto
  Posted 6 years ago. Direct link to Ian Pulizzotto's post “Interesting question! Th...”
  Interesting question!
  
  This could occur if f(x) fails to be 1-to-1 (and so fails to be invertible).
  For example, if f(x) = x^2 and g(x) = sqrt(x), then f(g(x)) = [sqrt(x)]^2 = x but g(f(x)) = sqrt(x^2) = |x|.
  
  Have a blessed, wonderful day!
  Comment on Ian Pulizzotto's post “Interesting question! Th...”
  (39 votes)
Abhishek Adari
Posted 5 months ago. Direct link to Abhishek Adari's post “Is the third question cor...”
Is the third question correct?
because im getting x-4 for h(f(x)) no matter how many times i try
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- sheldonlynn97
  Posted 5 months ago. Direct link to sheldonlynn97's post “h(f(x)) = (3/2)( f(x) ...”
  h(f(x)) = (3/2)( f(x) +8) [write function]
  h(f(x)) = (3/2)( ((2/3)x-8) +8) [put f(x) in]
  h(f(x)) = (3/2)( ((2/3)x) [-8, +8 cancels out]
  h(f(x)) = x [here (3/2)*(2/3)=1]
  
  f(h(x)) = (2/3)(h(x))-8 [write function]
  f(h(x)) = (2/3)((3/2)(x+8))-8 [put h(x) in]
  f(h(x)) = (2/3)((3/2)(x)(8))-8 [expand, noting that the 3/2 applies to the x and the +8]
  f(h(x)) = (2/3)(((3/2)x)+12)-8 [the (3/2)8 simplifies to 12 BUT is still enclosed with the 2/3 bracket]
  f(h(x) = x+8-8 [here (2/3)*(3/2) is 1, (2/3)12 is 8]
  f(h(x) = x [simplify]
  
  h(f(x)) = x
  f(h(x)) = x
  
  The two functions are invertible.
  The main point I think you have got caught on is the conversion of +8 to 12 due to the 3/2, but then exiting that out of all of the brackets so that it is then not affected by the 2/3, which would result in x+4, instead of x.
  
  Hope this helps :D
  Button navigates to signup page
  (12 votes)
Marc LaVincci
Posted 7 years ago. Direct link to Marc LaVincci's post “What happens if both simp...”
What happens if both simplify down to x+3 or any other values?
Are they still inverse of each other?
Button navigates to signup pageComment on Marc LaVincci's post “What happens if both simp...”
(6 votes)
Answer
kalyaa.pradeep
Posted 5 months ago. Direct link to kalyaa.pradeep's post “Would the functions still...”
Would the functions still be inverses if their result was equal but not x? For example f(g(x)) = x - 24 and g(f(x)) = x - 24. Would these functions still be inverses of each other?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Kim Seidel
  Posted 5 months ago. Direct link to Kim Seidel's post “No. If you use "x" as th...”
  No. If you use "x" as the input to either f(g(x)) or g(f(x)), you must get "x" as the final out. To be inverse functions, each must completely reverse the operations performed by the other function. In your case, your final result is different than "x", so some operations is not being reversed.
  Button navigates to signup page
  (7 votes)
Yuta Takeda
Posted 4 years ago. Direct link to Yuta Takeda's post “Can anyone SHOW me why f(...”
Can anyone SHOW me why f(g(x)) = g(f(x)) = x?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- amydylee
  Posted 6 months ago. Direct link to amydylee's post “This might be a bit late....”
  This might be a bit late...
  So, the reason why functions are inverses if f(g(x)) = x and g(f(x)) = x is because of the fundamental definition of inverse functions: that is, the input and output values are switched.
  So, let's take f(g(x)) = x. What is it saying? It is saying that if you INPUT the output of function g into function f, you get the input of function g as the OUTPUT. To make it more clear: x is the input of g, and g(x) is the output. However, inputting the output of g into f causes f to output x, which is the input of g.
  Now, for g(f(x)) = x, it is essentially the same thing. f(x) = output of f and x = input of f. Now, inputting f(x) - the output of f, into g gets you the output x - the input of f.
  See, it's just inverse functions in play.
  Hope this helped!
  Button navigates to signup page
  (4 votes)
ashley franklin
Posted 7 years ago. Direct link to ashley franklin's post “I NEED HELP! I'm working ...”
I NEED HELP! I'm working on a problem and have the g(f(x))=2(x+1/2)-1 as a problem I'm just trying to figure it out with the fraction. i stink writing stuff out so if you can understand please help me;(
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Meerah Tahir
  Posted 5 years ago. Direct link to Meerah Tahir's post “Your working is wrong bok...”
  Your working is wrong bokeum. It will be 2x+2/2 which wil become 2x/2 +2/2 to simplify it. It will become x+1-1 the (-1) from the original question. so it will equal x.
  Button navigates to signup page
  (2 votes)
Casanova Matthew
Posted 2 months ago. Direct link to Casanova Matthew's post “How are ever gonna use th...”
How are ever gonna use this in the real world?
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(2 votes)
Answer
- JAYSYN
  Posted 21 days ago. Direct link to JAYSYN's post “you might not but if you ...”
  you might not but if you go into a career that needs math then probably
  Button navigates to signup page
  (2 votes)
Lily Martin
Posted 7 years ago. Direct link to Lily Martin's post “in 2, how did they get fr...”
in 2, how did they get from =4(4/1 x−10)+10 to =x-40+10
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(2 votes)
Answer
- Edison Elder
  Posted 7 years ago. Direct link to Edison Elder's post “Multiply everything in br...”
  Multiply everything in brackets by 4. 4*x/4 = x 4*-10 = -40
  4(x/4 - 10) + 10 = x-40+10. You stated the problem in the wrong form.
  It should be 4(x/4 - 10) + 10 instead of 4(4/x - 10) +10
  Comment on Edison Elder's post “Multiply everything in br...”
  (2 votes)
Behrad Ansari
Posted 2 months ago. Direct link to Behrad Ansari's post “can we find f & g such th...”
can we find f & g such that f(g(x))=x but g(f(x) does not equal x?
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(2 votes)
Answer
- Kim Seidel
  Posted 2 months ago. Direct link to Kim Seidel's post “Both must = x to have an ...”
  Both must = x to have an inverse function. You may just have a math error on the one that doesn't = x. Check to see.
  Button navigates to signup page
  (1 vote)

Algebra (all content)

Course: Algebra (all content) > Unit 7

The inverse composition rule

Example 1: Functions f‍ and g‍ are inverses

Example 2: Functions f‍ and g‍ are not inverses

Check your understanding

1) f(x)=2x+7‍ and h(x)=x−72‍

2) f(x)=4x+10‍ and g(x)=14x−10‍

3) f(x)=23x−8‍ and h(x)=32(x+8)‍

Want to join the conversation?

Example 1: Functions $f$ ‍ and $g$ ‍ are inverses

Example 2: Functions $f$ ‍ and $g$ ‍ are not inverses

1) $f (x) = 2 x + 7$ ‍ and $h (x) = \frac{x - 7}{2}$ ‍

2) $f (x) = 4 x + 10$ ‍ and $g (x) = \frac{1}{4} x - 10$ ‍

3) $f (x) = \frac{2}{3} x - 8$ ‍ and $h (x) = \frac{3}{2} (x + 8)$ ‍