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.

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

if i take 2 functions f(x) and g(x) and composite of only one of them results in x , can they be said as inverse functions ?

For 𝑓 and 𝑔 to be inverse functions, the domain of either function must be equal to the range of the other function. If 𝑓(𝑎) = 𝑏, but 𝑔(𝑏) ≠ 𝑎, then 𝑓 maps 𝑎 to 𝑏, but 𝑔 does not map 𝑏 to 𝑎. This means that the range of 𝑔 is not equal to the domain of 𝑓, and thereby 𝑓 and 𝑔 are not inverse functions.

What if one of the functions is f(x) = x^2 and the other is g(x) = x^(1/2)? Square and square root seem to be inverse functions, but this isn't the case if x is negative. In that case, should I use specific values?

Yes. Your functions are only inverses if you take [0, ∞) as their domains (or appropriate subsets). x² is not invertible if its domain is all of ℝ, since most outputs are hit by two inputs.

Main content

Get ready for AP® Calculus

Course: Get ready for AP® Calculus > Unit 3

Lesson 7: Verifying inverse functions by composition

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 to b, then the inverse must take b to a.

Let's take functions f and g for example: f, left parenthesis, x, right parenthesis, equals, start fraction, x, plus, 1, divided by, 3, end fraction and g, left parenthesis, x, right parenthesis, equals, 3, x, minus, 1.

Notice how f, left parenthesis, 5, right parenthesis, equals, 2 and g, left parenthesis, 2, right parenthesis, equals, 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, left parenthesis, f, left parenthesis, 5, right parenthesis, right parenthesis, equals, 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, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, equals, x for all x in the domain of g
g, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis, equals, 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, left parenthesis, x, right parenthesis, equals, start fraction, x, plus, 1, divided by, 3, end fraction and g, left parenthesis, x, right parenthesis, equals, 3, x, minus, 1.

Let's find f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis and g, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis.

f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis	g, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis
$\begin{aligned} f(\greenD{g(x)})&=\dfrac{\greenD{g(x)}+1}{3}\\\\&=\dfrac{\greenD{3x-1}+1}{3}\\\\&=\dfrac{3x}{3}\\\\&=x\\\end{aligned}$	$\qquad\qquad \begin{aligned}g(\purpleC{f(x)})&=3\left(\purpleC{f(x)}\right)-1\\\\&=3\left(\purpleC{\dfrac{x+1}{3}}\right)-1\\\\&=x+1-1\\\\&=x\\\end{aligned}$

So we see that functions f and g are inverses because f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, equals, x and g, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis, equals, x.

Example 2: Functions f and g are not inverses

If f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis or g, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis is not equal to x, then f and g cannot be inverses.

Let's try this for f, left parenthesis, x, right parenthesis, equals, 5, x, minus, 7 and g, left parenthesis, x, right parenthesis, equals, start fraction, x, divided by, 5, end fraction, plus, 7.

f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis	g, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis
$\begin{aligned} f(\greenD{g(x)})&=5(\greenD{g(x)})-7\\\\&=5\left(\greenD{\dfrac{x}{5}+7}\right)-7\\\\&=x+35-7\\\\&=x+28\end{aligned}\qquad$	$\qquad\begin{aligned} g(\purpleC{f(x)})&=\dfrac{\purpleC{f(x)}}{5}+7\\\\&=\dfrac{\purpleC{5x-7}}{5}+7\\\\&=x-\dfrac75+7\\\\&=x+\dfrac{28}{5}\\\end{aligned}$

So functions f and g are not inverses because f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, does not equal, x and g, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis, does not equal, x.

(Note here, that we could have concluded that f and g were not inverses after showing that f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, equals, x, plus, 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, left parenthesis, x, right parenthesis, equals, 2, x, plus, 7 and h, left parenthesis, x, right parenthesis, equals, start fraction, x, minus, 7, divided by, 2, end fraction

Write simplified expressions for f, left parenthesis, h, left parenthesis, x, right parenthesis, right parenthesis and h, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis in terms of x.

f, left parenthesis, h, left parenthesis, x, right parenthesis, right parenthesis, equals

h, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis, equals

Are functions f and h inverses?

[I need help!]

2) f, left parenthesis, x, right parenthesis, equals, 4, x, plus, 10 and g, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, 4, end fraction, x, minus, 10

Write simplified expressions for f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis and g, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis in terms of x.

f, left parenthesis, g, left parenthesis, x, right parenthesis, right parenthesis, equals

g, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis, equals

Are functions f and g inverses?

[I need help!]

3) f, left parenthesis, x, right parenthesis, equals, start fraction, 2, divided by, 3, end fraction, x, minus, 8 and h, left parenthesis, x, right parenthesis, equals, start fraction, 3, divided by, 2, end fraction, left parenthesis, x, plus, 8, right parenthesis

f, left parenthesis, h, left parenthesis, x, right parenthesis, right parenthesis, equals

h, left parenthesis, f, left parenthesis, x, right parenthesis, right parenthesis, equals

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 7 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...”
  (35 votes)
alexander erlichherzog
Posted 5 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
(19 votes)
Answer
- Ian Pulizzotto
  Posted 5 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...”
  (37 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 3 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?
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(3 votes)
Answer
- Kim Seidel
  Posted 3 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)
Abhishek Adari
Posted 3 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 3 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
  (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?
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(3 votes)
Answer
- amydylee
  Posted 4 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
  (3 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)
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)
J R Adhwaith
Posted 3 years ago. Direct link to J R Adhwaith's post “if i take 2 functions f(x...”
if i take 2 functions f(x) and g(x) and composite of only one of them results in x , can they be said as inverse functions ?
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(1 vote)
Answer
- Jerry Nilsson
  Posted 3 years ago. Direct link to Jerry Nilsson's post “For 𝑓 and 𝑔 to be inver...”
  For 𝑓 and 𝑔 to be inverse functions, the domain of either function must be equal to the range of the other function.
  
  If 𝑓(𝑎) = 𝑏, but 𝑔(𝑏) ≠ 𝑎,
  then 𝑓 maps 𝑎 to 𝑏, but 𝑔 does not map 𝑏 to 𝑎.
  This means that the range of 𝑔 is not equal to the domain of 𝑓,
  and thereby 𝑓 and 𝑔 are not inverse functions.
  Button navigates to signup page
  (3 votes)
proudsea1503
Posted 10 days ago. Direct link to proudsea1503's post “What if one of the functi...”
What if one of the functions is f(x) = x^2 and the other is g(x) = x^(1/2)? Square and square root seem to be inverse functions, but this isn't the case if x is negative. In that case, should I use specific values?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- kubleeka
  Posted 8 days ago. Direct link to kubleeka's post “Yes. Your functions are o...”
  Yes. Your functions are only inverses if you take [0, ∞) as their domains (or appropriate subsets). x² is not invertible if its domain is all of ℝ, since most outputs are hit by two inputs.
  Button navigates to signup page
  (3 votes)

Get ready for AP® Calculus

Course: Get ready for AP® Calculus > Unit 3

The inverse composition rule

Example 1: Functions ffff and gggg are inverses

Example 2: Functions ffff and gggg are not inverses

Check your understanding

1) f(x)=2x+7f(x)=2x+7f(x)=2x+7f, left parenthesis, x, right parenthesis, equals, 2, x, plus, 7 and h(x)=x−72h(x)=\dfrac{x-7}{2}h(x)=2x−7​h, left parenthesis, x, right parenthesis, equals, start fraction, x, minus, 7, divided by, 2, end fraction

2) f(x)=4x+10f(x)=4x+10f(x)=4x+10f, left parenthesis, x, right parenthesis, equals, 4, x, plus, 10 and g(x)=14x−10g(x)=\dfrac{1}{4}x-10g(x)=41​x−10g, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, 4, end fraction, x, minus, 10

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Example 1: Functions f and g are inverses

Example 2: Functions f and g are not inverses

1) f, left parenthesis, x, right parenthesis, equals, 2, x, plus, 7 and h, left parenthesis, x, right parenthesis, equals, start fraction, x, minus, 7, divided by, 2, end fraction

2) f, left parenthesis, x, right parenthesis, equals, 4, x, plus, 10 and g, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, 4, end fraction, x, minus, 10