• Support PF! Buy your school textbooks, materials and every day products Here!

What is the condition for fucntion fg^(-1) to exist ?

  • #1
438
0

Homework Statement



what is the condition for fucntion fg^(-1) to exist ?

Homework Equations





The Attempt at a Solution



i understand that g^(-1) only happens when its one-one , how about fg^(-1) ?
 

Answers and Replies

  • #2
317
0


Homework Statement



what is the condition for fucntion fg^(-1) to exist ?

Homework Equations





The Attempt at a Solution



i understand that g^(-1) only happens when its one-one , how about fg^(-1) ?
Are you thinking of [tex](f \circ g)^{-1} = g^{-1} \circ f^{-1}[/tex] ?

Then I think this is what you are looking for
http://planetmath.org/encyclopedia/InverseOfCompositionOfFunctions.html [Broken]
 
Last edited by a moderator:
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,794
925


[itex]fg^{-1}[/itex] is only defined on the intersection of the domains of f and g.
 
  • #4
317
0


[itex]fg^{-1}[/itex] is only defined on the intersection of the domains of f and g.
We know that HallsoftIvy, but you think since this is a pre-Calculus question that the original poster is simply looking for and try to understand the definition of

[tex]f \circ g^{-1}[/tex]?
 
  • #5
438
0


We know that HallsoftIvy, but you think since this is a pre-Calculus question that the original poster is simply looking for and try to understand the definition of

[tex]f \circ g^{-1}[/tex]?
thanks for helping , i mean [tex]fg^{-1}[/tex] , because i had this previous example , where i am supposed to prove that function gf exist , so the range of f is a subset or equal the domain of g . I wonder if i could applying the same reasoning for this function .
 
  • #6
33,173
4,858


Homework Statement



what is the condition for fucntion fg^(-1) to exist ?

Homework Equations





The Attempt at a Solution



i understand that g^(-1) only happens when its one-one , how about fg^(-1) ?
You aren't being clear in what you're asking. Which of the following are you asking about?
  • fg-1, the product of f and g-1
  • (fg)-1, the inverse of the product of f and g
  • f [itex]\circ[/itex] g-1, the composition of f and g-1
  • (f [itex]\circ[/itex] g)-1, the inverse of the composition of f and g
 
  • #7
438
0


You aren't being clear in what you're asking. Which of the following are you asking about?
  • fg-1, the product of f and g-1
  • (fg)-1, the inverse of the product of f and g
  • f [itex]\circ[/itex] g-1, the composition of f and g-1
  • (f [itex]\circ[/itex] g)-1, the inverse of the composition of f and g
sorry, i meant number 3 .
 
  • #8
33,173
4,858


thereddevils said:
i understand that g^(-1) only happens when its one-one
When who is one-to-one? See if you can restate your question in a more precise way.
 
  • #9
438
0


When who is one-to-one? See if you can restate your question in a more precise way.
ok i will look for an example .

The function f and g are defined by f(x)= ln x , x are all positive real numbers and g(x)=x^2-1 , x is all positive real numbers . Determine whether the composite function g^(-1) o f exists . Find the restricted domain of f such that the function exists .
 
  • #10
33,173
4,858


For this example, the restricted domain for g makes it a one-to-one function, so the inverse, g-1, exists.

For g-1 o f, you need to look at the domain of f (x such that x > 0), and the range of f. The outputs from f are going to be the inputs to g-1. Are there any numbers in the range of the ln function that are not allowed in the domain for g-1?
 
  • #11
438
0


For this example, the restricted domain for g makes it a one-to-one function, so the inverse, g-1, exists.

For g-1 o f, you need to look at the domain of f (x such that x > 0), and the range of f. The outputs from f are going to be the inputs to g-1. Are there any numbers in the range of the ln function that are not allowed in the domain for g-1?
the range of f is y such that [tex]y\in R [/tex]

the domain of g^(-1) is the range of g which is (-1 , infinity)

since the [tex]R_f[/tex] is not equal or a subset of [tex]D_{g^{-1}}[/tex] , [tex]g^{-1} \circ f[/tex] doesn't exist .

for the function g^(-1) o f to exist , range of f has to be (-1 , infinity)

Am i correct ?
 
  • #12
33,173
4,858


Yes. Now, can you figure out how to restrict the domain of f so that the range of f is {y | y > -1}?

IOW, what restrictions can you place on the domain of ln(x) so that the range is {y | y > -1}?
 
  • #13
438
0


Yes. Now, can you figure out how to restrict the domain of f so that the range of f is {y | y > -1}?

IOW, what restrictions can you place on the domain of ln(x) so that the range is {y | y > -1}?
domain of f is (1/e , infinity) ?
 
  • #14
33,173
4,858


Yes, good!
 

Related Threads for: What is the condition for fucntion fg^(-1) to exist ?

Replies
3
Views
511
Replies
34
Views
1K
Replies
2
Views
12K
  • Last Post
Replies
4
Views
1K
Replies
2
Views
1K
Replies
10
Views
2K
Replies
7
Views
3K
Replies
2
Views
6K
Top