- #1

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## 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) ?

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- Thread starter thereddevils
- Start date

- #1

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

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

- #2

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## 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]

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- #3

HallsofIvy

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

- #4

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[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

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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

Mark44

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You aren't being clear in what you're asking. Which of the following are you asking about?## 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) ?

- 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

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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

Mark44

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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

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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

Mark44

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

For g

- #11

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^{-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

Mark44

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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

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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

Mark44

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Yes, good!

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