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

## Homework Statement

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

## The Attempt at a Solution

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

## Homework Statement

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

## The Attempt at a Solution

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

Are you thinking of $$(f \circ g)^{-1} = g^{-1} \circ f^{-1}$$ ?

Then I think this is what you are looking for
http://planetmath.org/encyclopedia/InverseOfCompositionOfFunctions.html [Broken]

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HallsofIvy
Homework Helper

$fg^{-1}$ is only defined on the intersection of the domains of f and g.

$fg^{-1}$ 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

$$f \circ g^{-1}$$?

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

$$f \circ g^{-1}$$?

thanks for helping , i mean $$fg^{-1}$$ , 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 .

Mark44
Mentor

## Homework Statement

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

## 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 $\circ$ g-1, the composition of f and g-1
• (f $\circ$ g)-1, the inverse of the composition of f and g

• fg-1, the product of f and g-1
• (fg)-1, the inverse of the product of f and g
• f $\circ$ g-1, the composition of f and g-1
• (f $\circ$ g)-1, the inverse of the composition of f and g

sorry, i meant number 3 .

Mark44
Mentor

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.

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 .

Mark44
Mentor

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?

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 $$y\in R$$

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

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

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

Am i correct ?

Mark44
Mentor

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}?

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) ?

Mark44
Mentor

Yes, good!