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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] ?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) ?
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[itex]fg^{-1}[/itex] is only defined on the intersection of the domains of f and g.
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 .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]?
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) ?
sorry, i meant number 3 .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
When who is one-to-one? See if you can restate your question in a more precise way.thereddevils said:i understand that g^(-1) only happens when its one-one
ok i will look for an example .When who is one-to-one? See if you can restate your question in a more precise way.
the range of f is y such that [tex]y\in R [/tex]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}?
domain of f is (1/e , infinity) ?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}?