Undergrad Domain of the identity function after inverse composition

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Domain restriction for the composition of inverse functions operates similarly to other composite functions, requiring careful consideration of the domains involved. In the given example, the functions f(x) and g(x) are inverses, but the compositions f(g(x)) and g(f(x)) have specific domain restrictions: f(g(x)) is defined on (0,1] and g(f(x)) on [0,∞). It is essential to ensure that all steps in the composition are defined, as further restrictions may apply based on the codomains of the functions. Graphically, the compositions can be represented as line segments, illustrating the relationship between the functions and their inverses. Understanding these domain restrictions is crucial for accurately analyzing the behavior of composite functions.
Derrick Palmiter
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Hi, I'm struggling to understand something. Does domain restriction work the same way for composition of inverse functions as it does for other composite functions? I would assume it does, but the end result seems counter-intuitive. For example:

If I have the function f(x) = 1/(1+x), with the domain restriction x > 0 and
g(x) = (1-x)/x, with the domain restriction 0 < x < 1 ...

Then these functions are inverses. If I compose them to test this, then of course, I get the identity function in both cases...but do I need to specify a domain restriction on f(g(x)) and g(f(x))? I believe, that if I do so, then the domain of both composites should be 0 < x < 1. Firstly, am I correct, and secondly, what does this actually mean...in terms of the functions being inverses? If we were to graph the composites, would we just have a line segment? If the answer is yes, does this have any affect on the inverses operation upon each other?

Thanks for any help or insight anyone can give, or any other similar examples I could investigate.
 
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A function can only be applied where it is defined. So ##f \circ g## is only defined on ##(0,1]## and ##g \circ f## is only defined on ##[0,\infty)##. If - as in your example - all domains are subsets of one common space and ## f\, : \,A \rightarrow f(A)## and ##g\, : \,B \rightarrow g(B)## then ##f \circ g## is defined on ##B## and ##g \circ f## is defined on ##A##. However, there can be further restrictions to the codomains, as ##f## is restricted to ##g(B) \cap A## in the composition ##f \circ g## and ##g## is restricted to ##f(A) \cap B## in the composition ##g \circ f##. In your example we had ##f(A)=B## and ##g(B)=A##, so no further restrictions to the compositions.

The simple rule is, that all steps have to be defined.
 
Thank you very much. The precalculus book I'm using doesn't explain this question of the composition of domains very clearly. So, just to make sure I understand, the domain of f ° g in my example would be (0,1] and of g ° f would be [0, ∞). (I know that's how you prefaced your explanation, forgive me, I'm just a sucker for certainty.)

Is there any geometric/graphical application of this fact that I should be aware of that I may be missing? Or perhaps a geometric explanation of the same reality?
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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