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Domain and Range of an Inverse Function
Date: 03/18/98 at 19:56:44
From: K.T Min
Subject: Inverse function
Hi. I have a question about inverse function.
1) Explain why Df = Rf-1 and Rf = Df-1
You can use concrete examples to illustrate your concepts.
D = R and R = D
f f-1 f f-1
the question is not d*f. It is D and D is a domain and R is a range.
f
Thank you.
Date: 03/20/98 at 07:50:56
From: Doctor Jerry
Subject: Re: Inverse function
Hi K. T. Min,
You are considering a function f, with domain D. I'll assume that you
are talking about real-valued functions of a real-variable. Let Re
denote the set of real numbers. From what you have said, the set R is
the set:
f(D) = {y in Re : there is an x in D for which y = f(x)}
I think that I should mention that when one discusses a function f
with domain D and range R, the set R need not be equal to f(D); it
must, however, contain f(D).
Also, you are assuming that f is one-to-one, so that f^{-1} exists.
You are asking why the domain of f^{-1} is R and the set:
{x in Re : there is a y in R for which x = f^{-1}(y)} is D.
We think of a one-to-one function as a rule for associating with each
point x of D an element of R. By our assumptions, different elements
of D go to different elements of R and every element of R is the image
of an element of D. Think about it as typing strings from the elements
of D to the elements of R. No two strings from D are tied to the same
object in R; also, there is a string attached to every element in R.
If we wish, we can regard R as the starting place of the strings,
instead of D. The rule for typing strings in this case is f^{-1}.
So, the domain of f is the range of f^{-1}, and the range of f is the
domain of f^{-1}.
A good example is y = ln(x). This function is defined for all x>0.
Its range is the set of all real numbers. The inverse of ln is the
exponential function exp. The domain of exp is the set of all real
numbers (which is the range of ln); the range of exp is the set of
positive numbers, which is the domain of ln.
-Doctor Jerry, The Math Forum
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