Domain and Range of an Inverse FunctionDate: 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 Check out our web site! http://mathforum.org/dr.math/ |
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