PurpleNight
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I know how to find f(x), but I don't know what f(1/x) means. If f(x)=(3x+2)/(3-2x) how do I find f(1/x)
Function f(1/x)
- Thread starter PurpleNight
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I know how to find f(x), but I don't know what f(1/x) means. If f(x)=(3x+2)/(3-2x) how do I find f(1/x)
Harry_the_cat
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Find f(x) first. Then find f(1/x) in the same way you would find f(a).
HERE IS A USEFUL THEOREM
(that I am not going to bother to prove because it takes too much time).
[MATH]g(x) \text { is the inverse of continuous function } f(x) \iff f(x) \text { is the inverse of continuous function } g(x).[/MATH]
That means that if you are given the definition of the inverse of f(x), you can determine the definition of f(x) by finding the inverse of the inverse.
So I would first solve for y in
[MATH]x = \dfrac{3y - 2}{2y + 3}[/MATH] to get y = f(x).
(that I am not going to bother to prove because it takes too much time).
[MATH]g(x) \text { is the inverse of continuous function } f(x) \iff f(x) \text { is the inverse of continuous function } g(x).[/MATH]
That means that if you are given the definition of the inverse of f(x), you can determine the definition of f(x) by finding the inverse of the inverse.
So I would first solve for y in
[MATH]x = \dfrac{3y - 2}{2y + 3}[/MATH] to get y = f(x).
Last edited:
Cubist
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I almost made a post like the above.
But then I spotted that OP has already calculated the "inverse of the inverse" without explicitly saying so...
The text in green made me think that you knew how to do it, but that you hadn't done it yet. And initially I skimmed over the later equation in the text because I assumed it was the same as the image. And obviously finding f(x) is a pre-requisite. Anyway, I think that your f(x) is correct. Please post back with your attempt at f(1/x) and hopefully someone will check it.
It is obvious that I was confused. Sorry.
We start with [MATH]g(x) = \dfrac{3x - 2}{2x + 3} \text { and } g(x) = f^{-1}(x).[/MATH]
You deduced that [MATH]f(x) = \dfrac{3x + 2}{3 - 2x}.[/MATH]
If you are correct that g(x) is the inverse of f(x), then g(f(x)) = x by definition.
Let's check.
[MATH]g(f(x)) = \dfrac{3 * \dfrac{3x + 2}{3 - 2x} - 2}{2 * \dfrac{3x + 2}{3 - 2x} + 3} = \dfrac{\dfrac{9x + 6 - 6 + 4x}{3 - 2x}}{\dfrac{6x +4 +9 - 6x}{3 - 2x}} = \implies[/MATH]
[MATH]g(f(x)) = \dfrac{13x}{3 - 2x} * \dfrac{3 - 2x}{13} = \dfrac{13x}{13} = x.[/MATH]
So you are correct about the definition of f(x).
Where are you having trouble in computing f(1/x)?
We start with [MATH]g(x) = \dfrac{3x - 2}{2x + 3} \text { and } g(x) = f^{-1}(x).[/MATH]
You deduced that [MATH]f(x) = \dfrac{3x + 2}{3 - 2x}.[/MATH]
If you are correct that g(x) is the inverse of f(x), then g(f(x)) = x by definition.
Let's check.
[MATH]g(f(x)) = \dfrac{3 * \dfrac{3x + 2}{3 - 2x} - 2}{2 * \dfrac{3x + 2}{3 - 2x} + 3} = \dfrac{\dfrac{9x + 6 - 6 + 4x}{3 - 2x}}{\dfrac{6x +4 +9 - 6x}{3 - 2x}} = \implies[/MATH]
[MATH]g(f(x)) = \dfrac{13x}{3 - 2x} * \dfrac{3 - 2x}{13} = \dfrac{13x}{13} = x.[/MATH]
So you are correct about the definition of f(x).
Where are you having trouble in computing f(1/x)?
As the feline said back in post 2. It depresses me to be out-thought by a cat.
Hi Jeff, I want to know more about the theorem, what is the name?
I am not sure the theorem has a generally agreed upon name. It takes a number of steps. I shall give an outine of the steps.
First, do you know that we can define an inverse of f(x) only on an open interval where f(x) is strictly increasing or strictly decreasing? Only in such an interval is the inverse well defined, but in every such interval, the inverse is well defined. Do you see why? This is why the inverse trig functions are defined over restricted domains although the sine and cosine functions are defined for all real x.
Let’s consider the strictly increasing case.
Second, given that f(x) is strictly increasing over (a, b) and that g(x) is the inverse of f(x) over (a, b),
then g(x) is strictly increasing and so has an inverse over (f(a), f(b)). Call that inverse h(x).
Third, consider ARBITRARY c such that a < c < b and f(c) = d.
But g(x) is the inverse of f(x)
Thus, by definition, g(f(c)) = c.
So g(d) = c.
But h(x) is the inverse of g(x).
Thus, by definition, h(g(d)) = d.
So h(c) = d.
Therefore h(c) = f(c).
But c was an arbitrary number between a and b.
Therefore h(x) = f(x) for all numbers between a and b.
In conclusion if g(x) is the inverse of f(x), then f(x) is the inverse of g(x).
First, do you know that we can define an inverse of f(x) only on an open interval where f(x) is strictly increasing or strictly decreasing? Only in such an interval is the inverse well defined, but in every such interval, the inverse is well defined. Do you see why? This is why the inverse trig functions are defined over restricted domains although the sine and cosine functions are defined for all real x.
Let’s consider the strictly increasing case.
Second, given that f(x) is strictly increasing over (a, b) and that g(x) is the inverse of f(x) over (a, b),
then g(x) is strictly increasing and so has an inverse over (f(a), f(b)). Call that inverse h(x).
Third, consider ARBITRARY c such that a < c < b and f(c) = d.
But g(x) is the inverse of f(x)
Thus, by definition, g(f(c)) = c.
So g(d) = c.
But h(x) is the inverse of g(x).
Thus, by definition, h(g(d)) = d.
So h(c) = d.
Therefore h(c) = f(c).
But c was an arbitrary number between a and b.
Therefore h(x) = f(x) for all numbers between a and b.
In conclusion if g(x) is the inverse of f(x), then f(x) is the inverse of g(x).
@Jomo I am not arguing. I am curious. I restricted my more formal proof to intervals that were monotonically increasing or decreasing. So what I view as a theorem applicable to certain domains, you view as a definition. In other words, your definition denies that it is proper to say that the square root function has an inverse at all.
I suspect that we are both missing something.
I suspect that we are both missing something.
Just a note:
A function [MATH]f = \{ (x,f(x)) \}[/MATH] always has an inverse (as a relation).
[MATH](g = ) \,f^{-1} := \{ (f(x),x) \}[/MATH], whether or not this is a function itself
and therefore
[MATH]g^{-1}=(f^{-1})^{-1} = \{ (x,f(x)) \} = f[/MATH]
So for any function f, the inverse of the inverse is the original function.
Consider the function [MATH] f= \{ (1,4), (2,8), (3,-1)\}[/MATH][MATH]g=f^{-1}= \{(4,1), (8,2), (-1,3)\}[/MATH]In fact the inverse of f is itself a function.
Neither f nor g are defined on an interval (and trivially therefore not an open one).
Neither f nor g are strictly increasing or strictly decreasing.
A function [MATH]f = \{ (x,f(x)) \}[/MATH] always has an inverse (as a relation).
[MATH](g = ) \,f^{-1} := \{ (f(x),x) \}[/MATH], whether or not this is a function itself
and therefore
[MATH]g^{-1}=(f^{-1})^{-1} = \{ (x,f(x)) \} = f[/MATH]
So for any function f, the inverse of the inverse is the original function.
I wonder is this what you meant to say?
Consider the function [MATH] f= \{ (1,4), (2,8), (3,-1)\}[/MATH][MATH]g=f^{-1}= \{(4,1), (8,2), (-1,3)\}[/MATH]In fact the inverse of f is itself a function.
Neither f nor g are defined on an interval (and trivially therefore not an open one).
Neither f nor g are strictly increasing or strictly decreasing.
Cubist
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I like the elegance of this approach, but isn't there something to add about non-bijective functions?
If [MATH] f= \{ (1,0), (2,0), (3,0)\}[/MATH][MATH]g=f^{-1}= \{ (0, \{1,2,3\}) \} [/MATH][MATH]g^{-1}=(f^{-1})^{-1}= \{ (\{1,2,3\}, 0) \} \ne f[/MATH]?
I guess that you're going to say that g is actually...
[MATH]g=f^{-1}= \{ (0,1), (0,2), (0,3) \} [/MATH]but I can't get my head around what this means
@lex
Thank you for the attempt to clarify my thinking and terminology.
When I started out, I was talking about continuous functions and their inverse functions (see post 4).
But I think I get your point. A function (whether continuous or not), always links an element in one set to an element in another set. So, we can always find an inverse for for any element in the set of f(x). But that inverse relationship may not be a function. And that motivates jomo’s definition
[MATH]g(x) \text { is an inverse function of } f(x) \iff g(f(x)) = x \text { and } f(g(x)) = x.[/MATH]
So what I called a theorem in post 4 is a definition of the term “inverse function,” and I should have said ”inverse function” rather than “inverse.”
Am I finally grasping what you and jomo are trying to say to me?
So what I was really proving, which I do think is correct, is this: if f(x) is a continuous function on the interval (a, b) and is strictly increasing (or strictly decreasing) on that interval, then f(x) has an inverse function defined on the interval (f(a), f(b)) (or (f(b), f(a))). Moreover, if f(x) is continuous on an interval and has an inverse function, f(x) is either strictly increasing or strictly decreasing on that interval.
And does that obviate the idea that the square root of non-negative x has no inverse function?
Thank you for the attempt to clarify my thinking and terminology.
When I started out, I was talking about continuous functions and their inverse functions (see post 4).
But I think I get your point. A function (whether continuous or not), always links an element in one set to an element in another set. So, we can always find an inverse for for any element in the set of f(x). But that inverse relationship may not be a function. And that motivates jomo’s definition
[MATH]g(x) \text { is an inverse function of } f(x) \iff g(f(x)) = x \text { and } f(g(x)) = x.[/MATH]
So what I called a theorem in post 4 is a definition of the term “inverse function,” and I should have said ”inverse function” rather than “inverse.”
Am I finally grasping what you and jomo are trying to say to me?
So what I was really proving, which I do think is correct, is this: if f(x) is a continuous function on the interval (a, b) and is strictly increasing (or strictly decreasing) on that interval, then f(x) has an inverse function defined on the interval (f(a), f(b)) (or (f(b), f(a))). Moreover, if f(x) is continuous on an interval and has an inverse function, f(x) is either strictly increasing or strictly decreasing on that interval.
And does that obviate the idea that the square root of non-negative x has no inverse function?
@Cubist
You guessed correctly.
Your second line is not correct, as the definition of the inverse is just to swap the two values in each ordered pair (which you have not done). You can of course define a function g mapping 0 to a (single) set, containing 3 numbers, but that is not a map from [MATH]\mathbb{R} \text{ to }\mathbb{R}[/MATH] and is definitely not the inverse of f.
I am guessing that having several 0's in the x position is what is unnerving you.
You should think of f as a binary relation on [MATH]\mathbb{R}[/MATH], i.e. simply a subset of [MATH]\mathbb{R} \times \mathbb{R}[/MATH].
(This relation may also happen to be a function from [MATH]\mathbb{R} \text{ to } \mathbb{R}[/MATH]).
Now think of your unnerving inverse: [MATH]f^{-1}= \{ (0,1), (0,2), (0,3) \}[/MATH] simply as the inverse relation of the relation f (i.e. just a subset of [MATH]\mathbb{R} \times \mathbb{R}[/MATH] and your difficulty should disappear. It is only if we would like to call this subset a function, that we would require it to have only one ordered pair beginning with 0.
In short, I suppose, think of these as relations, i.e. subsets, rather than as functions.
A function [MATH]f[/MATH] from [MATH]\mathbb{R} \text{ to }\mathbb{R}[/MATH] is simply a relation on [MATH]\mathbb{R}[/MATH] (the function [MATH]f[/MATH] is a subset of [MATH]\mathbb{R} \times \mathbb{R}[/MATH]), with one extra property, that [MATH](x,y)\in f \text{ and } (x,y') \in f[/MATH], implies [MATH]y=y'[/MATH].
I like the elegance of this approach, but isn't there something to add about non-bijective functions?
If [MATH] f= \{ (1,0), (2,0), (3,0)\}[/MATH][MATH]g=f^{-1}= \{ (0, \{1,2,3\}) \} [/MATH][MATH]g^{-1}=(f^{-1})^{-1}= \{ (\{1,2,3\}, 0) \} \ne f[/MATH]?
I guess that you're going to say that g is actually...
[MATH]g=f^{-1}= \{ (0,1), (0,2), (0,3) \} [/MATH]but I can't get my head around what this means
You guessed correctly.
Your second line is not correct, as the definition of the inverse is just to swap the two values in each ordered pair (which you have not done). You can of course define a function g mapping 0 to a (single) set, containing 3 numbers, but that is not a map from [MATH]\mathbb{R} \text{ to }\mathbb{R}[/MATH] and is definitely not the inverse of f.
I am guessing that having several 0's in the x position is what is unnerving you.
You should think of f as a binary relation on [MATH]\mathbb{R}[/MATH], i.e. simply a subset of [MATH]\mathbb{R} \times \mathbb{R}[/MATH].
(This relation may also happen to be a function from [MATH]\mathbb{R} \text{ to } \mathbb{R}[/MATH]).
Now think of your unnerving inverse: [MATH]f^{-1}= \{ (0,1), (0,2), (0,3) \}[/MATH] simply as the inverse relation of the relation f (i.e. just a subset of [MATH]\mathbb{R} \times \mathbb{R}[/MATH] and your difficulty should disappear. It is only if we would like to call this subset a function, that we would require it to have only one ordered pair beginning with 0.
In short, I suppose, think of these as relations, i.e. subsets, rather than as functions.
A function [MATH]f[/MATH] from [MATH]\mathbb{R} \text{ to }\mathbb{R}[/MATH] is simply a relation on [MATH]\mathbb{R}[/MATH] (the function [MATH]f[/MATH] is a subset of [MATH]\mathbb{R} \times \mathbb{R}[/MATH]), with one extra property, that [MATH](x,y)\in f \text{ and } (x,y') \in f[/MATH], implies [MATH]y=y'[/MATH].
@JeffM I should have added that often people use the phrase 'the inverse of a function' to mean 'the inverse function', which I'm sure is perfectly acceptable if the context allows.
However e.g. in allowing my trivial proof that for a function [MATH]f, \,(f^{-1})^{-1} = f[/MATH] it is very convenient to use 'inverse' to mean inverse as a relation, since we do not care and it doesn't matter whether the inverse of [MATH]f[/MATH] is itself a function.
However e.g. in allowing my trivial proof that for a function [MATH]f, \,(f^{-1})^{-1} = f[/MATH] it is very convenient to use 'inverse' to mean inverse as a relation, since we do not care and it doesn't matter whether the inverse of [MATH]f[/MATH] is itself a function.