Inverting Functions Homework: Find f^{-1}(S) for Sets A, B and S

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Homework Statement



for sets A, B and S with S being a subset of B, and a function f: A --> B we define:

f[tex]^{-1}[/tex](S) = { a [tex]\in A[/tex] : f(a) [tex]\in S[/tex] }

find f[tex]^{-1}[/tex](S) for:

1) f(x) = x - floor(x) where S = { y: 0 < y < 1 }

2) f(x) = x[tex]^{3}[/tex]-7x+16 where S = { y: 10 <= y <= 22 }

3) f(t) = (cos(t),sin(t)) where S = { (x,y): x < 0, y > 0 }

My attempted work:

I can't find any questions similar to this in my notebook and textbook! I don't even know how to begin. I know that (x - floor(x)) is the same as (x % 1). Am i supposed to find the inverse of the function such that it satisfies the boundaries of S?
 
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Yes, that is exactly what the problem says. If y= x- floor(x) the x= what in terms of y?
If [itex]y= x^3- 7x+ 16[/itex], then x= what in terms of y?

But your last problem is written incorrectly. f(t)= cos(t)sin(t) is a numerical function. Did you mean f(t)= (cos(t), sin(t))?
 
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Ah sorry, yes i meant to have a comma there.

Also, when S is a set, f[tex]^{-1}[/tex](S) typically means "the set of things that f sends to S", a meaning of f[tex]^{-1}[/tex] distinct from "the inverse function of f". So i think i need to find the set that satisfies the constraints of S, right?

For question 1, the numbers that will satisfy are all nonzero numbers which are not whole numbers.

Someone told me that i am supposed to write my answer like: f[tex]^{-1}[/tex](S) = { stuff in thing | stuff not in thing and stuff > 0 } although i find that somewhat vague. I assume "thing" means S.

So my attempt at an answer is f[tex]^{-1}[/tex](S) = { Q [tex]\cap[/tex] R\Q | Z } although i doubt that's right. :confused:

My attempt at question 2 isn't getting far either. x[tex]^{3}[/tex]-7x = y-16

I can't isolate x though. :(
 
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rooski said:
Ah sorry, yes i meant to have a comma there.

Also, when S is a set, f[tex]^{-1}[/tex](S) typically means "the set of things that f sends to S", a meaning of f[tex]^{-1}[/tex] distinct from "the inverse function of f". So i think i need to find the set that satisfies the constraints of S, right?
Absolutely! For example, the function [itex]f(x)= x^2[/itex] is not invertible as a function on the real numbers but you can still ask about [itex]f^{-1}(-1, 4)[/itex], say. That would be the set of all real numbers, x, such that [itex]x^2[/itex] is in (-1, 4). That is just (-2, 2).

The very first time I got up before a class in graduate school to do a proof, it was precisely a problem dealing with "[itex]f^{-1}(A)[/itex]" where A was a set- I stupidly did the whole thing assuming f was invertible!

For question 1, the numbers that will satisfy are all nonzero numbers which are not whole numbers.
Yes, that is correct.

Someone told me that i am supposed to write my answer like: f[tex]^{-1}[/tex](S) = { stuff in thing | stuff not in thing and stuff > 0 } although i find that somewhat vague. I assume "thing" means S.
I have no idea what that means. Was it your teacher who told you that? If it was, ask your teacher for clarification. If not, ignore it.

[/quote]So my attempt at an answer is f[tex]^{-1}[/tex](S) = { Q [tex]\cap[/tex] R\Q | Z } although i doubt that's right. :confused:[/quote]
Q? Rational numbers? Why mention rational numbers at all? [itex]Q\cap R[/itex] is just R.

My attempt at question 2 isn't getting far either. x[tex]^{3}[/tex]-7x = y-16

I can't isolate x though. :(
I suggest graphing the function. Also look at [math]y'= 3x^2- 7[/math] to determine where the max and min values of y are.
 
I suggest graphing the function. Also look at [math]y'= 3x^2- 7[/math] to determine where the max and min values of y are.

So i get x = sqrt(7/3) and x = -1 * sqrt(7/3)

Not sure how that will help me though. Plugging that into the original equation just gives me an obscure result.
 
So using the boundaries 10 and 22 i noticed that the x intercepts are -3, 3, -2, 2, -1 and 1.

So i can say f-1(S) = { y E R : ( -3 <= y <= -2 V -1<= y <= 1 V 2 <= y <= 3 ) } right?