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All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Advanced Calc homework - Inverse functions

I'm having trouble understanding inverse functions in Advanced Calculus. Specifically, if a function is not bijective (in particular if it is not injective), is the inverse undefined or is it defined, but as a relation but not a function? Something else?

Here is one example from several similar problems on my homework:

Prove or provide a counterexample:

Let f:A ⇾ B. If C ⊂ A, then C ⊂ f ⁻¹( f( C)).

The professor stated that we cannot assume f is bijective.

The book (and my notes) are both unclear regarding the effect on an inverse when the function is not bijective.

Thanks.

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You can shear a sheep many times but skin him only once.

Lazernugget

Member

Registered: 2011-01-22

Posts: 47

Re: Advanced Calc homework - Inverse functions

Sorry, I wish I could help, but my calculus knowledge is limited to F(x) and factoring curved destination point equations... Sorry! I'm only 10 anyway... XD

-Lazernugget

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MATH......that is all.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Advanced Calc homework - Inverse functions

hi All_Is_Number

This link may help you here.

http://en.wikipedia.org/wiki/Injective

If the function is not injective then values in the range come from more than one value in the domain.  See the non-injection diagram on this page.

So a counter example would be

with real domain!  (this statement is a key one, because if you only allowed + numbers ........)

Choose your inverse relation to be the multi valued square root function.

If C = 2, f-1(f(2)) may be -2.

General hint for other questions: draw a set to set diagram each time.

Bob

Last edited by Bob (2011-01-23 10:55:22)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Advanced Calc homework - Inverse functions

Thanks Bob.

It seems my post wasn't clear regarding what it is that I don't understand. I'll try to give a concrete example to better illustrate my problem.

Let f = {(x, y) | y = x², x ∈ R, y ∈ R}.

What is f ⁻¹([0, ∞))? Is it defined? Is f ⁻¹([0, ∞)) = ∅? Is f ⁻¹([0, ∞)) defined as a relation (not a function) equal to the set of real numbers? Something else?

Also, is f(∅) = ∅?

At first I thought that the notation f ⁻¹(A) for some set A implied that f ⁻¹ is a function, which would imply that f is a bijection. But, being told that I can't assume f is a bijection leads me to believe that my initial belief is incorrect.

Last edited by All_Is_Number (2011-01-23 13:22:34)

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You can shear a sheep many times but skin him only once.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,627

Re: Advanced Calc homework - Inverse functions

hi All_Is_Number

Any rule that you make that 'connects' values in the domain to values in the range is a relation.

These become functions when they are 'well defined';  ie. from a start point in the domain you have one path only leading to the range, so you know where you're going.

An inverse is something that takes you back from the co-domain to the domain.

Notice I didn't use the word range this time; I deliberately said co-domain.  There's a difference.

If the domain is all positive integers and the co-domain is all positive integers and the function is square, then the range is only part of the co-domain because for example you cannot get to 17 in the co-domain by squaring an integer.

If I try to create an inverse function on the co-domain, I cannot make it 'well defined' because there's no √17 in the domain.

But if I change the co-domain to, say, the square numbers, then the function square root becomes 'well defined'. 
(because I've made the co-domain = the range)

So whether an inverse function is possible doesn't just depend on the function; it also depends on the domain and co-domain you choose.  By setting a limit as I did in my example, I can turn a relation into a function or not.  That's why I took care to specify my sets in the earlier post.

Does that help to clear this up?

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Advanced Calc homework - Inverse functions

Does that help to clear this up?

Bob

No, but thanks anyway for trying.

I already understood all that you've written in your previous post, but that isn't what was giving me problems. I apologize for not presenting my problem in a manner more easily understood.

What I needed to know is if f ⁻¹(A), for A ⊂ Domain of f exists when f is not injective, and if so, how is it defined. I've since found out that f ⁻¹ is indeed defined as a relation in such a case.

Thanks again.

Last edited by All_Is_Number (2011-01-24 14:17:33)

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You can shear a sheep many times but skin him only once.

All_Is_Number

Member

Registered: 2006-07-10

Posts: 258

Re: Advanced Calc homework - Inverse functions

Just in case someone does a search and finds this thread, the solution for the example problem from post 1:

Prove or provide a counterexample:

Let f : A ⇾ B. If C ⊂ A. Then C ⊂ f ⁻¹(f (C)).

Proof:

Let f : A ⇾ B = f (a) = a², C ⊂ A, and x ∈ C.

f (x) ∈ f (C) and f (C) ⊂ B.

f ⁻¹(f (C)) = {α | α² ∈ f (C)}.

f (x) ∈ f (C), so x ∈ f ⁻¹(f (C)).

∴ C ⊂ f ⁻¹(f (C)). ∎

Last edited by All_Is_Number (2011-01-24 19:16:03)

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You can shear a sheep many times but skin him only once.