prove a function is invertible using Calculus?by thinkgreen95 Tags: calculus, function, invertible, prove 

#1
Aug1809, 09:30 AM

P: 4

How do you prove a function is invertible using calculus? I have a question on my review packet and it says to show that f(x) is invertible using calculus...apparently it is important that it is called an injection?




#2
Aug1809, 09:41 AM

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Yes. A function f: A > B is injective (or an injection) when two function values being equal implies that they are the image of the same point.
That is: for all a, b in A: f(a) = f(b) implies a = b. Why this is a necessary condition is easy to see. Suppose that you have two values a, b that are different, but f(a) = f(b) = y. Now, you are writing down the inverse function f^{1}... what value do you assign for f^{1}(y)? It should be both a and b, because f(a) = y and f(b) = y, but f^{1}(y) can only have one value. It is also a sufficient condition: after all, all you need to define an inverse function is to say for every y in B, which x in A to map it to such that f(x) = y again. If it is not clear, think about f(x) = x^{2}. If you only define the function for x > 0 (you can include 0 if you like) then there is no problem to write down the inverse function: f^{1}(y) = sqrt(y). But if you define f(x) for all x (also negative numbers) it is no longer injective. Indeed, 2 and 2 are completely different numbers, but f(2) = f(2) = 4. So you cannot consistently assign a value for f^{1}(4), and always get f(f^{1}(x)) = x. 



#3
Aug1809, 09:51 AM

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To zoom in on how calculus may help you in proving invertibility:
On what point A on the graph of a function can it occur that two neighbouring points have the SAME function value? Clearly, this can only happen if A is either a maximum or minimum! But this means that if a point is NOT a max or min, then the function is invertible in a neighbourhood of A. Transform this into a requirement upon the derivative of f!! 



#4
Aug1809, 10:29 AM

P: 4

prove a function is invertible using Calculus?
Thanks a bunch, guys!




#5
Aug1809, 11:24 AM

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Ah, I forgot to say this in my last post, but you've already replied so let me add it in a new message instead of editing the old one...
From a practical point of view, injectivity is very useful to prove invertibility. If you want to show that a function is invertible, it is sufficient to show that it is injective. The latter is usually rather straightforward, just start the proof with "Suppose that f(a) = f(b)" and try to get to "... so a = b", or by contradiction: start with "Suppose that a is not equal to b" and try to get to "... so f(a) is not equal to b". For example: f(x) = x^{2} is not injective on the real numbers. Proof: The simplest is a proof by counterexample: f(2) = f(2) although clearly, 2 is not equal to 2. Another one: f(x) = 2x + 1 is injective. Proof: Suppose that f(a) = f(b). Then by definition of the function, 2a + 1 = 2b + 1. Subtract 1 from both sides and divide by 2, it follows that a = b. So if f(a) = f(b), then a = b: the function is injective. 



#6
Aug1809, 02:05 PM

P: 608

If you do not recognize the term "injective" maybe your text calls it "onetoone" instead.




#7
Aug1809, 02:38 PM

P: 810

One additional point.
If a function is an injection, its inverse's domain is the original functions range (and NOT it's codomain). So, take f(x) = e^x. It's R to R  so it's domain and codomain are both the real numbers. But for any real x, e^x is always positive, so it's range is the positive reals, R+. Thus, its inverse, the logarithm, is a function R+ to R (and not R to R). Now, if a function is a bijection, things are a bit more ideal. The range is equal to the codomain, so these functions are perfectly invertible, without having to worry about what the range of the function is. Bijections are pretty magical in math. 


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