# Prove a function is invertible using Calculus?

1. Aug 18, 2009

### thinkgreen95

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. Aug 18, 2009

### CompuChip

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) = x2. 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. Aug 18, 2009

### arildno

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. Aug 18, 2009

### thinkgreen95

Thanks a bunch, guys!

5. Aug 18, 2009

### CompuChip

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) = x2 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. Aug 18, 2009

### g_edgar

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

7. Aug 18, 2009