# Prove injectivity of function.

1. Oct 27, 2012

### stefan10

1. The problem statement, all variables and given/known data

Prove that f is injective.

2. Relevant equations

$f:(- \infty, 3] \rightarrow [-7,\infty) \vert f(x) = x^2 -6x+2$

3. The attempt at a solution

I wish to prove this by calculus. I know that the maximum is three, and this is the only way the quadratic can be one-to-one. However; I'm having problems formalizing this into a proper proof.

This is what I have so far:

Definition of injective function: $f(x_{2}) = f(x_{1}) \Rightarrow x_{2}=x_{1}$

$f(x)' = 2x-6$

When $f(x)' = 0, x=3$

This is true, only if $(3, f(3))$ is a maximum or a minimum.

In this case $(3,f(3))$ is a minimum, and $f(3) = -7.$

Since 3 is the only critical point, f(x) is always increasing when x < 3.

This implies injectivity, because ..... (how do I conclude this, it makes sense intuitively, but I don't know how to write it?)

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How can I explain that since 3 is a minimum and also the only critical number, the function is injective upon these domain constraints? Must I use images and pre-images? Also I know there is certainly a way to prove this without calculus, since calculus isn't a pre-requisite for the course this is for, but I'd like to use differentiation, since I have the tool and it's the first thing I had thought of upon noting that this function is quadratic and the only way for it to be injective is if f(x)' = 0 when x=3 and that it is the only place where f(x)'=0. Thank you!

Last edited: Oct 27, 2012
2. Oct 27, 2012

### LCKurtz

If you want a calculus argument, you have f'(x) = 2(x-3) < 0 if x < 3. That tells you the function is decreasing on that interval, hence injective.