Injective Function

So I'm going back to the basics and the book gives an example stating that if I want to show f(x)= 3x-5 is injective I do the following:
f(x1)= f(x2)
3x1-5=3x2-5
3x1=3x2
x1=x2....thus it is injective.

Now let's say we didn't know the following is a parabola and thus could not apply the horizontal line test. f(x)=x^2
x^2(1)=x^2(2), then we take the square root of both sides so we have..
x1=x2.

According to that the parabola is one-to-one, but we know it isn't. What gives?

(x1)^2 = (x2)^2 =/=> x1 = x2.
Proof: let x1 = 1 , x2 = -1
However:
x1 = x2 ==> (x1)^2 = (x2)^2

I see, so I would just stop at (x1)^2 = (x2)^2?

>>3
Yes. It would be even better to provide a counterexample; in that case you know that there's no point in finding further proof. If you can't find a counterexample, you may just be stuck.

(x1)^2 = (x2)^2
x1 = x2

Note that this is wrong because sqrt(x^2) is not equal to x; instead, it is equal to abs(x). When you end up with abs(x1) = abs(x2) finding a counterexample suddenly becomes easy.

Aren't all even functions over the domain of real numbers non-injective? Isn't that sort of a given?

>>6
Correct.

>>6

Of course it's a given, but what if you didn't know that?

>>8
Then take the definition of an even function.

if f(x) = f(-x) for all x, that implied in particular (in very particular) that f(1) = f(-1) and so f isn't injective