Why is it that $f(x)$ is even if $f(-x) = f(x)$? I found this definition in this PDF file on page 17: http://math.byu.edu/home/sites/default/files/u107/proofs_crash_course.pdf
I couldn't quite grasp it. I tried substituting x for a value in both f(x) and f(-x) but the result wasn't equal. So then, how was this proven? Is it correct?
Please understand that I'm requesting an explanation of the definition : f(x) is even if f(-x) = f(x). I'm not asking for an answer to the question asked in the PDF reference.
 A: You also might be asking why we would call a function like $f(x) = x^2$ "even" - that is, why would we choose that word.  Well, it turns out that $f(x) = x^k$ is an even function when $k$ is an even number, and is an odd function when $k$ is an odd number. So although we could have chosen other words for these types of functions (for example, "symmetric" and "antisymmetric"), choosing "even" and "odd" as the names has some mnemonic value.
A: A definition is a definition, you are not allowed to request a proof, or to question it, you are expected to memorize it and be able to use it. 
But, you may ask for an intuitive justification and a motivation for a definition. Consider the function $g(x)=3x^4-6x^2-5$. Notice things like
$g(-1)=3(-1)^4-6(-1)^2-5 = 3 (1)^4-6(1)^2-5 = g(1)$, and
$g(-4)=3(-4)^4-6(-4)^2-5 = 3 (4)^4-6(4)^2-5 = g(4)$, and more generally
$g(-x)=3(-x)^4-6(-x)^2-5 = 3 (x)^4-6(x)^2-5 = g(x)$ for all $x$.
This function has the property that $g(-x)=g(x)$ for all $x$. It seems this 
has something to do with the powers of $x$ that appear in it: $x^4$, $x^2$, and, you may say, $x^0$ (like $5x^0=5$). It happens more generally that if all powers of $x$ in a function are even, then the values of that function at $-x$ and at $x$ are the same. But, this also happens for some other functions, like $\cos(-x)=\cos(x)$ for all $x$ even if you see no powers of $x$ here. It is convenient to give a name to functions that satisfy this property. Since the easiest examples involved even powers, we call such functions even functions. 
But, we do not want to be restricted to only consider functions in which the powers of $x$ are even: It turns out to be convenient to use this same name whenever a function $f$ has the property that $f(-x)=f(x)$ for all $x$. All such functions are by definition called even (this is just a convention, nothing to prove here, like you need no proof to justify why the color red is called red), but once this convention, or definition, if accepted you could use it to prove that one function or another is even, i.e. that it satisfied the defining property of even functions, that $f(-x)=f(x)$ for all $x$. For example the function $f(x)=|x|+5$ is even (even if it looks like $x$ appears to the first power which is odd), since $f(-x)=|-x|+5=|x|+5=f(x)$ for all $x$. 
Another example $h(x)=x^3-4x$. Here if you start with $h(-x)$ we get
$h(-x)=(-x)^3-4(-x)= -(x^3)+4x= -\Bigl(x^3-4x\Bigl)=-h(x)$.
So $h(-x)=-h(x)$ for all $x$.  It is convenient to call all functions with this property odd functions. That is $f$ is an odd function if, by definition (convention, agreement) $f(-x)=-f(x)$ for all $x$ (in the domain of $f$).
There are many examples of such functions where all powers of $x$ are odd, and this justifies the name, but there are other examples too, like $\sin(-x)=-\sin(x)$ for all $x$, so $\sin$ is an odd function. 
The graph of each even function is symmetric about the $y$-axis. This is something one could prove, using the definition. Indeed, if the point $(x,f(x))$ is on the graph, the so is the point $(-x,f(x))=(-x,f(-x))$,
but $(x,f(x))$ and $(-x,f(x))$ are symmetric about the $y$-axis.
Like, the graph of $y=x^2$ is symmetric about the $y$ axis, and so is the graph of $y=\cos x$.  
The graph of each odd function is symmetric about the origin. This is something one could prove, using the definition. Indeed, if the point $(x,f(x))$ is on the graph, the so is the point $(-x,-f(x))=(-x,f(-x))$,
but $(x,f(x))$ and $(-x,-f(x))$ are symmetric about the origin.
Like, the graph of $y=x^3$ is symmetric about the origin, and so is the graph of $y=\sin x$.  
Finally, there are functions that are neither even nor odd. Usually that happens when one mixes (additively) even and odd powers of $x$, like in $p(x)=x^3-5x^2$. Here $p(1)=-4$ and $p(-1)=-6$. So, $p(-1)\not=p(1)$ and this function $p$ is not even. But we also have $-p(1)=-(-4)=4$ and $4\not=-6$, that is $p(-1)\not=-p(1)$ so this function is not odd either. So, it is neither. 
One alternative way to see this, if to start with $p(-x)$ and see if one obtains $p(x)$ (in that case $p$ would be even), or one obtains $-p(x)$
(in that case $p$ would be odd), or one obtains neither $p(x)$ nor $-p(x)$. 
In this example $p(-x)=(-x)^3-5(-x)^2 = -x^3-5x^2$ which is different from  both
$p(x)=x^3-5x^2$ and from $-p(x)=-x^3+5x^2$, so $p$ is neither an even nor an odd function. 
Other examples of functions that are neither even not odd are $2-\sin(x)$ and $x+\cos(x)$. On the other hand $\dfrac{x^2-1}{x^3}$ is odd, regardless that we mix even and odd powers, because this time we do not add/subtract them, but instead miltiply/divide them (the numerator has only even powers, and the denominator has only odd powers). You may use the definition to verify (that is, prove) that $\dfrac{x^2-1}{x^3}$ is indeed an odd function. You may also try to find out things like whether the product of two odd functions is odd or even (it is even), and whether the product of an odd function and an even function is even or odd. 
A: There's a chance you might be getting confused with the word "even". It doesn't mean that the function equals 2,4,6, etc. We use the word "even" for that too (when we are talking about numbers), but when people say that a function is even, they mean that it is mirrored over the y-axis. 
For example, $f(x)=x^{2}$ is "even", not because it equals 2 or 4 or 6, but because it is symmetric across the y-axis. This mirror property is a result of the fact that $f(x)=f(-x)$. Why is it that way? Well, if you plug in 1, it's the same as if you had plugged in -1. Same with 2 and -2. Thus $f(x)$ is the same for positive and negative x; in other words, $f(x)=f(-x)$.  $cos(x)$ has the property too, and tons of other functions. I'll show you some pics for clarity, credit to wolfram alpha: 



If you're asking proof that $f(x)$ is even if $f(x)=f(-x)$, it's not something that you can "prove", at least not in a mathematical way. Let me say a bit more:
It's not like we have a concept of what "even" means in our head, and we have a concept of what it means that $f(x)=f(-x)$ as well, and we can show that those are equivalent. Instead, "even" is just a label that we put on functions for which $f(x)=f(-x)$, rather than saying that outright each time. We could never use the word even, but it's just easier. We observed a trait and then we gave it a name. 
An analogous example: you can't "prove" that we call those round edible fruits with stems "apples", that's just the name we gave them. The apples came first, and then came the name. I guess the only "proof" that those things are called apples would be to take a survey and see what percent of people actually DO call them apples. You could do that for this too, and I'm confident that you'd find that when it came to mathematicians, in English, at least 95% would call such a function even. Though this wouldn't be a proof, it would just be strong evidence if you had a large sample size. Proofs are, in theory, reasoning which makes it IMPOSSIBLE for the statement to be false.
A: The definition of an even function is that it satisfies f(x)=f(-x). Likewise, an odd function satisfies f(-x)=-f(x). An example of an even function is cos(x).
A: I'm trying to upload some images, but my computer is going very slow. Consider: $$f(x)=\sin(x)$$ If you were to do $$f\big({\pi\over2}\big)=\sin\big({\pi\over2}\big)$$ and also were to do $$f\big(-{\pi\over2}\big)=\sin\big(-{\pi\over2}\big)$$ you could see that $$f\big({\pi\over2}\big)=1\ \text{and} \ f\big(-{\pi\over2}\big)=-1$$ So $$f({\pi\over2})=-f(-{\pi\over2})$$ making it an odd function. You could do the same argument for even functions using $$f(x)=\cos(x)$$ Does that help any?
A: You want to proof this: If $f$is even, then $f(x)$ is not one-to-one. You can proof this by deduction. Let be any $x$, by definition, since $f$ is even, then $f(x)=f(-x)$, therefore $f(x)$ is not unique, then $f(x)$ is not one-to-one. 
You must remember this: $f$ is one-to-one if $f(x)=f(y)$ then $x=y$.
