Applying a function to both sides of an equation doesn't change it? Why is it that applying a function to both sides of an equation doesn't change it? Can this be proven? Can you point to some material to read more about this?
 A: One definition of a function is, given an element $x$ of a set X, it is a rule that gives one element of another set, $Y$. So if you choose $p=x$ (of course $p \in X$), $f(p)=f(x)$. 
However, note that $f(p)=f(x)$ does not imply $p=x$, as multiple elements of the set $X$ may be mapped into the same element of $Y$.  
A: Applying a function to both sides of an equation does not change the equality because of the definitions of equality and a function.
A function has a unique output for every input. That means that if you input $x$ to a function and get $a$ as the output, then $a$ will always be the output of the function with input $x$. You can't put $x$ into the same function later and get $b$ out. Or rather, the only way you get $b$ out of the same function with the same input is if $a$ and $b$ are equal.
Since the two sides of your equation are equal, they are effectively the same thing, they just look different. So when you put them as inputs into a function, you should get the same thing out each time.
A: $x=y$ means that the symbols $x$ and $y$ are somehow 'labels' of exactly the same mathematical object. Any expression (so not only functions) in wich these symbols are encountered will not be affected if these symbols are interchanged.
If e.g. one of your beloved ones carries two names: Judy and Aphrodite, then the following statements are equivalent:


*

*I love Judy.

*I love Aphrodite.

A: I will expand on the answers to this important question, which give in my opinion, a rough draft to a more complete answer. I will also give examples to make sense of the first part of this answer, which will be the theory. To follow along, it will help to draw pictures of the sets described below just to organize the letters in your head. 
In the following definitions, I use the convention that the codomain is always the range of the function.
$$ f: D \to R  \\
g: U \to V
$$
The equation
$$ f(x) = g(x) \tag{1}$$
only makes sense if there exists elements $x \in D $ and $x \in U$. If solutions to this equation exist, the domains must overlap $C = D \cap U$, with $C$ a nonempty set.  To be clear, even if $C$ is nonempty, $(1)$ may not have solutions. But to give us a fighting chance, $C$ better be nonempty (by the way, if you are following along with pictures, you should also draw the range sets overlapping). 
Lets call the solution set of $(1)$ $S$ and we know it must lie in the common domain $S \subseteq C$. Consider a function $h$ which we wish to compose with, or apply to, $(1)$
$$
\begin{align*}
&h: K \to L \\[0.5em]
&h(f(x)) = h(g(x)) \tag{2}
\end{align*}
$$
Before we think about applied functions $h$ that preserve solutions sets [does $(2)$ have the same solution set $S$ as $(1)$? ], we need to make sure that the composition makes sense in the first place. $h(f(x))$ only makes sense if the range of $f$ overlaps with the domain of $h$ so that $R \cap K$ is nonempty. That is, $h(f(x))$ only makes sense if there is some subset of $D$ such that the images under $f$ are in the domain of $h$. Let's call this subset of $D$, which is the domain of the composition, $D_h$.
Likewise for the right hand side, $h(g(x))$ only makes sense if the range of $g$ overlaps with the domain of $h$ so that $V \cap K$ is nonempty. That is, $h(g(x))$ only makes sense if there is some subset of $U$ such that images under $g$ are in the domain of $h$. Let's call this subset of $U$, which is the domain of the composition, $U_h$.
The common domain $D_h \cap U_h$ needs to be nonempty in order to apply $h$ in the first place. Otherwise, $(2)$ turns into an equation with no solutions. [If $(1)$ has a nonempty solution set, and our goal is producing an equation $(2)$ with the same solution set, then it better have some solutions. The common domain condition $D_h \cap U_h$ nonempty gives us a fighting change for some solutions]. In particular, the overlap $D_h \cap U_h$ lies entirely in the common domain $C$. All the solutions $S \subseteq C$ of $(1)$, are also solutions of $(2)$ if $S \subseteq (D_h \cap U_h) \subseteq C$, by definition of a function [ if $f(S) = g(S)$ is a true statement, so too is $h(f(S)) = h(g(S))$ ]. However, $S \subseteq (D_h \cap U_h)$ need not be satisfied. Sometimes $(D_h \cap U_h)$ will include some, but not all, of $S$. Therefore $(2)$ will definitely produce missing solutions. And sometimes the territory outside of $S$ but inside of $(D_h \cap U_h)$ has solutions to $h(f(x)) = h(g(x))$. These are extraneous solutions or extra solutions which aren't solutions to the original problem. For instance, $h(f(2)) = h(g(2))$, where $2$ is not in $S$ but in $(D_h \cap U_h)$, might produce $h(4) = h(7)$, which might be a true statement depending on the nature of $h$. I think you see where "injective" now comes into play. But first off, let me write down the first criteria of 2, both of which must be satisfied in order to preserve the solution set $S$:


*

*The common domain of the composition $(D_h \cap U_h)$ must contain the solution set $S \subseteq (D_h \cap U_h)$ of $(1)$


This is too abstract for me to remember (I'm writing a lot of words - this is not what goes on in my head when I'm solving problems). The examples below will make sense of this point. At the very least, you need to make sure that the composition is well defined in the first place (this should be obvious - this is what goes on in my head). Criteria 1. above will be satisfied if it's rewriting below (a weakening of the statement) is satisfied:


*

*The range of $f$ is fully contained in the domain of $h$ and the range of $g$ is fully contained in the domain of $h$. This guarantees we are safe in applying $h$. Furthermore, $D_h = D$ and $U_h = U$, making $D_h \cap U_h = C$, guaranteeing that the common domain of the composition contains the full solution set.


Most functions that we apply have a domain that takes all possible inputs, and therefore we never have to worry about criteria 1 (a function with any range will always work because $h$'s domain is as large as can be). For instance "add 2 to each side of the equation" is represented as $h(x) = x + 2$, which has a domain of all real numbers. Or the cubing function $h(x) = x^3$, also has a domain of all real numbers. 
All we have to worry about now are the extraneous solutions! How do we get rid of those? Not every $x \in C$ is a solution to $(1)$, only those $x \in S \subseteq C$ are solutions to $(1)$. Therefore to prevent $(2)$ from having solutions such as $h(4) = h(7)$ coming from say $h(f(3)) = h(g(3))$ we need to make $h$ injective so that one output corresponds to only one input. But here is a very important question. Injective over what domain? Over $h$'s full domain? The answer is no. $h$ only needs to be injective over $(K \cap R) \cup (K \cap V)$. Or rewriting this by pulling out a $K$/undoing the distributiion, $h$ only needs to be injective over $K \cap (R \cup V)$. It only needs to be injective over part of it's domain - that part which intersects with the ranges. This is very important. $h(x) = x^2$ is not injective over it's full domain. But consider $\sqrt{x} = 7$. I can apply the squaring function to this equation. Why? $\sqrt{x}$ has a range of $0$ and the positive numbers. Over this, $h = x^2$ is injective! Therefore $\sqrt{x} ^2 = 7^2$ or $x = 49$. This is very important. Usually when you square an equation, you can sometimes produce extraneous solutions and therefore have to check your answers. I never want to do math where I have to check my answers. I want to minimize this as much as possible - and you do this by following the two criteria I that have written down (the 2nd one is coming). I don't have to check $x = 49$ in $\sqrt{x} = 7$, because I knew exactly what I was doing when I applied the squaring function. I never changed the solution set.


*$h$ needs to be injective ideally over it's full domain. At the very least, it needs to be injective over all possible inputs it can receive. That is, $h$ needs to be injective over $K \cap (R \cup V)$


Examples:
$$ x^2 = 100$$
Can I? 
$$
\begin{align}
\sqrt{x^2} = \sqrt{100} \\
|x| = 10
\end{align}
$$
Yes! The square root here has a domain which fully contains the range of $x^2$. It's injective over it's entire domain. Both criteria are satisfied. I don't need to check my answers. I never changed the solution set.
$$ \sqrt{3 -x} = 10$$
Can I?
$$ 
\begin{align}
\sqrt{3 - x}^2 = 10^2 \\
3 - x = 100\\
x = -97
\end{align}
$$
Yes, and I don't need to check my solutions. The squaring function's domain encompasses the range of the square root. It's injective, over the range of the square root.
$$ e^x = 15 $$
Can I?
$$ \ln(e^x) = \ln(15)$$
Yes! The domain of the natural logarithm matches the range of the exponential function. The natural log is injective over it's full domain.
$$ \sqrt{3-x} = x - 1
$$
Can I?
$$\sqrt{3-x}^2 = (x-1)^2 $$
No, I can't (I will need to check for possible extraneous solutions - I already know, which I can explain if needed, that squaring doesn't create missing solutions). Although the squaring function's domain encompasses the range of the square root and the range of the function $x - 1$, the squaring function is injective over the range of the square root function, but not injective over the range of $(x-1)$, which is the reals. Criteria 2. is not satisfied
$$ x^2 - 10 = 2 $$
Can I?
$$ 
\begin{align}
\sqrt{x^2 - 10} = \sqrt{2}
\end{align}
$$
No by what I've recently said. The domain of the square root is zero and the positives. Range of $x^2 - 10$ is everything $-10$ and bigger. The domain does not contain this range so (weak version) criteria 1. is not satisfied. In reality, this is a valid operation. I can tell by looking at the equation that, first, it will have solutions. This is important. Rewrite it as $x^2 = 12$ to see this. Next, because I know that solutions exist, they have to be in an interval which produces a positive image (the left hand side is set equal to $2$, a positive number). Therefore, while the domain of the square root does not include the full range of $x^2 -10$, we know that the solution set of the original equation has to be in an interval which produces a positive image. The solution set can't be in $-\sqrt{10} < x < \sqrt{10}$. We have no interest in this interval. Therefore, we only have interest in the range of $x^2 - 10$ which is $0$ or positive!!! The domain of the square root function contains this range of interest. (Strong version) criteria 1. is satisfied. The function is injective over the range it can receive. Criteria 2. is satisfied. We are good to go. (It's all about where the solution set is located and making sure that $D_h\cap U_h$ captures this entire solution set). 
$$ \sqrt{x - 3} = \sqrt{-x + 1}$$
Can I?
$$ \sqrt{x-3}^2 = \sqrt{-x+1}^2$$
And proceed to get $x = 2$? No! I haven't touched on this yet, but the starting equation has no solution because it's junk. It' doesn't even have a common domain. $x$ cannot be both $x \geq 3$ and $x \leq 1$.
I don't like to memorize things, and there are a lot of words here. I'm still getting used to this because I had to study this problem very recently due to another problem that called for it (I should have studied it all the way back in algebra!). But it's pretty easy, and with time, I think it will become second nature for me. When we have to check solutions and don't know why we have to check solutions, or if we don't need to check solutions (our goal) and don't know why, then we aren't in control of the math. I hope this helps
A: Assuming $=$ denotes equivalence, if $x=y$, then $x$ and $y$ are the same object. Therefore, $f\left(x\right)=f\left(y\right)$ for any $f$ whose domain includes $x$ (or equivalently, $y$).
That is, the function cannot distinguish between $x$ and $y$.
