Determine whether each of these functions is a bijection from R to R 
Determine whether each of these functions is a bijection from R to R.

I have a few questions in my text book asking me that. An example of one of them is:
$f(x) = −3x + 4$
From my understanding a bijection is a one-to-one function. With this in mind, how exactly can I determine if an equation like the one above is a bijection or not from R to R?
 A: This problem is quite simple, so I respond it in a more guiding fashion:
When determining whether a function is bijective (or, a bijection), one should check if it is one-to-one (or, injective) AND onto (or, surjective).
Is it one to one? Yes. How so? A natural question to ask is whether two different elements are mapped to a same element? In fact, this is to check the definition of one-to-one!
Let us consider if there exist two such elements. Suppose $f(x) = f(y)$. Then $-3x + 4 = -3y + 4$, which implies $x = y$. So here we've get an implication: $"f(x) = f(y)" \;\Rightarrow\; "x = y" .$
So we've known that there are no two such elements unless they are the same! Hence the function is one-to-one.
For the onto-ness, we check if all elements in the codomain, namely $\mathbb{R}$, are all been mapped. We start the argument:
For each y $\in \mathbb{R}$, is there an x $\in \mathbb{R}$, such that $f(x) = y$ ? Definitely yes! Since we can take $x = (y-4)/(-3)$ to make $f(x) = (-3)*(y-4)/(-3) + 4 = y $.
A: A function $f:X\to Y$ is a bijection if it is injective (one-to-one) and surjective (onto).
We say $f$ is injective if $f(x)=f(y)$ implies $x=y$. We say $f$ is surjective if for every $y\in Y$ there exists an $x\in X$ such that $f(x)=y$.
Now, we are given $f:\Bbb R\to\Bbb R$ defined by $f(x)=-3\,x+4$. 
To see that $f$ is injective, suppose $f(x)=f(y)$. Then $-3\,x+4=-3\,y+4$ so that $-3\,x=-3\,y$. It follows that $x=y$. Hence $f$ is injective.
To see that $f$ is surjective, suppose $y\in\Bbb R$. Can you find an $x\in\Bbb R$ such that $f(x)=y$?
A: Another way to show bijection which is worth mentioning is that a function $f:A\to B$ is bijective if and only if it has an inverse $f^{-1}:B\to A$.  In your example, setting $x=-3y+4$ gives $y=-\frac13x+\frac43$. So,
$$f^{-1}(x)=-\frac13x+\frac43.$$
Hence, your function is injective.
