How to prove if a function is bijective? I am having problems being able to formally demonstrate when a function is bijective (and therefore, surjective and injective). Here's an example:
How do I prove that $g(x)$ is bijective?
\begin{align}
f &: \mathbb R \to\mathbb R \\
g &: \mathbb R \to\mathbb R \\
g(x) &= 2f(x) + 3
\end{align}
However, I fear I don't really know how to do such. I realize that the above example implies a composition (which makes things slighty harder?). In any case, I don't understand how to prove such (be it a composition or not).
For injective, I believe I need to prove that different elements of the codomain have different preimages in the domain. Alright, but, well, how?
As for surjective, I think I have to prove that all the elements of the codomain have one, and only one preimage in the domain, right? I don't know how to prove that either!
EDIT
f is a bijection. Sorry I forgot to say that.

You haven't said enough about the function $f$ to say whether $g$ is bijective.
"Injective" means different elements of the domain always map to different elements of the codomain.
"Surjective" means every element of the codomain has at least one preimage in the domain.
Later edit: What you've now added---that $f$ is a bijection---bring us to the point where we can answer the question.  However, maybe you should look at what I wrote above.  Since both definitions that I gave contradict what you wrote, that might be enough to get you there.
