$f(x)=3x+4$ - Injective and Surjective? As a follow-up to Understanding why $f(x)=2x$ is injective, I'm working on proving/disproving that $$f(x)=3x+4,$$ where inputs/outputs live on real numbers, is injective and surjective.
Supposing that $$f(a)=f(b),$$ then $$3a+4=3b+4.$$
Solve for $0$:
$$3a+4-4-3b=0$$
$$3(a-b)=0$$
So, $a$ must equal $b$. Therefore, $f$ is injective.
With respect to whether it's surjective, I looked at its graph:

Since $$3x + 4$$ is linear, then it's continuous, I believe. As a result, is that proof enough that it's surjective, i.e. for all $x$ in $$f(x),$$ the output will cover all real numbers?
 A: I would structure your proofs like so.
Claim: The mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=3x+4$ is injective. 
Proof. Let $x_1,x_2\in\mathbb{R}$ and suppose $f(x_1)=f(x_2)$. Then
$$
f(x_1) = f(x_2)\\[0.5em]
3x_1+4 = 3x_2+4\\[0.5em]
3x_1=3x_2\\[0.5em]
x_1=x_2.
$$
Thus, the mapping is injective. $\blacksquare$
Claim: The mapping $f\colon\mathbb{R}\to\mathbb{R}$ defined by $f(x)=3x+4$ is surjective. 
Proof. Suppose $y\in\mathbb{R}$. Then let $x=\frac{y-4}{3}$. We have the following:
\begin{align}
f(x) &= 3x+4\\[0.5em]
&= 3\left(\frac{y-4}{3}\right)+4\\[1em]
&= (y-4)+4\\[0.5em]
&= y.
\end{align}
Thus, the mapping is surjective. $\blacksquare$
A: You can prove directly that $f$ is surjective.
Suppose $y\in\mathbb R$. Can we find an $x\in\mathbb R$ with $f(x)=y$?
$$3x+4 = y \iff 3x = y-4 \iff x = \tfrac13 y - \tfrac43$$
and that's certainly a real number if $x$ is, so we're done.
A: Suppose to the contrary that some real number $r$ was not in its image.
But $x = \frac{r-4}{3}$ is equal to that number $r$.
A: To show surjectivity, reason as follows:
Suppose $f(x)$ is some arbitrary number $y$.  Is there an $x$ I can plug in to $f$ that will produce that number $y$ ?
In your case this amounts to solving $y=3x+4$ for $x$, which will show that the answer is yes.
Since $y$ was an arbitrary number, you have shown that any real number will be hit by $f$, which means it is surjective.
