Proving if $\frac{3x+1}{x-1}$ is onto? So, I have this function: $f(x)=\frac{3x+1}{x-1}$.
So, in proving if it is onto, then by definition, for every b in B, there exists an a in A such that $f(a)=b$. So, let's solve or a.
We get: $b=\frac{3a+1}{a-1}$, $b(a-1)=3a+1$, $3a-b(a-1)=-1$, $(3a-ba)-b=-1$,$a(3-b)-b=-1$, $a(3-b)=b-1$, $a=\frac{b-1}{3-b}$.
So, as I see it, b is undefined for -3 and 3, and therefore, the function cannot be onto.
Is this correct? Usually when I do these problems, it's always the variable minus of plus a number (eg. $\frac{x+1}{x-5}$, where it is undefined for 5.)

Stop at the step before the last. You proved that 
$$
f(a) = b \iff a(3-b) = b-1
$$
As a consequence:


*

*if $b\neq 3$, there is a solution (obtained with the division)

*if $b=3$ this is equivalent to
$$
0 = 2
$$which is always false. Hence there is no solution and the function is not onto on $\Bbb R$ (but onto $\Bbb R - \{3\}$)


Do not divide by 0!
