Question about an inverse function I need to find the rule of the inverse of the function $f(x) = 6x + 4$.
Should I factorize is to $\frac{x}{6} - \frac{2}{3}$ or should I leave it as $\frac{x-4}{6}$.
Thanks 
 A: $$ y= 6x+4$$ Swap $x$ and $y$: $$x=6y+4$$Now express $y$ and you are done...
A: Both are fine, but if you want to make it obvious that it is the inverse, you can put it as $\displaystyle \frac{x-4}{6}$, since it is apparent when put next to $6x+4$. 
A: Basically for inverse function you can swap $x$ and $f(x)$, which will result in $x = 6f^{-1}(x)+4 \ \implies f^{-1}(x)=\frac{x-4}{6}$ which is actually the same as $\frac{x}{6}-\frac{2}{3}$.
Note: You solved it correctly. Both ways work, but often when you work with graphs of the linear equation it is better to keep (linear) functions in a form of $ax+b$, so I would prefer having $f^{-1}(x)=\frac{1}{6} \cdot x -\frac{2}{3}$
A: The inverse function of a function " undoes" what the original function has done. 
In other words, the composition of $f$ and of its inverse , $f^{-1}$ , is the identity function : 
$ (f \circ f^{-1}) (x) = (f^{-1}\circ f) (x)=$  Id$(x) = x$. 
Now in our case, what is the original function actually doing? 
$ (x)_{(input)} \rightarrow_{\times 6 } (6x) \rightarrow_{+4 } (6x+4)_{(output)}$


What should do a function that takes $6x+4$ as input in order to  give
  back $x$ as output


$(6x+4)_{(input)}  \rightarrow_{-4} (6x) \rightarrow_{ /6} (x)_{(output)}$. 
So, the inverse function , let's call it $g$ , first substracts $4$ from its input and then divides the difference by 6 . In other words : 

$g(x) = \frac {x-4} {6}$. 

Let's check. If $g$ is actually the inverse function of $f$, then $(g\circ f) (x) =x$. 
$(g\circ f) (x) = g(f(x))= \frac {(6x+4) - 4} {6}= \frac {6x} {6} = x.$. 
So : $g = f^{-1}$. 
The composition in reverse order should also be checked. 
