I'm try to solve this last part of my question and I just can't prove why the functions are inverses of each other on a graph.
be prepared to do a little algebra
also, your inverse is wrong
some how you change \(x-4\) in to \(4-x\)
x - 4 is not the same as 4 - x
so lets start at the beginning
ok
or just agree it should be \[f^{-1}(x)=\frac{x-4}{-7}\] which would be more natural to write as either \[f^{-1}(x)=-\frac{x-4}{7}\] or \[f^{-1}(x)=\frac{4-x}{7}\]
you did all the steps correct, but when you subtracted \(4\) from both sides you wrote \(4-x\) instead of \(x-4\)
btw you should do this in your head for a line
\[f(x)=-7x+4\] means "multiply by \(-7\) then add \(4\) inverse will say "subtract 4, then divide by -7"
oh you have to do something else for sure!!
plugging in 1 for the first one accomplishes nothing other than telling you that \(f(1)=-7+4=-3\) it does mean that \(f^{-1}(-3)=7\) if we did it right
what you have to do is to show that \[f(f^{-1}(x))=x\]
that is where the algebra comes in
it is not as hard as it looks
\[f(f^{-1}(x))=f(\frac{4-x}{7})\]is the first step
then since \[f(\spadesuit)=-7\spadesuit+4\] you cut and paste and get \[f(\frac{4-x}{7})=-7\times \frac{4-x}{7}+4\]
since you know you will end up with just \(x\) you should expect a raft of cancellation
the 7 goes
to give \[-(4-x)+4\] then a tiny bit more algebra gives \(x\)
if you want to be ambitious you can also show that \[f^{-1}(f(x))=x\] as well
ok now i just read your question and it says "on a graph" you can graph both \(y=-7x+4\) and \(y=\frac{1}{7}x+\frac{4}{7}\) and see that they are symmetric with respect to \(y=x\)
ok i made a mistake there one is \[y=-7x+4\] the other is \[y=\frac{1}{7}x-\frac{4}{7}\]
plug in multiples of 7 it is still going to suck though here is a nice picture http://www.wolframalpha.com/input/?i=inverse+-7x%2B4
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