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No, the first one is not correct, didn't check the rest.

ok then would the first one be a?

One check you can do yourself for checking inverse functions is to plot a few points of the original function and a few of the inverse. The inverse function will be an exact replica of the original reflected across the line \(y = x\) (just a line extending up and to the right at a \(45^\circ\) angle).
I have plotted your answer here so you can see that it is not correct:

Here's what it would look like if correct:

No, (a) is not correct either. Can you show me how you are doing this?

ok what did u use to get that? that can help me

my teacher just told us that you would need to make everything opposite?

Oh, it's an expensive program called Mathematica. There are undoubtedly many cheaper or free options, this just happens to be the one I have at hand. You could also use graph paper and pencil :-)
Uh, that's probably not exactly how they put it...
Let's say you want to find the inverse of \[f(x)=3x+2\]First, replace \(f(x)\) with \(y\):
\[y = 3x+2\]now switch the two variables. anywhere you see \(y\), you put \(x\) and vice versa.
\[x = 3y+2\]
now solve that for \(y\) and you have your inverse function
\[x-2=3y\]\[y = \frac{x-2}{3}\]

soo i dont really know what to do she not a very good teacher

wow ok i understand now thanks can u check the rest imma try to solve it real quick

ok, I'll the check the others and you do that one

would it be umm d?

1+x/4=y is what i got
but it can also be the same thing as d right?

|dw:1443542618922:dw|

by the rules of operator precedence, \[1+x/4 = y\] is the same as\[1 + \frac{x}{4} = y\]

(to be clear, 3rd one is referring to Question #3)

yeah i realized that can i solve them and u check them for me real quick?
yeah i know what u meant