f(x)= x-7/x+2 g(x)=-2x-7/x-3 find f(g(x)) and g(f(x))

- anonymous

f(x)= x-7/x+2 g(x)=-2x-7/x-3 find f(g(x)) and g(f(x))

- chestercat

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- anonymous

G(x)= -2x-7/x-1 (correction)

- anonymous

prepare to do a raft of algebra

- anonymous

Ok

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## More answers

- anonymous

\[f(x)=\frac{x-7}{x+2}\\
g(x)=\frac{-2x-7}{x-1}\]

- anonymous

that right?

- anonymous

Yes :)

- anonymous

ok lets to \[f(g(x))\] first

- anonymous

what i am going to do is use cut and paste to put \[\frac{-2x-7}{x-1}\] everywhere i see an \(x\) in \[\frac{x-7}{x+2}\]

- anonymous

\[\large f(g(x))=f(\frac{-2x-7}{x-1})=\frac{\frac{-2x-7}{x-1}-7}{\frac{-2x-7}{x-1}+2}\]

- anonymous

clear what i did?
that is the first step, bunch of algebra comes next

- anonymous

Yes I understand so far :)

- anonymous

ok now before we do the algebra, i want to let you know that a miracle will occur
the reason you were given this problem is that \(f\) is the inverse of \(g\) and vice versa
that means when we get done we will find \[f(g(x))=x\]

- anonymous

Ok :D

- anonymous

staring here \[\frac{\frac{-2x-7}{x-1}-7}{\frac{-2x-7}{x-1}+2}\] get rid of the compound fraction by multiplying top and bottom by \(x-1\) (carefully)

- anonymous

\[\frac{-2x-7-7(x-1)}{-2x-7+2(x-1)}\] is step one

- anonymous

notice the judicious use of parentheses

- anonymous

now remove the parentheses using the distributive property to get \[\frac{-2x-7-7x+7}{-2x-7+2x-2}\]

- anonymous

combine like terms, and since they are inverses you will get an orgy of cancellation \[\frac{-9x}{-9}=x\]finished

- anonymous

method for doing the second one is similar

- anonymous

That makes sense

- anonymous

I am a bit confused for the second one

- anonymous

same as the first, only inside out

- anonymous

I put..g(f(x))= -2(x-7/x+2)-7/(x-7/x+2)-1

- anonymous

try it and see what happens
it will be easy enough to know if you are right or not, because the answer will be \(x\)

- anonymous

yeah that looks right

- anonymous

However after this step I wasn't positive if I should multiply both by x+2 or distribute

- anonymous

yes multiply top and bottom by \(x+2\) like i did before, carefully, using parentheses

- anonymous

Ok I'll try that :)

- anonymous

I don't think I'm doing this correctly ::

- anonymous

:/

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