Question

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

Answer 1

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

Answer 2

prepare to do a raft of algebra

Answer 3

Ok

Answer 4

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

Answer 5

that right?

Answer 6

Yes :)

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

Yes I understand so far :)

Answer 12

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 \[f(g(x))=x\]

Answer 13

Ok :D

Answer 14

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)

Answer 15

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

Answer 16

notice the judicious use of parentheses

Answer 17

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

Answer 18

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

Answer 19

method for doing the second one is similar

Answer 20

That makes sense

Answer 21

I am a bit confused for the second one

Answer 22

same as the first, only inside out

Answer 23

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

Answer 24

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

Answer 25

yeah that looks right

Answer 26

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

Answer 27

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

Answer 28

Ok I'll try that :)

Answer 29

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

Answer 30

:/