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\[f(x) = \frac{ x-7 }{ x+3 } ; g(x)=\frac{ -3x-7 }{ x-1 }\]

Well? If it were f(3), what would you do?

plug in 3 to every (x)

Do that with g(x) and you will have f(g(x)). It's just what the notation means.

so for every (x) in f(x) i plug in -3x - 7 / x - 1 ?

You have it. Let's see it.

\[\frac{ -3x - 7/ x - 1 - 7}{ -3x - 7/ x - 1 +3 } \]

my apologies. can i cross out ( -3x - 7 / x - 1 ) ?

or do i multiply ( x - 1 ) for numerator and denominator?

for the numerator do i distribute?

from that i got \[\frac{ -10x - 6 }{ x - 1 }\]

Okay, now tackle the denominator.

could you start me off so i know where to begin?

be prepared to do a raft of algebra
ready?

very ready!

first we compute \[f(g(x)) = x\]

\[f(g(x))=f(\frac{ -3x-7 }{ x-1 })\]

THERE'S that denominator!

you need the first step?

(I would rather just find the inverse, without doing f(g(x))=x and vv, but ... \(-:(\) )

uh, i got \[\frac{ -10x - 6 }{ 8 }\]

ok lets go slow

now multiply out using the almighty distributive law
then combine like terms

then \[\frac{ -3x - 7 - 7x + 1 }{ -3x - 7 + 3x - 1 }\]

forgot that distributive law already huh?

distribute the \(-7\) up top and the \(3\) below

starting here \[\frac{-3x-7-7(x-1)}{-3x-7+3(x-1)}\]

(-3x - 7 ) ( -7x + 1)
(-3x - 7 ) ( 3x - 1)

oh no

what can you get with
\[-7(x-1)\]when you distribute?

-7x + 7 OMG -.- im so disappointed with myself right now.

yay

\[(-3 (\frac{ x-7 }{ x+3 }) - 7 (numerator) ; (\frac{ x-7 }{ x+3}) - 1 (denominator)\]

let me try to write it

should have used parentheses around the first term in the top

multiply by \(x+3\) top and bottom
what do you get before distributing etc

this might be trickier because of the \(-3\) up top so be careful with parentheses

-3(x-7) - 7(x+3) / (x-7) - 1(x+3) ?

yes

now get rid of the parentheses