Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.

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Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.

Mathematics
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"Confirm" -- This is nice. Just follow the instructions. That's what "confirm" is all about.
\[f(x) = \frac{ x-7 }{ x+3 } ; g(x)=\frac{ -3x-7 }{ x-1 }\]
Well? If it were f(3), what would you do?

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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.
\[\frac{ -3x - 7/ x - 1 - 7}{ -3x - 7/ x - 1 +3 } \]
Horrid notation. Please use parentheses to clarify intent. f(x) = (x-7)/(x+3) g(x) = (-3x-7)/(x-1) f(g(x)) = ([(-3x-7)/(x-1)]-7)/([(-3x-7)/(x-1)]+3) It's not pretty, but it's complete and accurate. Now, for your best algebra skills.
my apologies. can i cross out ( -3x - 7 / x - 1 ) ?
or do i multiply ( x - 1 ) for numerator and denominator?
You simplify. Let's take the numerator. \(\dfrac{-3x-7}{x-1} - 7 = \dfrac{-3x-7}{x-1} - \dfrac{7(x-1)}{x-1} = \dfrac{(-3x-7)-7(x-1)}{x-1}\) Keep going. We're just adding fractions.
for the numerator do i distribute?
You do what it takes to simplify it. If the Distributive Property is appropriate, then do that. \(= \dfrac{-3x - 7 - 7x + 1}{x-1}\) One step at a time.
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 \[f(g(x)) = x\]
\[f(g(x))=f(\frac{ -3x-7 }{ x-1 })\]
now we are going to replace all \(x\) in \(f(x)\) by \(\frac{ -3x-7 }{ x-1 }\) that is actually very easy for me to do here, by cutting and pasting
You already have the denominator. You MUST show some algebra. ([(-3x-7)/(x-1)]+3) = \(\dfrac{-3x-7}{x-1}+3\) Go!
\[f(x) = \frac{ x-7 }{ x+3 }\]so \[f(g(x))= \frac{ \frac{ -3x-7 }{ x-1 }-7 }{ \frac{ -3x-7 }{ x-1 }+3 }\]
THERE'S that denominator!
to get rid of that annoying compound fraction, multiply the numerator and denominator by \(x-1\) (carefully using parentheses)
then multiply out, combine like terms since you know the answer will just be \(x\) you should expect an orgy of cancellation at the last couple steps
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
\[ \frac{ \frac{ -3x-7 }{ x-1 }-7 }{ \frac{ -3x-7 }{ x-1 }+3 }\] the \(x-1\) will cancel top and bottom to get \[\frac{-3x-7-7(x-1)}{-3x-7+3(x-1)}\]
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.
ok so lets back up to \[\frac{-3x-7-7(x-1)}{-3x-7+3(x-1)}\] and see what we get when we remove the parentheses by "we" i mean "you"
i got, -10x / - 10 which would equal to (x) so we have f(g(x)) out of the way, i started g(f(x)), coud you guide me with that to please?
yay
ok sure lets start just as before with an annoying compound fraction but try it yourself first, it is going to work almost exactly like this one
\[(-3 (\frac{ x-7 }{ x+3 }) - 7 (numerator) ; (\frac{ x-7 }{ x+3}) - 1 (denominator)\]
let me try to write it
yeah actually that looks good now multiply top and bottom but this time by \(x+3\) instead of \(x-1\)
\[ g(x)=\frac{ -3x-7 }{ x-1 }\] \[ g(f(x))=\frac{ -3 \frac{ x-7 }{ x+3 } -7 }{ \frac{ x-7 }{ x+3 } -1 }\]
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
you should of course just be left with some x up top and a number in the bottom that will cancel with it
-10x / - 10 like before we get (x) for g(f(x)). whew! long problem lol. but thank you sooooo much for all of your time & help!! i wish i could give you more than just a medal & a fan!
glad to help, hope you learned something (at least learned how to do these!) who studies math in late august?

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