anonymous
  • anonymous
Confirm that f and g are inverses by showing that f(g(x)) = x and g(f(x)) = x.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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tkhunny
  • tkhunny
"Confirm" -- This is nice. Just follow the instructions. That's what "confirm" is all about.
anonymous
  • anonymous
\[f(x) = \frac{ x-7 }{ x+3 } ; g(x)=\frac{ -3x-7 }{ x-1 }\]
tkhunny
  • tkhunny
Well? If it were f(3), what would you do?

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anonymous
  • anonymous
plug in 3 to every (x)
tkhunny
  • tkhunny
Do that with g(x) and you will have f(g(x)). It's just what the notation means.
anonymous
  • anonymous
so for every (x) in f(x) i plug in -3x - 7 / x - 1 ?
tkhunny
  • tkhunny
You have it. Let's see it.
anonymous
  • anonymous
\[\frac{ -3x - 7/ x - 1 - 7}{ -3x - 7/ x - 1 +3 } \]
tkhunny
  • tkhunny
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.
anonymous
  • anonymous
my apologies. can i cross out ( -3x - 7 / x - 1 ) ?
anonymous
  • anonymous
or do i multiply ( x - 1 ) for numerator and denominator?
tkhunny
  • tkhunny
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.
anonymous
  • anonymous
for the numerator do i distribute?
tkhunny
  • tkhunny
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.
anonymous
  • anonymous
from that i got \[\frac{ -10x - 6 }{ x - 1 }\]
tkhunny
  • tkhunny
Okay, now tackle the denominator.
anonymous
  • anonymous
could you start me off so i know where to begin?
anonymous
  • anonymous
be prepared to do a raft of algebra ready?
anonymous
  • anonymous
very ready!
anonymous
  • anonymous
first we compute \[f(g(x)) = x\]
anonymous
  • anonymous
\[f(g(x))=f(\frac{ -3x-7 }{ x-1 })\]
anonymous
  • anonymous
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
tkhunny
  • tkhunny
You already have the denominator. You MUST show some algebra. ([(-3x-7)/(x-1)]+3) = \(\dfrac{-3x-7}{x-1}+3\) Go!
anonymous
  • anonymous
\[f(x) = \frac{ x-7 }{ x+3 }\]so \[f(g(x))= \frac{ \frac{ -3x-7 }{ x-1 }-7 }{ \frac{ -3x-7 }{ x-1 }+3 }\]
tkhunny
  • tkhunny
THERE'S that denominator!
anonymous
  • anonymous
to get rid of that annoying compound fraction, multiply the numerator and denominator by \(x-1\) (carefully using parentheses)
anonymous
  • anonymous
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
anonymous
  • anonymous
you need the first step?
SolomonZelman
  • SolomonZelman
(I would rather just find the inverse, without doing f(g(x))=x and vv, but ... \(-:(\) )
anonymous
  • anonymous
uh, i got \[\frac{ -10x - 6 }{ 8 }\]
anonymous
  • anonymous
ok lets go slow
anonymous
  • anonymous
\[ \frac{ \frac{ -3x-7 }{ x-1 }-7 }{ \frac{ -3x-7 }{ x-1 }+3 }\] the \(x-1\) will cancel top and bottom to get \[\frac{-3x-7-7(x-1)}{-3x-7+3(x-1)}\]
anonymous
  • anonymous
now multiply out using the almighty distributive law then combine like terms
anonymous
  • anonymous
then \[\frac{ -3x - 7 - 7x + 1 }{ -3x - 7 + 3x - 1 }\]
anonymous
  • anonymous
forgot that distributive law already huh?
anonymous
  • anonymous
distribute the \(-7\) up top and the \(3\) below
anonymous
  • anonymous
starting here \[\frac{-3x-7-7(x-1)}{-3x-7+3(x-1)}\]
anonymous
  • anonymous
(-3x - 7 ) ( -7x + 1) (-3x - 7 ) ( 3x - 1)
anonymous
  • anonymous
oh no
anonymous
  • anonymous
what can you get with \[-7(x-1)\]when you distribute?
anonymous
  • anonymous
-7x + 7 OMG -.- im so disappointed with myself right now.
anonymous
  • anonymous
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"
anonymous
  • anonymous
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?
anonymous
  • anonymous
yay
anonymous
  • anonymous
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
anonymous
  • anonymous
\[(-3 (\frac{ x-7 }{ x+3 }) - 7 (numerator) ; (\frac{ x-7 }{ x+3}) - 1 (denominator)\]
anonymous
  • anonymous
let me try to write it
anonymous
  • anonymous
yeah actually that looks good now multiply top and bottom but this time by \(x+3\) instead of \(x-1\)
anonymous
  • anonymous
\[ g(x)=\frac{ -3x-7 }{ x-1 }\] \[ g(f(x))=\frac{ -3 \frac{ x-7 }{ x+3 } -7 }{ \frac{ x-7 }{ x+3 } -1 }\]
anonymous
  • anonymous
should have used parentheses around the first term in the top
anonymous
  • anonymous
multiply by \(x+3\) top and bottom what do you get before distributing etc
anonymous
  • anonymous
this might be trickier because of the \(-3\) up top so be careful with parentheses
anonymous
  • anonymous
-3(x-7) - 7(x+3) / (x-7) - 1(x+3) ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
now get rid of the parentheses
anonymous
  • anonymous
you should of course just be left with some x up top and a number in the bottom that will cancel with it
anonymous
  • anonymous
-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!
anonymous
  • anonymous
glad to help, hope you learned something (at least learned how to do these!) who studies math in late august?

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