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

- anonymous

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

- Stacey Warren - Expert brainly.com

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- schrodinger

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- tkhunny

"Confirm" -- This is nice. Just follow the instructions. That's what "confirm" is all about.

- anonymous

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

- tkhunny

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

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## More answers

- anonymous

plug in 3 to every (x)

- tkhunny

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

- anonymous

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

- tkhunny

You have it. Let's see it.

- anonymous

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

- 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

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

- anonymous

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

- 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

for the numerator do i distribute?

- 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

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

- tkhunny

Okay, now tackle the denominator.

- anonymous

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

- anonymous

be prepared to do a raft of algebra
ready?

- anonymous

very ready!

- anonymous

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

- anonymous

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

- 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

You already have the denominator. You MUST show some algebra.
([(-3x-7)/(x-1)]+3) = \(\dfrac{-3x-7}{x-1}+3\)
Go!

- 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

THERE'S that denominator!

- anonymous

to get rid of that annoying compound fraction, multiply the numerator and denominator by \(x-1\) (carefully using parentheses)

- 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

you need the first step?

- SolomonZelman

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

- anonymous

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

- anonymous

ok lets go slow

- 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

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

- anonymous

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

- anonymous

forgot that distributive law already huh?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

oh no

- anonymous

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

- anonymous

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

- 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

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

yay

- 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

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

- anonymous

let me try to write it

- anonymous

yeah actually that looks good
now multiply top and bottom but this time by \(x+3\) instead of \(x-1\)

- 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

should have used parentheses around the first term in the top

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

yes

- anonymous

now get rid of the parentheses

- 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

-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

glad to help, hope you learned something (at least learned how to do these!)
who studies math in late august?

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