Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

plug in 3 to every (x)

Answer 5

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

Answer 6

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

Answer 7

You have it. Let's see it.

Answer 8

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

Answer 9

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.

Answer 10

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

Answer 11

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

Answer 12

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.

Answer 13

for the numerator do i distribute?

Answer 14

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.

Answer 15

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

Answer 16

Okay, now tackle the denominator.

Answer 17

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

Answer 18

be prepared to do a raft of algebra
ready?

Answer 19

very ready!

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

You already have the denominator. You MUST show some algebra.

([(-3x-7)/(x-1)]+3) = \(\dfrac{-3x-7}{x-1}+3\)

Go!

Answer 24

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

Answer 25

THERE'S that denominator!

Answer 26

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

Answer 27

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

Answer 28

you need the first step?

Answer 29

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

Answer 30

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

Answer 31

ok lets go slow

Answer 32

\[ \frac{ \frac{ -3x-7 }{ x-1 }-7 }{ \frac{ -3x-7 }{ x-1 }+3 }\] the \(x-1\) will cancel top and bottom to get
\[\frac{-3x-7-7(x-1)}{-3x-7+3(x-1)}\]

Answer 33

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

Answer 34

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

Answer 35

forgot that distributive law already huh?

Answer 36

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

Answer 37

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

Answer 38

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

Answer 39

oh no

Answer 40

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

Answer 41

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

Answer 42

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"

Answer 43

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?

Answer 44

yay

Answer 45

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

Answer 46

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

Answer 47

let me try to write it

Answer 48

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

Answer 49

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

Answer 50

should have used parentheses around the first term in the top

Answer 51

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

Answer 52

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

Answer 53

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

Answer 54

yes

Answer 55

now get rid of the parentheses

Answer 56

you should of course just be left with some x up top and a number in the bottom that will cancel with it

Answer 57

-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!

Answer 58

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