Question

ignore

Answer 1

\(\sf f(x)=\dfrac{x-7}{x+2}\)

\(\sf g(x)=\dfrac{-2x-7}{x-1}\)

Answer 2

\(\sf f(x)=\dfrac{\dfrac{-2x-7}{x-1}-7}{\dfrac{-2x-7}{x-1}+2}\)

Answer 3

\(\sf f(x)=({\dfrac{-2x-7}{x-1}-7})*({\dfrac{-2x-7}{x-1}+2})\)

Answer 4

@ganeshie8 , can you check my work so far and help me? I got stuck :/

Answer 5

I have to prove that they are inverses by doing f(g(x)) and g(f(x))

Answer 6

Ah, I just realized I messed up on the last part, actually on most of it :/

Answer 7

\(\sf f(x)=({\dfrac{-2x-7}{x-1}-7})*({\dfrac{x-1}{-2x-7}+2})\)

Answer 8

Okie do you mean you're given
\[\sf f(x)=\dfrac{x-7}{x+2}\]
\[\sf g(x)=\dfrac{-2x-7}{x-1}\]

and you're trying to find :
\[\sf f(g(x))=\dfrac{\dfrac{-2x-7}{x-1}-7}{\dfrac{-2x-7}{x-1}+2}\]
?

Answer 9

Yes, then I also have to do g(f(x))

Answer 10

got you :)

Answer 11

To prove that the functions are inverses of each other.

Answer 12

simplify f(g(x)) first and show that it gives you "x" back

Answer 13

\[\sf f(g(x))=\dfrac{\dfrac{-2x-7}{x-1}-7}{\dfrac{-2x-7}{x-1}+2}\]


multiply top and bottom by \(x-1\) so that the fractions disappear :
\[\sf f(g(x))=\dfrac{-2x-7-7(x-1)}{-2x-7+2(x-1)}\]

Answer 14

simplify and see if you really get x

Answer 15

With that, couldn't I cancel everything other than the -7 and +2?

Answer 16

thats not that obvious yet
start by simplifying numerator and denominator separately first

Answer 17

\[\sf f(g(x))=\dfrac{\dfrac{-2x-7}{x-1}-7}{\dfrac{-2x-7}{x-1}+2}\]


multiply top and bottom by \(x-1\) so that the fractions disappear :
\[\sf f(g(x))=\dfrac{-2x-7-7(x-1)}{-2x-7+2(x-1)}\]

\[\sf f(g(x))=\dfrac{-2x-7-7x+7}{-2x-7+2x-2}\]

Answer 18

Mmmm... ok
\(\sf f(g(x))=\dfrac{-2x-7-7(x-1)}{-2x-7+2(x-1)}=\dfrac{-2x-7-7x-7}{-2x-7+2x-2}\)

Answer 19

\(\sf -9x-14\)

Answer 20

no, there is a sign mistake on numerator
check once

Answer 21

in ur second to last reply

Answer 22

Yes, I see now :) \(\sf f(g(x))=\dfrac{-2x-7-7(x-1)}{-2x-7+2(x-1)}=\dfrac{-2x-7-7x+7}{-2x-7+2x-2}\)

Answer 23

So numerator is just -9x

Answer 24

Then denominator would be \(\sf -9\) then \(\sf\dfrac{-9x}{-9}=x\)

Answer 25

Perfect!
so we're done wid f(g(x))

Answer 26

start g(f(x)) now

Answer 27

\(\sf g(f(x))=\dfrac{-2\dfrac{x-7}{x+2}-7}{\dfrac{x-7}{x+2}-1}\)

Answer 28

Yes! same story :)
start by multiplying x+2 top and bottom so that fractions disappear

Answer 29

\[\sf g(f(x))=\dfrac{-2\dfrac{x-7}{x+2}-7}{\dfrac{x-7}{x+2}-1}\]

\[\sf g(f(x))=\dfrac{-2(x-7)-7(x+2)}{x-7-1(x+2)}\]

Answer 30

\(\sf g(f(x))=\dfrac{-2x+14-7x-14}{x-7-1x-2}=\dfrac{-9x}{-9}=x\)

Answer 31

Looks good! since g(f(x)) = f(g(x)) = x, the given functions are inverses of each other

Answer 32

Yay! Thank you! :D

Answer 33

yw:)