Question

I'm having problem confirming they are inverses can someone please help me and explain each step?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

Okay. Have you computed f(g(x)) yet?

Answer 5

Yes and i got stuck at the bottom

Answer 6

i got up to here:

Answer 7

\[\frac{ -3x ^{2}-11x+14 }{ -3x ^{2}-x+4 }\]

Answer 8

I would recheck your computation.

Answer 9

what is g(0) and f(0)?

Answer 10

@amistre64 im not sure what you're asking
@RBauer4 okay give me a sec

Answer 11

@RBauer4 i don't know what i did wrong

Answer 12

was just wondering something, let x=0

prolly confusing reciprocal for inverse

an easy check is to take say f(x) and swap out x for y, and solve for y to see if we get to g(x)

Answer 13

Can you set it up please. I don't understand what you mean

Answer 14

\[y=f(x)=\frac{ x-7 }{ x+3 }\]
\[x=f(y)=\frac{y-7 }{y+3 }\]
see if f(y) = g(x) by solving for y

\[x=\frac{y-7 }{y+3 }\]
\[x(y+3)=y-7\]
\[yx+3x=y-7\]
\[yx-y=-3x-7\]
\[y(x-1)=-3x-7\]
\[y=\frac{-3x-7}{x-1}\]

Answer 15

@amistre64 I understand now, but I have to plug it in the other way because I have to show the thought process

Answer 16

Note, the question does state to show f(g(x)) = x and that g(f(x)) = x.

Answer 17

yeah :) but this was just an idea .... the rest is just algebra methods

Answer 18

actually, this method shows that f(f(y)) = f(g(x)) = x in a longer format prolly

Answer 19

Certainly, and I feel that the method of computing an inverse function is more meaningful than showing f(g(x)) = x and g(f(x)) = x; but, that's just my opinion.

Answer 20

Can either of you help me doing it the other way... PLEASE

Answer 21

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

Answer 22

Multiply -7 by (x-1)/(x-1) and 3 by (x-1)/(x-1)

Answer 23

or, mutliply top and bottom by (x-1) to clear out the fractiony looking messes :)

Answer 24

i