Question

Show that f and g are inverse functions.

f(x)=X+3/X-2 g(x)=2x+3/x-1

Answer 1

f(g(x)) = ? , g(f(x)) = ?

Answer 2

this algebra is going to stink, but it will be ok

Answer 3

or you can do algebra as well..

Answer 4

I tried to solve it but for the answer im not getting x. For inverse function you should always get x

Answer 5

\[f(g(x))=f(\frac{2x+3}{x-1})\]
\[=\frac{\frac{2x+3}{x-1}+3}{\frac{2x+3}{x-1}-2}\]
if it is correct, when you simplify this there will be a tone of cancellation and you will get \(x\)

Answer 6

yes, it is correct
http://www.wolframalpha.com/input/?i= \frac{\frac{2x%2B3}{x-1}%2B3}{\frac{2x%2B3}{x-1}-2}

Answer 7

I have this part done...after this, you multiple 3 by x-1

Answer 8

3 would have a denominater 1 so it will be 3/1

Answer 9

multiply top and bottom by \(x-1\) to clear the compound fraction

Answer 10

you can do this as well :
f(x) = (x+3)/(x-2)

let f(x) = y
y(x-2) = x+3
yx -2y = x+3
yx-x = 2y + 3
x = 2y + 3 / y -1 = g(x)

Answer 11

you get
\[\frac{2x+3+3(x-1)}{2x+3-2(x-1)}\]

Answer 12

which will give you \(x\) lickety split

Answer 13

yeh I got

Answer 14

sorry its a little crooked

Answer 15

first step :
\[\frac{ \frac{ 2x+3 }{ x-1 } + 3}{ \frac{ 2x+3 }{ x-1 } -2 }\]

Answer 16

\[\frac{ \frac{ 2x + 3 + 3(x-1) }{ x-1 } }{ \frac{ 2x+3 - 2(x-1) }{ x-1 } }\]

Answer 17

thanks guyz

Answer 18

\[\frac{ 2x + 3 + 3x - 3 }{ 2x + 3 -2x + 2 }\] = x