Verifying functions are inverses using composition, HELP!

[kristopher0123]

Alright, I was given the following,
f(x) = x-1/x+5
g(x) = -5x-1/x-1

I thought I was supposed to plug in f(g(x) = and plug in -5x-1/x-1 where ever I saw an x in the f(x) equation, but I don't get f(x) = x, but I KNOW it does. Please help!

 

[kristopher0123]

Would this be anywhere near being on the right track? Because that's what I came up with when I put in g(x) for all x's on f(x)

f(g(x)) =
[-2(3x+1)/(x-1)]
________________
[-2(2x+1)/(x-1)]

(sorry if there's formatting issues, the underscore is supposed to indicate top line divided by bottom line)

 

[hongsgirl]

Hi kristopher0123

Would this be anywhere near being on the right track? Because that's what I came up with when I put in g(x) for all x's on f(x)

f(g(x)) =
[-2(3x+1)/(x-1)]
________________
[-2(2x+1)/(x-1)]

(sorry if there's formatting issues, the underscore is supposed to indicate top line divided by bottom line)

You actually will get x. I'm not sure how you got above equation but that certainly won't give you an x.
As you've said, just subs. g(x) for all x in f(x) to find f(g(x)).

$\displaystyle f(g(x))=\frac{\frac{-5x-1}{x-1}-1}{\frac{-5x-1}{x-1}+5}$

Just simplify it

 

[Deleted member 4993]

Alright, I was given the following,
f(x) = x-1/x+5
g(x) = -5x-1/x-1

I thought I was supposed to plug in f(g(x) = and plug in -5x-1/x-1 where ever I saw an x in the f(x) equation, but I don't get f(x) = x, but I KNOW it does.

Your statement is WRONG. You did not state g(x) = f[sup:3971dbpi]-1[/sup:3971dbpi](x) - and in that case f[g(x)] = x [not f(x) = x]

Please help!

First of all, you need to write your problem properly - with grouping symbols (using PEMDAS) as guide.

The way you wrote it - translates to:

$\displaystyle f(x) \, = \, x \, - \, \frac{1}{x} \, + \, 5$

is certainly not correct. You should have written it as:

f(x) = (x-1)/(x+5)

g(x) = (-5x-1)/(x-1)

Assuming my interpretation of your problem is correct - then

$\displaystyle f[g(x)] \, = \, \frac{\frac{-5x - 1}{x-1} - 1}{\frac{-5x-1}{x-1} + \, 5}$

$\displaystyle f[g(x)] \, = \, \frac{\frac{-5x - 1 - x + 1}{x-1}}{\frac{-5x-1 + 5x -5}{x-1}}$

$\displaystyle f[g(x)] \, = \, \frac{\frac{-6x }{x-1}}{\frac{-6}{x-1}}$

$\displaystyle f[g(x)] \, = \, x$

 

[kristopher0123]

Thanks both of you, and yeah, sorry about formatting issues haha, need to get used to writing this stuff on a computer!

 

[kristopher0123]

Ooooookay haha, so I got the first part, now to verify g(f(x)), I'm just as confused. Plugging in the f(x) equation for the x's in the g(x) equation, I get this...

g(f(x)) =

(-5[(x-1)/(x+5)]-1)
_________________
([(x-1)/(x+5)]-1)

I have no clue where to go from there though...haha, I'm so bad at this! :?

 

[Deleted member 4993]

Ooooookay haha, so I got the first part, now to verify g(f(x)), I'm just as confused. Plugging in the f(x) equation for the x's in the g(x) equation, I get this...

g(f(x)) =

(-5[(x-1)/(x+5)]-1)
_________________
([(x-1)/(x+5)]-1)


([(-5x+5)/(x+5)]-1)
_________________
([(x-1)/(x+5)]-1)

and continue....
I have no clue where to go from there though...haha, I'm so bad at this! :?