Question

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

Answer 1

f(x)=

Answer 2

g(x)=

Answer 3

I can help with these because these are my favorite things to work on

Answer 4

What I usually do first is put one of the functions in mixed form.

Answer 5

I understand the concept of what im supposed to do, substitute the x and make sure its equals the same thing on both sides, but i get mixed up and end up confusing myself.

Answer 6

Yeah, I know, but if you just plug it in like that, it will make it more difficult to reduce.

Answer 7

Okay, show me how you do it your way then :)

Answer 8

\[f(x) \\= \frac{x - 9}{x + 5} \\= \frac{x + 5 - 14}{x + 5} \\= \frac{x + 5}{x + 5} - \frac{14}{x + 5} \\= 1 - \frac{14}{x + 5}\]

Answer 9

Oh look, now we only have one x

Answer 10

Now we can insert g(x) in to the one x rather than insert g(x) in to two x's

Answer 11

how does it look once you add g(x) in?

Answer 12

using your way.

Answer 13

But let's see if we can make g(x) easier to deal with:

\[g(x) = \frac{-5x - 9}{x - 1} = \frac{-(5x + 9)}{x - 1}\]

Answer 14

I had it completely typed out and I messed up and accidentally erased everythign I typed

Answer 15

Now you can insert g(x) into f(x) and get what you need

Answer 16

Let me demonstrate for you

Answer 17

Im sorry but im confused now. I would appreciate it if you could do everything at once? If not its fine. I know its alot of work

Answer 18

@Hero ?

Answer 19

Sorry, I had to re-start my computer. I will show you

Answer 20

Okay thank you, this is my last problem on my review

Answer 21

Are you there? it wont let me tag you

Answer 22

I'm here. I think I made a mistake with one of these. Hang on

Answer 23

Okay, take your time

Answer 24

Okay, I see my mistake

\[g(x) = -5 - \frac{14}{x-1}\]

Answer 25

And now we can do f(g(x)) and g(f(x)) relatively easy

Answer 26

\[f(g(x)) = 1 - \frac{14}{-\frac{14}{x-1} - 5 + 5} \\= 1 - \frac{14}{-\frac{14}{x-1}} \\= 1 + \frac{14}{\frac{14}{x - 1}} \\= 1 + \left(14 \div \frac{14}{x - 1}\right) \\= 1 + \left(14 \times \frac{x - 1}{14}\right) \\=1 + (x - 1) \\= 1 + x - 1 \\=1 - 1 + x \\ = x\]

Answer 27

I know it doesn't appear that way, but this is the easy way.

Answer 28

You can do the same thing with \(g(f(x))\)

Answer 29

I only need to prove that one side equals x because then the other side obviously does as well. Thank you so much, im going to look over the way you do it now

Answer 30

I should have color coded the g(x) I inserted into f(x)

Answer 31

Basically

\(f(x) = 1 - \dfrac{14}{x + 5}\)

\(g(x) = -5 - \dfrac{14}{x - 1}\)

Answer 32

When you have fractional functions of this type you only want to insert one into the other ONCE.

Answer 33

If you have any questions, NOW is the time to ask.

Answer 34

I suppose you understand it now?