Uncertain about composite functions.

[Biojar]

So I'm not so great at mathematics and also I am ashamed.

Let f(x) = (1/x+1)
and g(x) = (x/x-1)

Help me find f(g(f(x)))? f(g(x+h))?

Thank you, escape artists.

 

[ggroy]

Assuming g(x) = (x/x-1) is not a misprint, g(x) = 0 for x != 0.

So g(f(x)) = 0 for all f(x) != 0.

f(x) = 0, when x = -1.

Hence f(g(f(x))) = (1/0 + 1) = infinity for all x != -1.

 

[Deset Gled]

Assuming g(x) = (x/x-1) is not a misprint

I'm going to assume it was a misprint, and what Biojar really meant was:

Let f(x) = 1/(x+1)
and g(x) = x/(x-1)

Then g(f(x) = [1/(x+1)]/[1/(x+1) + 1]
=(x+1)^-1 * [(x+1)^-1 - 1]^-1
=[(x+1)((x+1)^-1 - 1)]^-1
=[1-(x+1)]^-1
=-1/x
For x=!-1, x=!0.

Then f(-1/x) = [-x^-1 - 1]^-1
=-1/(1/x + 1)
=x/(x+1)
For x=!-1, x=!0.

Assuming I didn't make any typo's, or course. It's late. Too late to be doing someone else's math homework

 

[DumbPaladin]

But see, my question is ... you posted on a D&D board.

HOW did you know there would be some people who would be able to understand, and correctly answer, your question!?

 

[Mustrum_Ridcully]

But see, my question is ... you posted on a D&D board.

HOW did you know there would be some people who would be able to understand, and correctly answer, your question!?

How could you not know?

 

[DumbPaladin]

Well, I'll tell you my math, even though I am admittedly terrible at math.

D&D player = smart
but D&D player =/= geek
Geek =/= good at math
Smart =/= good at math

Ergo ... there was a good chance that a board full of D&D players would all be bad at math.

I will admit to being a geek, but math is NOT an area of expertise. I survived college-level algebra and will never go there again. (I got straight A's in geometry, though.)

Anyway, to the OP, I'm glad you found some help ... I'm just surprised you would think to seek it here.

 

[Kaledor]

...

Then f(-1/x) = [-x^-1 - 1]^-1
=-1/(1/x + 1)
=x/(x+1)
For x=!-1, x=!0.
...

hmmm. Not being critical, but you've got some misplaced negatives in your work (easy to do when typing math problems on messageboards ).
Assuming g(x) = x/(x-1) and f(x) = 1/(x+1)

Your first line should be:
Then g(f(x) = [1/(x+1)]/[1/(x+1) - 1], not "+". You end up fixing it in your next line, so your final answer for g(f(x)) is still correct

(-1/x)

But in the first line of your second portion that sneaky negative comes back.
Shouldn't f(-1/x) be [-x^-1 + 1]^-1 , not "-" in this case.

I didn't use the negative exponents and instead wrote it as:
1 / [ (-1/x) + 1 ]
= 1 / [ (-1/x) + (x/x) ]
= 1 / [ (-1 + x) /x ]
= x/ (-1 + x)

or


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

(which can be confirmed graphically and by evaluating select points).


As for f(g(x+h))

g(x+h) = (x+h)/(x + h -1)

Then f(g(x+h)) = 1/[ (x+h)/(x + h -1) +1 ]
=1 / [ (x+h)/(x + h -1) + (x + h - 1)/(x + h -1) ]
=1 / [ (2x + 2h - 1)/(x + h -1) ]
= (x + h -1) / (2x + 2h - 1)

And it's too late for me to think of anything fancier to do with this answer... at first glance I don't see it reducing or simplifying any more. It's a curious problem, seems like there's a calculus extension being asked (using the definition of derivatives and limits). Good times

 

[Kaledor]

Uggg... And now I notice that the OP posted yesterday and my work is probably too late to help. Man, I've been lurking for years and the first thread that I get to post in with math and I'm a day late. lol. Figures.

I see math and I want to do it (or correct it... sorry)

Ah well.

 

[Deset Gled]

hmmm. Not being critical, but you've got some misplaced negatives in your work (easy to do when typing math problems on messageboards )...

You appear to be correct.

I hate writing out complicated rational equations in in-line text. Switching to the exponential notation was a quick method of trying to simplify it a bit. Obviously, it was only somewhat successful.

 

[Kaledor]

double post ?