If f(x)= sqrt (x), g(x)=x/(x-1), and h(x)=^3sqrt (x) find fogoh.

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- September 2nd 2008, 08:21 PMsjenkinsfind fogoh
If f(x)= sqrt (x), g(x)=x/(x-1), and h(x)=^3sqrt (x) find fogoh.

- September 2nd 2008, 08:25 PMJhevon
- September 2nd 2008, 08:30 PMsjenkins
I don't know, I'm so far behind in my knowledge of functions, here's my guess...

goh=^3sqrt(x)/^3sqrt(x)-1 and then I guess to plug that into f(x) would you just put the whole thing under another square root sign?

I'm sorry, I know I probably sound really stupid with this, but that is the way I feel. I'm just not comprehending this. - September 2nd 2008, 08:37 PMdamadama
theres no reason to feel stupid about it, functions are something that just eventually click and you get it....

lets talk basics

if you have f(x) = ln(x) and g(x) = x^2, then f(g(x)) = ln( (x^2) )

all u basically want to do is plugin g(x) as x in f(x).

so if we have h(x) as sqrt(x) then

f(g(h(x)))) = ln( ((sqrt(x)^2) ) rite? I didn't do your example cuz i wanted you to follow the process. - September 2nd 2008, 08:48 PMsjenkins
I really thought that's what I was doing. I took h(x) because that is in the middle I think that's where you're supposed to start. So h(x)=^3sqrt(x) (by the way, does this mean cubed root of x?)

Anyway, then I substitute g(x) into the h(x) so I get ^3sqrt(x/(x-1)

Then I take f(x) and substitute where the x's are so I would have ^3sqrt(sqrt(x)/(sqrt(x)-1)

Am I even on the right track? - September 3rd 2008, 06:42 AMdamadama
i think you're working backwards, i think that ^3sqrt(x) you are trying to say is cuberoot of x... lets call it (x)^1/3

so g(x) = x / (x - 1), thus g(h(x)) = x^1/3 / (x^1/3 -1)

and f(g(h(x))) is going to be that ^ in a square root .... sqrt ( x^1/3 / .....)