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