1. ## find fogoh

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

2. Originally Posted by sjenkins
If f(x)= sqrt (x), g(x)=x/(x-1), and h(x)=^3sqrt (x) find fogoh.
hint: $\displaystyle (f \circ g \circ h)(x) = f(g(h(x)))$

that is, find $\displaystyle g(h(x))$ and plug it into $\displaystyle f(x)$

can you continue?

3. 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.

4. 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.

5. 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?

6. 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 / .....)

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# f o g o h example

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