Composition of a function Fogoh

**M670** · January 13th 2013, 07:51 AM

Find

.

I understand the steps in doing this but this one I am having trouble when getting the to the last part...
I start by writing the h
$h(x)=\sqrt[3]8x$ then I get $\frac{\sqrt[3]8x}{\sqrt[3]8x-1}$
Buts it's adding in the last f(x) I am having trouble?

**Plato** · January 13th 2013, 08:09 AM

Originally Posted by M670

Find

.

I understand the steps in doing this but this one I am having trouble when getting the to the last part...
I start by writing the h
$h(x)=\sqrt[3]8x$ then I get $\frac{\sqrt[3]8x}{\sqrt[3]8x-1}$
Buts it's adding in the last f(x) I am having trouble?

If $A=\frac{\sqrt[3]8x}{\sqrt[3]8x-1}$ then lastly $\sqrt[3]{8A}$

**M670** · January 13th 2013, 08:25 AM

See this is what I thought I would get $\sqrt8$ with $\frac{\sqrt[3]8x}{\sqrt[3]8x-1}$ so something looking like $\sqrt8\frac{\sqrt[3]8x}{\sqrt[3]8x-1}$ only the whole equation under the root
Because I would replace the X of $\sqrt8x$ with $\frac{\sqrt[3]8x}{\sqrt[3]8x-1}$

**Plato** · January 13th 2013, 08:32 AM

Originally Posted by M670

See this is what I thought I would get $\sqrt8$ with $\frac{\sqrt[3]8x}{\sqrt[3]8x-1}$ so something looking like $\sqrt8\frac{\sqrt[3]8x}{\sqrt[3]8x-1}$ only the whole equation under the root
Because I would replace the X of $\sqrt8x$ with $\frac{\sqrt[3]8x}{\sqrt[3]8x-1}$

Within $f(x)$ we replace $x$ with $h\circ g(x)$ .

**M670** · January 13th 2013, 08:38 AM

That's what I did is it not?

**Plato** · January 13th 2013, 09:03 AM

Originally Posted by M670

That's what I did is it not?

NO it is not.

In post #1 it is $f(x)=\sqrt{8x~}\text{ NOT}f(x)=\sqrt{8}~x$

The $x$ is under the radical.

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