Composition of a function Fogoh

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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

$\displaystyle h(x)=\sqrt[3]8x$ then I get $\displaystyle \frac{\sqrt[3]8x}{\sqrt[3]8x-1}$

Buts it's adding in the last f(x) I am having trouble?

Re: Composition of a function Fogoh

Quote:

Originally Posted by

**M670**

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

Re: Composition of a function Fogoh

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

Because I would replace the X of $\displaystyle \sqrt8x$ with $\displaystyle \frac{\sqrt[3]8x}{\sqrt[3]8x-1}$

Re: Composition of a function Fogoh

Quote:

Originally Posted by

**M670** See this is what I thought I would get $\displaystyle \sqrt8$ with $\displaystyle \frac{\sqrt[3]8x}{\sqrt[3]8x-1}$ so something looking like $\displaystyle \sqrt8\frac{\sqrt[3]8x}{\sqrt[3]8x-1}$ only the whole equation under the root

Because I would replace the X of $\displaystyle \sqrt8x$ with $\displaystyle \frac{\sqrt[3]8x}{\sqrt[3]8x-1}$

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

Re: Composition of a function Fogoh

That's what I did is it not?

Re: Composition of a function Fogoh

Quote:

Originally Posted by

**M670** That's what I did is it not?

**NO it is not.**

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

The $\displaystyle x$ is under the radical.