Composite function

**Stroodle** · June 17th 2009, 05:02 PM

Can someone please show me the working for:

$f(x)=\frac{1}{2}\left (\frac{1}{x}+1\right )$

$g(x)=\frac{1}{2x-1}$

Find $f\circ g$

Cheers

**Chris L T521** · June 17th 2009, 05:08 PM

Originally Posted by Stroodle

Can someone please show me the working for:

$f(x)=\frac{1}{2}\left (\frac{1}{x}+1\right )$

$g(x)=\frac{1}{2x-1}$

Find $f\circ g$

Cheers

Note $f\circ g=f\!\left(g\!\left(x\right)\right)=\frac{1}{2}\le ft(\frac{1}{g\!\left(x\right)}+1\right).$

Now substitute what $g\!\left(x\right)$ is into this expression, and simplify. It will work out very nicely. (Check spoiler to see if you're right)

Spoiler:

**Stroodle** · June 17th 2009, 05:10 PM

Thanks. I'm just having trouble with simplifying it though.

**VonNemo19** · June 17th 2009, 05:11 PM

Originally Posted by Stroodle

Can someone please show me the working for:

$f(x)=\frac{1}{2}\left (\frac{1}{x}+1\right )$

$g(x)=\frac{1}{2x-1}$

Find $f\circ g$

Cheers

All this wants is for you to substitute g(x) everywhere you see x in f(x).

$f(x)=\frac{1}{2}\left (\frac{1}{\text{insert g(x) here}}+1\right )$

$f(x)=\frac{1}{2}\left (\frac{1}{\frac{1}{2x-1}}+1\right )=\frac{1}{2}(2x-1+1)=\frac{2x}{2}=x$

remember that $\frac{1}{\frac{1}{2x-1}}=\frac{1}{1}*\frac{2x-1}{1}=2x-1$ . (flip it and multiply)

It appears that these functions are inverses.

**Stroodle** · June 17th 2009, 05:13 PM

Awesome. Got it. THanks

**Chris L T521** · June 17th 2009, 05:15 PM

Originally Posted by Stroodle

Thanks. I'm just having trouble with simplifying it though.

No problem.

Note that when you do the substitution, you have $f\!\left(g\!\left(x\right)\right)=\frac{1}{2}\left (\frac{1}{\frac{1}{2x-1}}+1\right)$ .

Note that $\frac{1}{\frac{1}{2x-1}}=2x-1$ . Thus, we now have $f\!\left(g\!\left(x\right)\right)=\frac{1}{2}\left (2x-1+1\right)$ .

Can you finish the simplification process?

**Stroodle** · June 17th 2009, 05:50 PM

Yeah. I just didn't work out that $\frac{1}{\frac{1}{2x-1}}=2x-1$ , but I understand it now. Thanks

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