# Thread: Composite function

1. ## Composite function

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

2. 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:
$f\!\left(g\!\left(x\right)\right)=x$

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

4. 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.

5. Awesome. Got it. THanks

6. 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?

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