1. ## Functions

I really don't get this...

If anyone could solve the following for me and explain every step it would be great.

If $f(2x-1)=x$ then $f(f(x))=?$

A journey towards my math enlightenment has begun

2. Originally Posted by ikislav
If anyone could solve the following for me and explain every step it would be great.
If $f(2x-1)=x$ then $f(f(x))=?$
From the given $f(x)=f\left(2\left(\frac{x}{2}+\frac{1}{2}\right)-1\right)=\frac{x}{2}+\frac{1}{2}$.
Now YOU finish.

3. Denote $t=2x-1$ then, $f(t)=(t+1)/2$ or equivalently $f(x)=(x+1)/2$ .

Edited: Sorry, I didn't see the previous post.

4. Originally Posted by Plato
From the given $f(x)=f\left(2\left(\frac{x}{2}+\frac{1}{2}\right)-1\right)=\frac{x}{2}+\frac{1}{2}$.
Now YOU finish.
No, I simply don't understand this...
How did you know that $f(x)=f\left(2\left(\frac{x}{2}+\frac{1}{2}\right)-1\right)=\frac{x}{2}+\frac{1}{2}$?

What you showed me has just confused me even more (if that is possible).
Please explain to me step by step, maybe then I will figure it out.

5. Originally Posted by ikislav
Please explain to me step by step, maybe then I will figure it out.
That is not how it works here.
You should understand that this is not a tutorial service.
Perhaps you need to hire a tutor.

Or else you could post some of your own work on this problem or explain what you do not understand about the question.

You are responsible for your own learning.

6. Originally Posted by ikislav
No, I simply don't understand this...
How did you know that $f(x)=f\left(2\left(\frac{x}{2}+\frac{1}{2}\right)-1\right)=\frac{x}{2}+\frac{1}{2}$?

What you showed me has just confused me even more (if that is possible).
Please explain to me step by step, maybe then I will figure it out.
I think FernandoRevilla's post is likely to be the most transparent.

Let $t=2x-1$ then, $\displaystyle x = \frac{1}{2} \cdot (t + 1)$. Then either follow with FernandoRevilla's post or Plato's.

-Dan

7. Originally Posted by Plato
That is not how it works here.
You should understand that this is not a tutorial service.
Perhaps you need to hire a tutor.

Or else you could post some of your own work on this problem or explain what you do not understand about the question.

You are responsible for your own learning.
Ok, I understand.

I made up a simpler example that I tried to solve:

$f(x^2)=x$
$f(f(x))=?$

So, looking at other examples,
I got to $f(x)=\sqrt{x}$
And now I have $f(\sqrt{x})=?$
This is the stage where I completely freeze... I can't continue.

8. Originally Posted by ikislav
I made up a simpler example that I tried to solve:
$f(x^2)=x$
$f(f(x))=?$
So, looking at other examples,
I got to $f(x)=\sqrt{x}$
And now I have $f(\sqrt{x})=?$
This is the stage where I completely freeze... I can't continue.
The original is much easier to work with because $2x-1$ is a bijection.
But let’s modify your example, $f(x^3)=x$.
Then $f(x) = \sqrt[3]{x}$ will work.
Now $f(f(x))=\sqrt[3]{{\sqrt[3]{x}}} = \sqrt[9]{x}$.
That works because $\sqrt[3]{{\sqrt[3]{x}}} = \left( {\left( x \right)^{\frac{1}{3}} } \right)^{\frac{1}{3}}$

9. Originally Posted by Plato
The original is much easier to work with because $2x-1$ is a bijection.
But let’s modify your example, $f(x^3)=x$.
Then $f(x) = \sqrt[3]{x}$ will work.
Now $f(f(x))=\sqrt[3]{{\sqrt[3]{x}}} = \sqrt[9]{x}$.
That works because $\sqrt[3]{{\sqrt[3]{x}}} = \left( {\left( x \right)^{\frac{1}{3}} } \right)^{\frac{1}{3}}$
Ok... so:

$f(2x-1)=x$

$t=2x-1$

$x=\frac{t+1}{2}$

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

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

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

$=\frac{x+1}{2}*\frac{1}{2}+\frac{1}{2}=$

$=\frac{x+1}{4}+\frac{2}{4}=$

$=\frac{x+1+2}{4}=\frac{x+3}{4}$

$f(f(x))=\frac{x+3}{4}$

Is this right?

10. Originally Posted by ikislav
Ok... so:
$f(f(x))=f(\frac{x+1}{2})$
Yes it is correct.

11. Originally Posted by Plato
Yes it is correct.
Woohoo!

Thanks for not making this easy, I feel large ammount of sattisfaction now.