I really don't get this...
If anyone could solve the following for me and explain every step it would be great.
If $\displaystyle f(2x-1)=x$ then $\displaystyle f(f(x))=?$
A journey towards my math enlightenment has begun
No, I simply don't understand this...
How did you know that $\displaystyle f(x)=f\left(2\left(\frac{x}{2}+\frac{1}{2}\right)-1\right)=\frac{x}{2}+\frac{1}{2}$?
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.
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:
$\displaystyle f(x^2)=x$
$\displaystyle f(f(x))=?$
So, looking at other examples,
I got to $\displaystyle f(x)=\sqrt{x}$
And now I have $\displaystyle f(\sqrt{x})=?$
This is the stage where I completely freeze... I can't continue.
Actually your ‘made up’ has several difficulties with it.
The original is much easier to work with because $\displaystyle 2x-1$ is a bijection.
But let’s modify your example, $\displaystyle f(x^3)=x$.
Then $\displaystyle f(x) = \sqrt[3]{x}$ will work.
Now $\displaystyle f(f(x))=\sqrt[3]{{\sqrt[3]{x}}} = \sqrt[9]{x}$.
That works because $\displaystyle \sqrt[3]{{\sqrt[3]{x}}} = \left( {\left( x \right)^{\frac{1}{3}} } \right)^{\frac{1}{3}} $
Ok... so:
$\displaystyle f(2x-1)=x$
$\displaystyle t=2x-1$
$\displaystyle x=\frac{t+1}{2}$
$\displaystyle f(x)=\frac{x+1}{2}$
$\displaystyle f(f(x))=f(\frac{x+1}{2})$
$\displaystyle f(\frac{x+1}{2}) = \frac{\frac{x+1}{2}}{2}+\frac{1}{2}=$
$\displaystyle =\frac{x+1}{2}*\frac{1}{2}+\frac{1}{2}=$
$\displaystyle =\frac{x+1}{4}+\frac{2}{4}=$
$\displaystyle =\frac{x+1+2}{4}=\frac{x+3}{4}$
$\displaystyle f(f(x))=\frac{x+3}{4}$
Is this right?