Let $f(x) = x^2-2x$. How many distinct real numbers $c$ satisfy $f(f(f(f(n)))) = 3$?
I have not found any good way to do this problem. I have just resorted to start with brute force:
We start with $$f(f(f(f(x))))=f(f(f(x^2-2x)))=f(f((x^2-2x)^2-2(x^2-2x)))\cdots$$on and on and on, until we finish our composition. However, this will make our problem extremely messy and is very inefficient. Is there any slicker way to start with, and how would I do it. Thanks in advance.
Instead of composing the function on itself 4 times, let's unwrap the given equation one layer at a time.
Which values of $x$ satisfy $f(x)=3$? This is simply a quadratic, we can solve it to see that $f(x)=3$ has roots at $x=3$ and $x=-1$.
Which values of $x$ satisfy $f(f(x))=3$? We must have either $f(x)=3$ or $f(x)=-1$ by the above, so this means that $x\in\{1,-1,3\}$.
Which values of $x$ satisfy $f(f(f(x)))=3$? We must have $f(x)\in\{1,-1,3\}$ which gives $x\in\{1\pm\sqrt2,1,-1,3\}$.
Finally, which values of $x$ satisfy $f(f(f(f(x))))=3$? We know that $f(x)\in\{1\pm\sqrt2,1,-1,3\}$. $f(x)=3$ and $f(x)=1$ give 2 roots each. $f(x)=-1$ gives 1 double root. $f(x)=1+\sqrt2$ and $f(x)=1-\sqrt2$ also give 2 roots each.
Putting it all together, there are 9 distinct real numbers $x$ satisfying $f(f(f(f(x))))=3$.