If $f(x)= 4x(1-x)$, if $x \in [0,1]$. find number of solutions of equation $$f(f(f(x)))={x\over 3}$$
method: i tried finding the compostion but it is too lenghty. also i tried put $x= \sin ^2 a$ but no success.
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You have an excellent idea!
Putting $x=\sin^2a$ we have $$f(\sin^2 a)=\sin^22a\implies f\circ f=\sin^24a\implies f\circ f\circ f=\sin^28a$$ so we need to solve $$3\sin^28a-\sin^2a=0$$
ə̷̶̸͇̘̜́̍͗̂̄︣͟
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Of course you're not going to solve that explicitly, rather use the Intermediate Value Theorem as in Hagen's answer. – Robert Israel Sep 07 '18 at 18:07
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@RobertIsrael Yes, that's what I meant – ə̷̶̸͇̘̜́̍͗̂̄︣͟ Sep 07 '18 at 18:11
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$f(x)$ grows monotonously from $0$ to $1$ and falls back monotonously to $0$. It follows that $f\circ f$ goes $0\to 1\to 0\to1\to 0$ and $f\circ f\circ f$ goes $0\to1\to0\to1\to0\to1\to0\to 1\to 0$. I claim that in each of the $8$ monotonic subintervals, there is exactly one solution of $f(f(f(x)))=\frac x3$. Indeed, the IVT gives us at least one solution for all intervals but the first, and the first has the trivial solution $x=0$. As we have degree $8$, we are done
Hagen von Eitzen
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