Let f(x)=4x(1-x), 0<=x<=1. The number of solutions of f(f(f(f(x))=x/5 is

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f : [0, 1] \to [0, 1]

f (x) = \frac{1 - x}{1 + x}, 0 \leq x \leq 1

g : [0, 1] \to [0, 1]

g (x) = 4 x (1 - x), 0 \leq x \leq 1

f (x) = ⎡ ⎢ ⎢ ⎣ \begin{matrix} \frac{1}{4} (3 x^{2} + 1) - \infty ⟨ x \leq 1 5 - 4 x 1 ⟨ x ⟨ 4 4 - x 4 \leq x ⟨ \infty \end{matrix}

f (x) = ∣ ∣ {sin}^{- 1} (sin x) ∣ ∣ - (\frac{π - x}{2})

f (x) = 0

x \in [- π, π]

f (x) = \frac{1}{1 - x}

f (x) = 1 - (1 - x)^{2}, x \leq 1

f {f (x)} = x

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