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This leads to a high-degree polynomial equation. Instead, consider symmetry and degree.
Note that \( f(x) \) is odd and rational of degree asymptotically 1 (since \( \deg N - \deg D = 3 - 2 = 1 \)), so \( f(f(x)) \) is a composition of rational functions of degrees 1 and 1, but numerator degree â 3 + 2 = 5? Actually, degree of numerator in \( f(f(x)) \) is \( \deg(N^3) = 9 \), and denominator degree \( D(N^2 + D^2) \) is degree \( 2(5) = 10 \)? Wait, more carefully:
After clearing denominators, the equation \( f(f(x)) = x \) becomes a rational function equation, and the degree of the resulting polynomial can be estimated.
Since \( f(x) \) is a rational function of degree \( \max(\deg ext{num}, \deg ext{den}) = 3 \), then \( f(f(x)) \) has degree at most \( 3 imes 3 = 9 \). So \( f(f(x)) - x = 0 \) is a rational equation whose numerator is a polynomial of degree at most 9.
But since \( f(f(x)) \) is odd (as composition of odd functions), and \( x \) is odd, the equation \( f(f(x)) = x \) is equivalent to \( g(x) = 0 \) where \( g \) is a degree ⤠9 polynomial, odd.
Moreover, we know \( x = 0 \) is a solution.
Because \( f \) is odd and smooth, and \( f(f(x)) - x \) is odd, all solutions come in pairs \( \pm x \), except possibly \( x = 0 \).
Letâs check degree: the leading term of \( f(x) \) is \( x \), so \( f(f(x)) \) has leading term \( f(x) o x \), so \( f(f(x)) o x \), but as a rational function, \( f(f(x)) = x + o(1) \), but algebraically, the numerator leads to degree 9.
In fact, it is known in functional iterations that such rational functions of degree ⥠2 can have up to \( 2n \) solutions for \( f^n(x) = x \), but here we are solving \( f(f(x)) = x \), so up to 9 solutions (since numerator degree ⤠9, odd).
But due to odd symmetry, solutions are symmetric about origin, so if \( x \) is a solution, so is \( -x \), and \( 0 \) if odd.