If $f(x)=x^3-12x^2+Ax+B>0$

$f(f(f(3)))=3$, $f(f(f(f(4))))=4$

then what is the value of $f(7)$


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  • 1
    $\begingroup$ What did you try? $\endgroup$ – Mark Apr 23 at 19:44
  • $\begingroup$ I confused, how to solve there is a function of functions four times. with two unknown variables A and B. $\endgroup$ – Pankaj Solanki Apr 23 at 19:52
  • $\begingroup$ $f(3)$ and $f(4)$ are expressions in $A$ and $B$. You plug them in $f(x)$ and the result again inside $f(x)$, and again, and again the necessary number of times, and then equate to $3$ and $4$ respectively. It gives a system of two equations in the two unknowns $A,B$. Solving for $A,B$ tells you the explicit form of $f(x)$. Then you can evaluate it at $x=7$. $\endgroup$ – user647486 Apr 23 at 20:23
  • $\begingroup$ The strange part is the $>0$ at the beginning. What does that mean? For every $x$ one must have $f(x)>0$? If that is the case, then there are no solutions for all the conditions, since $f$ is a cubic polynomial, and therefore must take negative values. It would be nice if this is the case, that way you don't have to get your hands dirty with the system of equations. $\endgroup$ – user647486 Apr 23 at 20:24

$3, 4$ are orbits of $f$ of order $3, 4.$ You ask for "the value" of $f(7),$ so we may simplify by assuming that $3, 4$ are actually fixed points of $f.$

We get $3=f(3) = 3A+B-81, 4 = f(4) = 4A+B-128,$ which is a system of linear equations with solution $A = 48, B = -60.$

Finally, $f(7) = -245 + 7A + B = \boxed{31.}$

  • $\begingroup$ I wonder whether the answer is always 31 (if you plug in the expression for $f^{3}(3), f^{4}(4)$ naively, you'll get a very high order system of equations). Rigorously proving that $f(7)$ is unique would be on another level. $\endgroup$ – Display name Apr 24 at 5:49

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