Maths Olympiad Prep

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Problem 808

AMC 12 late, AIME early
Algebra Difficulty 4.5 Prove it HMMT November · United States · 2014

Let f(x)=x2+6x+7f(x) = x^{2} + 6x + 7. Determine the smallest possible value of f(f(f(f(x))))f(f(f(f(x)))) over all real numbers xx.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

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Official solution

Solution:

Consider that f(x)=x2+6x+7=(x+3)22f(x) = x^{2} + 6x + 7 = (x+3)^{2} - 2. So f(x)2f(x) \geq -2 for real numbers xx. Also, ff is increasing on the interval [3,)[-3, \infty).

Therefore
f(f(x))f(2)=1,f(f(f(x)))f(1)=2, \begin{gathered} f(f(x)) \geq f(-2) = -1, \\ f(f(f(x))) \geq f(-1) = 2, \end{gathered}
and
f(f(f(f(x))))f(2)=23 f(f(f(f(x)))) \geq f(2) = 23
Thus, the minimum value of f(f(f(f(x))))f(f(f(f(x)))) is 2323 and equality is obtained when x=3x = -3.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.