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Algebra Difficulty 4.8 AIME Find the answer

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

A number or a short expression. Spacing and $ signs are ignored.

Solution

Consider that f(x)=x2+6x+7=(x+3)22f(x)=x^{2}+6 x+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)=1f(f(x)) \geq f(-2)=-1 f(f(f(x)))f(1)=2f(f(f(x))) \geq f(-1)=2 and f(f(f(f(x))))f(2)=23f(f(f(f(x)))) \geq f(2)=23 Thus, the minimum value of f(f(f(f(x))))f(f(f(f(x)))) is 23 and equality is obtained when x=3x=-3.

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