Let f(x)=x2+6x+7. Determine the smallest possible value of f(f(f(f(x)))) over all real numbers x.
A number or a short expression. Spacing and $ signs are ignored.
Solution
Consider that f(x)=x2+6x+7=(x+3)2−2. So f(x)≥−2 for real numbers x. Also, f is increasing on the interval [−3,∞). Therefore f(f(x))≥f(−2)=−1f(f(f(x)))≥f(−1)=2 and f(f(f(f(x))))≥f(2)=23 Thus, the minimum value of f(f(f(f(x)))) is 23 and equality is obtained when x=−3.
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