Problem:
Let f(x) be a third-degree polynomial with real coefficients satisfying
∣f(1)∣=∣f(2)∣=∣f(3)∣=∣f(5)∣=∣f(6)∣=∣f(7)∣=12.
Find ∣f(0)∣.
Solution:
Without loss of generality assume that f(x) has a positive leading coefficient. Polynomial f(x) has degree 3, so each of f(x)+12 and f(x)−12 has at most three distinct roots. Because 1,2,3,5,6,7 are among these combined roots, each polynomial has precisely three of these as roots. Because f has a local maximum at a point a and a local minimum at a point b where a<b,f increases on the interval (−∞,a), decreases on the interval (a,b), and increases on the interval (b,∞). This shows that f(x) must be equal to −12 at x=1,5, and 6 and equal to 12 at x=2,3, and 7. Thus if f(x) has leading coefficient c, then f(x)−12=c(x−2)(x−3)(x−7), so f(x)=c(x−2)(x−3)(x−7)+12. Similarly f(x)=c(x−1)(x−5)(x−6)−12. Then f(0)=−42c+12=−30c−12 implying that c=2 and ∣f(0)∣=72.
The problems on this page are the property of the MAA's American Mathematics Competitions