Let P(x) be a polynomial with degree 3, consider the polynomial
Q(x)=(x^{3}2x+1P(x))(2x^{3}5x^{2}+4P(x)).
Assume that Q(x)≤0, ∀x and P(0)=3. Calculate Q(1).
Consider x=0
Q(0) = (1P(0))(4P(0)) = 2 since P(0)=3
Since Q(x) is always nonpositive, P(0) must be between 1 and 4.
Similarly if you plot out
f1(x) = x^32x+1 and
f2(x) = 2x^35x^2+4,
then P(x) must always be between those 2 functions.
One possible P(x) would be the average of f1 and f2, this would assure that P(x) is always between f1 and f2. Try g(x) to be the average of f1 and f2:
g(x) = (3/2)x^3  2.5x^2  x + 2.5
But g(x) doesn't quite work because g(0)=2.5 whereas P(0) = 3.
So let h(x) be a weighted average of f1 and f2 scaled so that
h(0)=3. Since 3 is 2/3 of the way from f1(0)=1 to f2(0)=4, try
h(x) = [f1(x) + 2f2(x)]/3
h(x) = (5/3)x^3  (10/3)x^2  (2/3)x + 3 now has h(0)=3
Now substitute h(x) as one possible valid candidate for P(x)
Q(x) = (f1(x)  h(x))(f2(x)  h(x))
f1(1) = 2
f2(1) = 3
h(1) = 4/3
Q(1) = (2  (4/3))(3  (4/3)) = (10/3)(5/3)
Q(1) = (50/9)
Notice that h(x) is cubic and passes through 3 specific points, (0,3) and the two points of intersection of f1 and f2. I believe that it takes 4 points to determine a unique cubic function, and since we only have 3 points, I am not convinced that h(x) is the only cubic function that P(x) could be.
Also, P(x) cannot be higher order than cubic or Q(x) would go above zero as x gets larger in at least one direction.

Posted by Larry
on 20230916 16:40:26 