(In reply to
solution (except for one detail) by Jer)
Starting from where Jer left off. We know there is at least one root of P(x) on any given half-open interval [n, n+1). (n is a positive integer).
On the interval [n, n+1): [x] = n, aka [x] is a constant on the interval. Then the derivative of [x] on [n, n+1) is zero.
Then P'(x) on [n, n+1) is 3x^2 - 2nx. I'll rewrite this as P'(x) = x*(2*(x-n) + x). x is positive and since x>n, then x-n is also positive. Then P'(x) is strictly positive on [n, n+1).
P'(x) being strictly positive then means P(x) can cross the x-axis at most once. Since Jer has shown it does cross at least once we can conclude that P(x) crosses the x-axis exactly once on the interval [n,n+1).
re: solution (finishing the one detail)
