The real polynomial x3 + px2 + qx + r has real roots a ≤ b ≤ c. Prove that f ' has a root in the interval [b/2 + c/2, b/3 + 2c/3]. What can we say about f if the root is at one of the endpoints?
Solution
p(x) = (x - a)(x - b)(x - c), so p'(x) = (x - a)(x - b) + (x - b)(x - c) + (x - a)(x - c).
We can write p'(x) = (x - b)(x - c) + (x - a)(2x - b - c), so p'(b/2 + c/2) = -1/4 (c - b)2 ≤ 0, with equality iff b = c.
p'(b/3 + 2c/3) = -2/9 (c - b)2 + (x - a) 1/3 (c - b) = 1/3 (c - b) (b - a) ≥ 0 with equality iff a = b or b = c.
If b = c, then b is a repeated root and p'(b) = 0. If a = b, then p'(b/3 + 2c/3) = 0. Otherwise, p'(x) is negative at b/2 + c/2 and positive at b/3 + 2c/3, so it has a zero in the interior of the interval.
© John Scholes
jscholes@kalva.demon.co.uk
5 Mar 2002