Let R be the real line. f : R → [-1, 1] is twice differentiable and f(0)2 + f '(0)2 = 4. Show that f(x0) + f ''(x0) = 0 for some x0.
Solution
Let k(x) = f(x)2 + f '(x)2. By the mean value theorem for some a in the interval (0, 2) we have f '(a) = 1/2 (f(2) - f(0) ). But |f(x)| ≤ 1 for all x, so |f '(a)| ≤ 1. Hence k(a) ≤ 1 + 1 = 2. Similarly, we can find b in the interval (-2, 0) with k(b) ≤ 2. We are given that k(0) = 4. Hence k(x) has a maximum at some interior point of (-2, 2). Let this point be c. Then certainly k(c) ≥ k(0) = 4, f(c)2 ≤ 1, so |f '(c)| > 0. We have k '(c) = 0. But k '(c) = 2 f '(c) ( f(c) + f ''(c) ). We have just shown that f '(c) is non-zero, so f(c) + f ''(c) = 0.
© John Scholes
jscholes@kalva.demon.co.uk
23 Jan 2001