Putnam 1994

Let R be the reals and R⁺ the positive reals. f : R → R⁺ is differentiable and f '(x) > f(x) for all x. For what k must f(x) exceed e^kx for all sufficiently large k?

Solution

Answer: k ≤ 1.

Let g(x) = e^-xf(x), then g'(x) = e^-x(f '(x) - f(x) ) > 0 for all x, so g is a strictly increasing function. Hence for any h < 0, g(x) > e^hx for all sufficiently large x. Hence if k < 1, then f(x) > e^kx for all sufficiently large x.

The tricky case is obviously k = 1. The argument above suggests that if g(x) keeps below 1, then we will have f(x) < e^x for all x. For example, we can take f(x) = e^{x - e-x}. Then f '(x) = f(x) (1 + e^-x) > f(x), but f(x) < e^kx for all x, provided k ≥ 1.

Putnam 1994

© John Scholes
jscholes@kalva.demon.co.uk
12 Dec 1998