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^{-x}f(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.

© John Scholes

jscholes@kalva.demon.co.uk

12 Dec 1998