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 ekx for all sufficiently large k?
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) > ehx for all sufficiently large x. Hence if k < 1, then f(x) > ekx 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) < ex for all x. For example, we can take f(x) = ex - e-x. Then f '(x) = f(x) (1 + e-x) > f(x), but f(x) < ekx for all x, provided k ≥ 1.
© John Scholes
12 Dec 1998