Ngô Quốc Anh

September 7, 2007

Bài tập nhỏ về bất đẳng thức đạo hàm

Filed under: Các Bài Tập Nhỏ — Ngô Quốc Anh @ 19:21

Let f be a differentiable real valued function with f(0) = 0 which satisfies f'(x) > f(x) for all x\in\mathbb{R}. Prove that f(x) > 0 for all x > 0.

Solution. We have

(e^{-x}f)' = e^{-x}f'-e^{-x}f > 0.

Thus e^{-x}f is strictly increasing. In particular, e^{-x}f(x) > e^{-0}f(0) = 0 for all x > 0. Thus f(x) > 0 for all x > 0.

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