Suppose that f\'(0) exists and f(x+y)=f(x)f(y) for all x and y. Prove that f\' e

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HERE IS f ' (0) = Lim (h->0) [f(h) - f(0)] / h. Claim: f(0) = 1 Proof of claim: f(0) = f(0 + 0) = f(0) * f(0) so divide both sides by f(0). Then f(0) = 1. Then f ' (0) = Lim (h->0) [f(h) - 1] / h First need to show that f ' (x) exists for all x. By definition we need to show that following limit exists. f ' (x) = Lim (h->0) [f(x+h) - f(x)] / h f ' (x) = Lim (h->0) [f(x)f(h) - f(x)] / h **** By the property f(x + y) = f(x) * f(y) f ' (x) = Lim (h->0) [f(x) (f(h) - 1)] / h f ' (x) = f(x) * Lim (h->0) [f(h) - 1] / h f ' (x) = f(x) * f ' (0) By assumption, f(x) exists for all x and f ' (0) exists since f is differentiable at 0. Therefore, f ' (x) exists for all x. Thus, f is differentiable everywhere and f ' (x) = f(x) * f ' (0). Now need to show f(x) = e^(cx) where c = f ' (0). Start with f ' (x) = f(x) * c. In calculus, this looks like a separable differential equation. Let's use the following notation: f ' (x) = dy/dx and f(x) = y. dy/dx = yc dy/dx * 1/y = c INT {1/y} dy = INT {c} dx ln y = cx e^(ln y) = e^(cx) y = e^(cx) Now change y back to f(x) f(x) = e^(cx)

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