# 今年交大應數所高微考古題

f : [a,b] 映至 |R, 在 [a,b] 上連續且在 (a,b) 可微,

設 f 在 c 的導數 ≧ f 在 x 的導數 for all x in (a,b) and some c,

且 f 在 c 附近的某個鄰域上非線性,

證明 f 在 c 的導數 ≠ [f(b) - f(a)]/(b-a) for all a≠b in [a,b]

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### 1 Answer

- EricLv 61 decade agoFavorite Answer
不成立。Counterexample. Define f:[-2,2]→R byf(x) = 1 if x ∈ [-2,-π/2)f(x) = -sin x if x ∈ [-π/2,π/2)f(x) = sin x - 2 if x ∈ [π/2,2]. which is continuous. Furthermore,f ʹ(x) = 0 if x ∈ (-2,-π/2)f ʹ(x) = -cos x if x ∈ [-π/2,π/2)f ʹ(x) = cos x if x ∈ [π/2,2). which is also continuous. f is nonlinear around c = π/2, and we observe that for all x ∈ (-2,2),f ʹ(π/2) = 0 ≥ f ʹ(x),yet [f(-π/2) - f(-2)]/(-π/2+2) = 0. ∎若限制子區間包含 c，則命題成立。Claim. Say f:[a,b]→R is continuous on [a,b] and differentiable on (a,b). Suppose there exists c ∈ (a,b) such that f ʹ(c) ≥ f ʹ(x) for all x ∈ (a,b), and furthermore that f is nonlinear in a neighborhood of c. Then[f(t)-f(s)]/(t-s) ≠ f ʹ(c)whenever s < t and c ∈ [s,t] .Proof. Suppose otherwise, and let s, t be so that s < c < t and[f(t)-f(s)]/(t-s) = f ʹ(c),or equivalently,f(t) = f(s) + f ʹ(c)(t-s),For each x ∈ [s,t], the mean value theorem asserts the existence of ξ ∈ (s,x) and η ∈ (x,t) such that[f(x) - f(s)]/(x-s) = fʹ(ξ) ≤ fʹ(c),[f(t) - f(x)]/(t-x) = fʹ(η) ≤ fʹ(c)hence f(x) ≤ f(s) + f ʹ(c)(x-s)f(x) ≥ f(t) - f ʹ(c)(t-x)= f(s) + f ʹ(c)(t-s) - f ʹ(c)(t-x)= f(s) + f ʹ(c)(x-s),and so f(x) = f(s) + f ʹ(c)(x-s) for all x ∈ [s,t], contradicting the fact that f was nonlinear in some neighborhood of c. ∎

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