Definition D1
f is differentiable at x if the limit lim (h→0) [f(x+h) − f(x)]/h exists, and its value is written f′(x).
Proof.
(1)a − b means a + (−b), and F/h means F · h⁻¹. The difference quotient is defined only for h ≠ 0.convention
(2)If a limit exists, its value is unique. Hence the notation f′(x) is justified.A5
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Lemma L1
If c is a point of an open interval I, f is differentiable at c, and f(c) is the maximum or minimum of f on I, then f′(c) = 0.
Proof.
(1)First suppose f(c) is the maximum. Since I is open, for small |h| we have c + h ∈ I and f(c+h) − f(c) ≤ 0.assumption, convention
(2)Since f is differentiable at c, the limit of the difference quotient exists, and both one-sided limits equal f′(c).assumption, D1, A4
(3)If h > 0, then [f(c+h) − f(c)]/h ≤ 0.(1), A6
(4)Taking the limit as h → 0+ gives f′(c) ≤ 0(2), (3), A2, A3
(5)If h < 0, then [f(c+h) − f(c)]/h ≥ 0.(1), A7
(6)Taking the limit as h → 0− gives f′(c) ≥ 0(2), (5), A2, A3
(7)Therefore f′(c) = 0.(4), (6), A8
(8)If f(c) is the minimum, the inequality in (1) reverses, all the inequalities in (3)–(6) reverse, and the same conclusion follows.(1), A6, A7, A2, A8
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Lemma L2
If f is continuous on [a, b], differentiable on (a, b), and f(a) = f(b), then there exists c ∈ (a, b) with f′(c) = 0.
Proof.
(1)Since f is continuous on the closed interval [a, b], it attains a maximum M and a minimum m.assumption, A1
(2)Case M = m: for every x, m ≤ f(x) ≤ M, so f is constant, and at c ∈ (a, b) the difference quotient is 0 · h⁻¹ = 0, hence f′(c) = 0.(1), A8, A18, A16, A13, A3, D1
(3)Case M ≠ m: since f(a) = f(b), at least one of M and m differs from f(a).(1), assumption
(4)The point attaining that value is neither a nor b, so it lies in (a, b). Call it c.(3), assumption
(5)c is a point of an open interval, f is differentiable at c, and f(c) is a maximum or minimum.(4), assumption
(6)Therefore f′(c) = 0.(5), L1
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Theorem T1
If a < b and f is continuous on [a, b] and differentiable on (a, b), then there exists c ∈ (a, b) with f′(c) = [f(b) − f(a)] / (b − a).
Proof.
(1)Since a < b, we have b − a ≠ 0, and we may set k = [f(b) − f(a)] · (b − a)⁻¹.assumption, A9
(2)Set φ(x) = f(x) − [f(a) + k(x − a)]. Then φ is continuous on [a, b] and differentiable on (a, b).(1), assumption, A11, A12, A10
(3)φ(a) = f(a) − [f(a) + k · 0] = 0.(2), A13, A18, A19
(4)Since k(b − a) = f(b) − f(a), φ(b) = f(b) − [f(a) + (f(b) − f(a))] = 0.(1), (2), A14, A15, A16, A17, A18, A19
(5)Therefore φ(a) = φ(b).(3), (4)
(6)By Rolle's theorem there exists c ∈ (a, b) with φ′(c) = 0.(2), (5), L2
(7)φ′(x) = f′(x) − k(2), A10, A11
(8)Therefore f′(c) = k = [f(b) − f(a)] / (b − a).(1), (6), (7), E, A18, A19
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