Definition D1
f is differentiable at c means that the limit lim (h→0) [f(c+h) − f(c)]/h exists, and its value is written f′(c).
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 the limit exists, its value is unique. Therefore the notation f′(c) is justified.A4
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Definition D2
On an open interval I containing c, if there is some δ > 0 with f(c+h) ≤ f(c) for every h with |h| < δ and c + h ∈ I, then c is called a local maximum point of f. Reversing the inequality gives a local minimum point.
Proof.
A local maximum is required only near c. It need not be the maximum of the whole interval.convention
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Lemma L1
If c is a point of an open interval I, f is differentiable at c, and c is a local maximum point of f, then f′(c) ≤ 0.
Proof.
(1)By the meaning of a local maximum, f(c+h) ≤ f(c) for every h with |h| < δ, and therefore f(c+h) − f(c) ≤ 0.assumption, D2, A8
(2)f is differentiable at c, so the limit of the difference quotient exists, and its right limit is also f′(c).assumption, D1, A3
(3)If 0 < h < δ, the numerator is at most 0 and the denominator is positive, so [f(c+h) − f(c)]/h ≤ 0(1), A5
(4)Taking the limit of both sides of (3) as h → 0+ gives f′(c) ≤ 0(2), (3), A1, A2
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Lemma L2
Under the same hypotheses, f′(c) ≥ 0.
Proof.
(1)For h with −δ < h < 0, also |h| < δ, so f(c+h) − f(c) ≤ 0.assumption, D2, A8
(2)f is differentiable at c, so the left limit of the difference quotient is also f′(c).assumption, D1, A3
(3)If h < 0, the numerator is at most 0 and the denominator is negative, so [f(c+h) − f(c)]/h ≥ 0(1), A6
(4)Taking the limit of both sides of (3) as h → 0− gives f′(c) ≥ 0(2), (3), A1, A2
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Theorem T1
If c is a point of an open interval I, f is differentiable at c, and c is a local maximum or local minimum point of f, then f′(c) = 0.
Proof.
(1)f is differentiable at c, so the left and right limits of the difference quotient are the same single number f′(c).assumption, D1, A3, A4
(2)Suppose c is a local maximum point. By L1, f′(c) ≤ 0.assumption, L1
(3)By L2, f′(c) ≥ 0.assumption, L2
(4)(2) and (3) are inequalities about the same single number of (1), so f′(c) = 0.(1), (2), (3), A7
(5)If c is a local minimum point, the inequality of D2 reverses, and the inequalities in (1)(3)(4) of L1 and L2 all reverse, giving f′(c) ≥ 0 and f′(c) ≤ 0. The conclusion is the same.D2, A8, A5, A6, A1, A7
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Corollary C1
The interior-point hypothesis cannot be dropped. Consider f(x) = x on [0, 1]: f attains its maximum at 1, yet f′(1) = 1.
Proof.
(1)If 0 ≤ x ≤ 1, then f(x) = x ≤ 1 = f(1), so f attains its maximum on [0, 1] at 1.convention
(2)If h < 0 and 1 + h ∈ [0, 1], then [f(1+h) − f(1)]/h = h · h⁻¹ = 1(1), A15, A16, A17, A10
(3)The difference quotient is the constant 1, so its left limit is 1.(2), A2, D1
(4)1 ≠ 0, so the conclusion fails. This is because 1 is not an interior point of [0, 1].(3), A19
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Corollary C2
The differentiability hypothesis cannot be dropped either. g(x) = −|x| has a local maximum at 0 but is not differentiable at 0.
Proof.
(1)|t| means t if t ≥ 0 and −t if t < 0. For every t, 0 ≤ |t|, so g(t) ≤ 0 = g(0).convention, A18, A8
(2)Therefore 0 is a local maximum point of g.(1), D2
(3)If h > 0, then [g(0+h) − g(0)]/h = (−h) · h⁻¹ = (−1)(h · h⁻¹) = −1(1), A15, A17, A11, A13, A10, A14
(4)If h < 0, then |h| = −h, so g(h) = −(−h) = h and [g(0+h) − g(0)]/h = h · h⁻¹ = 1(1), A12, A15, A17, A10
(5)Therefore the right derivative is −1, the left derivative is 1, and 1 ≠ −1.(3), (4), A2, A20
(6)Therefore g is not differentiable at 0.(5), A9
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