Definition D1
f is said to increase on an interval I when f(x₁) < f(x₂) for every two points of I with x₁ < x₂, and to decrease when f(x₂) < f(x₁).
Proof.
Increase and decrease are relations between two points. They are not properties determined at one point.convention
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Theorem T1
If f is continuous on [a, b], differentiable on (a, b), and f′(x) > 0 for every x in (a, b), then f increases on [a, b].
Proof.
(1)Take two points with a ≤ x₁ < x₂ ≤ b. f is continuous on [x₁, x₂] and differentiable on (x₁, x₂).assumption, convention
(2)By the mean value theorem, there is c ∈ (x₁, x₂) with f(x₂) − f(x₁) = f′(c)(x₂ − x₁).(1), A1
(3)c is a point of (a, b), so f′(c) > 0.(1), (2), assumption
(4)x₁ < x₂, so 0 < x₂ − x₁.(1), A6
(5)The product of two positives is positive, so 0 < f′(c)(x₂ − x₁) = f(x₂) − f(x₁)(2), (3), (4), A5
(6)Therefore f(x₁) < f(x₂), and f increases on [a, b].(5), A6, D1
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Theorem T2
Under the same hypotheses, if f′(x) < 0 for every x in (a, b), then f decreases on [a, b].
Proof.
(1)Take a ≤ x₁ < x₂ ≤ b; by the mean value theorem there is c ∈ (x₁, x₂) with f(x₂) − f(x₁) = f′(c)(x₂ − x₁).assumption, A1
(2)f′(c) < 0, so 0 < −f′(c); and x₁ < x₂, so 0 < x₂ − x₁.(1), assumption, A6, A18, A19
(3)Therefore 0 < (−f′(c))(x₂ − x₁) = −[f′(c)(x₂ − x₁)](2), A5, A11
(4)Thus f(x₂) − f(x₁) < 0 and f(x₂) < f(x₁).(1), (3), A6, A17, A18, A19
(5)Therefore f decreases on [a, b].(4), D1
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Corollary C1
If f is continuous on [a, b] and f′(x) = 0 for every x in (a, b), then f is constant on [a, b].
Proof.
(1)Take a ≤ x₁ < x₂ ≤ b; there is c ∈ (x₁, x₂) with f(x₂) − f(x₁) = f′(c)(x₂ − x₁).assumption, A1
(2)f′(c) = 0, so f(x₂) − f(x₁) = 0 · (x₂ − x₁) = 0(1), assumption, A14, A15
(3)Therefore f(x₂) = f(x₁), and the value is the same however the two points are chosen.(2), E, A17, A18, A19
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Corollary C2
The converse of T1 fails. f(x) = x³ increases on ℝ, yet f′(0) = 0.
Proof.
(1)f′(x) = 3x² and f′(0) = 3 · (0 · 0) = 0.A2, A15
(2)If x₁ < x₂, then x₂³ − x₁³ = (x₂ − x₁)(x₂² + x₂x₁ + x₁²).A3
(3)Let S = x₂² + x₂x₁ + x₁²; then 4S = (2x₂ + x₁)² + 3x₁².(2), A4, A12, A13, A14, A16, A17
(4)Case x₁ ≠ 0: 0 < 3x₁² and 0 ≤ (2x₂ + x₁)², so 0 < 4S, and therefore 0 < S.(3), A8, A10, A7, A9
(5)Case x₁ = 0: x₁ < x₂, so x₂ ≠ 0 and S = x₂² > 0.(3), A15, A19, A8
(6)In either case 0 < S and 0 < x₂ − x₁, so 0 < (x₂ − x₁)S = x₂³ − x₁³(2), (4), (5), A6, A5
(7)Therefore x₁³ < x₂³, and f increases on ℝ.(6), A6, D1
(8)The derivative of an increasing function need not be positive at every point.(1), (7)
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