Definition D1
If f(p) ≤ f(x) for every x, then f(p) is called the minimum of f and p a minimum point. If f(x) ≤ f(p) for every x, then f(p) is called the maximum and p a maximum point.
Proof.
(1)u ≤ v means u < v or u = v.convention
(2)The minimum is a value below which no x produces a smaller value, and a minimum point is a point actually attaining that value.convention
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Definition D2
If f(p) < f(x) for every x with x ≠ p, then p is called the unique minimum point. The unique maximum point is defined the same way.
Proof.
The definition of a minimum point allows equality; the definition of a unique minimum point requires the inequality to be strict at every other point.convention
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Lemma L1
If f(x) = ax² + bx + c (a ≠ 0) and p = −b/2a, q = c − b²/4a, then for every x, f(x) − f(p) = a(x − p)²
Proof.
(1)x − p means x + (−p).convention
(2)f(x) = a(x − p)² + qA1
(3)p − p = 0, so f(p) = a × 0 + q = 0 + q = q(2), A10, A11, A8, A9
(4)f(x) − f(p) = (a(x−p)² + q) + (−q) = a(x−p)² + (q + (−q))(2), (3), A7
(5)q + (−q) = 0, so f(x) − f(p) = a(x−p)²(4), A10, A9
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Lemma L2
If x ≠ p, then 0 < (x − p)²
Proof.
(1)If x − p = 0, adding p to both sides gives x = p.A7, A8, A9, A10
(2)x ≠ p, so x − p ≠ 0.assumption, (1)
(3)0 < (x − p)²(2), A2
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Theorem T1
If a > 0, then f(p) is the minimum of f, and p is the unique minimum point.
Proof.
(1)Suppose a > 0, and let x ≠ p.assumption
(2)0 < (x − p)²(1), L2
(3)0 < a, so 0 < a(x − p)²(1), (2), A3
(4)a(x − p)² = f(x) − f(p)L1
(5)0 < f(x) − f(p)(3), (4)
(6)Adding f(p) to both sides gives f(p) < f(x)(5), A5, A7, A8, A9, A10
(7)If x = p, then f(p) = f(x).convention
(8)f(p) ≤ f(x) for every x, so f(p) is the minimum and p is a minimum point.(6), (7), D1
(9)The inequality is strict at every x ≠ p, so p is the unique minimum point.(6), D2
(10)If r ≠ p were also a minimum point, then f(r) ≤ f(p), but (6) gives f(p) < f(r). This contradicts trichotomy, so p is the only minimum point.(6), D1, A6
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Corollary C
If a < 0, then f(p) is the maximum of f, and p is the unique maximum point.
Proof.
(1)Suppose a < 0, and let x ≠ p.assumption
(2)0 < (x − p)²(1), L2
(3)a < 0, so a(x − p)² < 0(1), (2), A4
(4)a(x − p)² = f(x) − f(p), so f(x) − f(p) < 0(3), L1
(5)Adding f(p) to both sides gives f(x) < f(p)(4), A5, A7, A8, A9, A10
(6)If x = p, then f(x) = f(p), so f(x) ≤ f(p) for every x, and f(p) is the maximum.(5), convention, D1
(7)The inequality is strict at every x ≠ p, so p is the unique maximum point.(5), D2
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