Extrema

A function f:𝒟→ℝ has an absolute (global) maximum at a point x=ω if

∀x∈𝒟, f(ω) ≥ f(x)

The value f(ω) is an absolute (global) maximum of f.

A function f:𝒟→ℝ has an absolute (global) minimum at a point x=ν if

∀x∈𝒟, f(ν) ≤ f(x)

The value f(ν) is an absolute (global) minimum of f.

Extreme value theorem

If f:𝒟→ℝ is continuous on a closed and bounded interval [𝒶,𝒷] then there exists ω∈ [𝒶,𝒷] such that f(ω) is a global maximum and there exists ν∈ [𝒶,𝒷] such that f(ν) is a global minimum.

A function f:𝒟→ℝ has a local (relative) maximum at x=ω if

∃ℎ∈ ℝ+: ∀x∈ ℬω (ℎ) ∩𝒟 , f(ω) ≥ f(x)

The value f(ω) is a local (relative) maximum of f.

A function f:𝒟→ℝ has a local (relative) minimum at x=ν if

∃ℎ∈ ℝ+: ∀x∈ ℬν (ℎ) ∩𝒟 , f(ν) ≤ f(x)

The value f(ω) is a local (relative) minimum of f.

A critical point of a function f:𝒟→ℝ is an interior point ω∈𝒟 such that

  • either f′(x) =0
  • or f′(x) is not defined at ω.

Geometry of local extrema.

If a function f:𝒟→ℝ has a local extremum at an interior point ω∈𝒟 and f′(ω) exists, then f′(ω) =0

Closed interval method

f(x) continuous on [𝒶,𝒷]

  1. find critical points in (𝒶,𝒷)
  2. compute f(𝒶) and f(𝒷)
  3. the largest and smallest values are global extrema

Explanations and Examples

  • try f1 (x) = x3
  • try f2 (x) = 1 x2+1
  • try f3 (x) =cosx
  • try f4 (x) = { 2-x2 x∈ [0,2) x-3 x∈[2,4]
  • try f5 (x) = x 2-x x∈[0,2)
f1 (x) = 13 x3 - 52 x2 +4x f2 (x) = ( x2 -1 ) 3 f3 (x) = 4x x2+1 f4 (x) = x23 (6-x) 13
f1 (x) = x2 +3x-2 on [1,3] f2 (x) = x2 -3 x23 on [0,2]