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Applications of Differentiation

3.1 Extrema on an Interval

Focus on closed-interval extrema, critical numbers, local versus absolute extrema.

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Ideas in this lesson

01Closed-interval extremaWorked steps · five questions02Critical numbersWorked steps · five questions03Local versus absolute extremaWorked steps · five questions
Idea 1 of 3

Closed-interval extrema

For a continuous function on a closed interval, compare values at endpoints and all interior critical numbers.

Worked example 1

For f(x)=(x−2)2f(x) = (x - 2)^{2} on [0,3][0, 3], find the absolute minimum.
  1. f′(x)=2(x−2)=0  ⟹  x=2f'(x) = 2(x - 2) = 0 \implies x = 2
  2. xx002233
    f(x)f(x)440011
  3. min⁡f=0\min f = 0

Answer: 00

Worked example 2

For f(x)=(x−3)2f(x) = (x - 3)^{2} on [0,4][0, 4], find the absolute maximum.
  1. f′(x)=2(x−3)=0  ⟹  x=3f'(x) = 2(x - 3) = 0 \implies x = 3
  2. xx003344
    f(x)f(x)990011
  3. max⁡f=9\max f = 9

Answer: 99

Idea 2 of 3

Critical numbers

Critical numbers occur within the domain where f′f' is zero or does not exist. A critical number need not be an extremum.

Worked example 1

For f(x)=x3−12xf(x) = x^{3} - 12x, find the positive critical number.
Coordinate graph-3-2-1123-20-101020xy
  1. f′(x)=3(x−2)(x+2)=0  ⟹  x=±2f'(x) = 3(x - 2)(x + 2) = 0 \implies x = \pm 2
  2. f′(x)>0f'(x) > 0 on (−∞,−2)(-\infty , -2) and (2,∞)(2, \infty ); f′(x)<0f'(x) < 0 on (−2,2)(-2, 2)
  3. xx−2-222
    f(x)f(x)1616−16-16
    requested value =2= 2

Answer: 22

Worked example 2

For f(x)=x3−27xf(x) = x^{3} - 27x, find the local minimum value.
Coordinate graph-4-3-2-11234-80-60-40-2020406080xy
  1. f′(x)=3(x−3)(x+3)=0  ⟹  x=±3f'(x) = 3(x - 3)(x + 3) = 0 \implies x = \pm 3
  2. f′(x)>0f'(x) > 0 on (−∞,−3)(-\infty , -3) and (3,∞)(3, \infty ); f′(x)<0f'(x) < 0 on (−3,3)(-3, 3)
  3. xx−3-333
    f(x)f(x)5454−54-54
    requested value =−54= -54

Answer: −54-54

Idea 3 of 3

Local versus absolute extrema

A local comparison concerns nearby inputs; an absolute comparison concerns the entire specified domain.

Worked example 1

For f(x)=(x−2)2f(x) = (x - 2)^{2} on [0,3][0, 3], find the absolute minimum.
  1. f′(x)=2(x−2)=0  ⟹  x=2f'(x) = 2(x - 2) = 0 \implies x = 2
  2. xx002233
    f(x)f(x)440011
  3. min⁡f=0\min f = 0

Answer: 00

Worked example 2

For f(x)=(x−3)2f(x) = (x - 3)^{2} on [0,4][0, 4], find the absolute maximum.
  1. f′(x)=2(x−3)=0  ⟹  x=3f'(x) = 2(x - 3) = 0 \implies x = 3
  2. xx003344
    f(x)f(x)990011
  3. max⁡f=9\max f = 9

Answer: 99

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §3.1,p3.1, p. 162162. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.