Unit 5 · Topic 5.2 Beta

Extreme Value Theorem, Global and Local Extrema, Critical Points

A function continuous on a closed interval reaches an absolute maximum and minimum there (Extreme Value Theorem).

Practice 2: Connecting RepresentationsPractice 3: Justification

Question set for this topic

Part 1 · Hook

Why this matters

A hiker walks a trail from the trailhead to a lookout. Somewhere on the way she is at her highest elevation and somewhere at her lowest. That sounds obvious, but it needs the trail to be unbroken. Calculus makes the idea exact and then shows where to look: at the ends of the trail and at points where the slope is zero or does not exist.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. ff is continuous on [a,b][a, b] when

  1. it is continuous at each point of (a,b)(a, b) and one-sided continuous at aa and bb
  2. it is defined at aa and bb
  3. it is differentiable at each point of [a,b][a, b]
  4. f(a)=f(b)f(a) = f(b)
Show the answer

Continuity on a closed interval: no breaks inside, and the one-sided limits at the ends match the endpoint values.

  • Correct: it is continuous at each point of (a,b)(a, b) and one-sided continuous at aa and bb:
  • it is defined at aa and bb:
  • it is differentiable at each point of [a,b][a, b]:
  • f(a)=f(b)f(a) = f(b):

2. For f(x)=|x|f(x) = |x|, what is f′(0)f'(0)?

  1. It does not exist
  2. 00
  3. 11
  4. −1-1
Show the answer

The graph has a corner at 0: slopes −1-1 on the left and 11 on the right.

  • Correct: It does not exist:
  • 00:
  • 11:
  • −1-1:

3. The Mean Value Theorem needs ff

  1. continuous on [a,b][a, b] and differentiable on (a,b)(a, b)
  2. differentiable on [a,b][a, b] and f(a)=f(b)f(a) = f(b)
  3. continuous on (a,b)(a, b) only
  4. increasing on [a,b][a, b]
Show the answer

Both hypotheses, every time.

  • Correct: continuous on [a,b][a, b] and differentiable on (a,b)(a, b):
  • differentiable on [a,b][a, b] and f(a)=f(b)f(a) = f(b):
  • continuous on (a,b)(a, b) only:
  • increasing on [a,b][a, b]:

Part 4 · See it

See it first

Left: f(x) = x cubed minus 3x on the closed interval from -1.8 to 2.3, with a relative maximum at (-1, 2), a relative and absolute minimum at (1, -2), and the absolute maximum, about 5.27, at the right endpoint. Right: g(x) = x for 0 ≤ x < 2 with g(2) = 0; the graph climbs toward an open circle at (2, 2) and never reaches 2, so g has no maximum value.
Left: continuous on a closed interval, so both extremes are reached; the absolute maximum is at an endpoint. Right: with a jump, the values climb toward 2 but never reach it. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. ff is continuous on a closed interval [a,b][a, b]the Extreme Value Theorem guarantees an absolute maximum value and an absolute minimum value on [a,b][a, b]
  2. An absolute extremum occurs inside the intervalit is also a relative extremum, where the tangent line is flat or does not exist
  3. So it happens where f′(x)=0f'(x) = 0 or f′(x)f'(x) does not existthose interior points, the critical points, plus the endpoints, are the only places to look
  4. A critical point is only a candidatesome critical points (x3x^3 at 0, a corner on a rising graph) are not extrema at all

Part 6 · Key ideas

Key ideas

  • Extreme Value Theorem: if ff is continuous on [a,b][a, b], then ff has at least one absolute maximum value and at least one absolute minimum value on [a,b][a, b].
  • Absolute (global) maximum: f(c)≥f(x)f(c) \ge f(x) for every xx in the domain or interval. Relative (local) maximum: f(c)≥f(x)f(c) \ge f(x) for every xx near cc, on an open interval around cc. Minimums the same way.
  • Critical point: an interior point cc of the domain where f′(c)=0f'(c) = 0 or f′(c)f'(c) does not exist.
  • Relative extrema happen only at critical points. Absolute extrema on [a,b][a, b] happen at critical points or at endpoints.
  • If a hypothesis fails (a jump, an open interval), the theorem promises nothing; extrema may or may not exist.

Part 7 · Misconception

A common mistake

The wrong idea: If f′(c)=0f'(c) = 0, then ff has a maximum or minimum at cc.

What actually happens: f′(c)=0f'(c) = 0 only makes cc a critical point, a candidate. For f(x)=x3f(x) = x^3, f′(0)=0f'(0) = 0 but the graph rises straight through. A corner can also be a critical point without an extremum. You must check what ff does around cc.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of ff on [−2.5,4][-2.5, 4]

f(x)=(x+1)2−2f(x) = (x + 1)^2 - 2 for −2.5≤x≤1-2.5 \le x \le 1 and f(x)f(x)=2+(x−1)−(x−1)22= 2 + (x - 1) - \frac{(x - 1)^2}{2} for 1<x≤41 < x \le 4. The pieces meet at (1,2)(1, 2), where the slope jumps from 4 to 1.
−3−2−11234−2−1123xyf(1, 2)

1. How many critical points does ff have in the open interval (−2.5,4)(-2.5, 4)?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

f′(−1)=0f'(-1) = 0 and f′(2)=0f'(2) = 0 (flat tangent lines), and f′(1)f'(1) does not exist (corner). Three critical points: x=−1x = -1, 11, 22.

  • Answer: 3

2. Which statement about ff at x=1x = 1 is true?

  1. ff has a relative maximum at x=1x = 1, because the graph has a corner there
  2. x=1x = 1 is not a critical point, because ff is continuous there
  3. ff has a relative minimum at x=1x = 1, because f′(1)f'(1) does not exist
  4. x=1x = 1 is a critical point, but f(1)f(1) is not a relative extremum
Show the answer

A critical point is only a candidate. At x=1x = 1, f′f' does not exist, but ff is rising on both sides (slopes near 4 on the left, near 1 on the right), so f(1)=2f(1) = 2 is neither a relative maximum nor a relative minimum.

  • ff has a relative maximum at x=1x = 1, because the graph has a corner there: A corner can be a peak, but this one is not: the graph keeps rising through it.
  • x=1x = 1 is not a critical point, because ff is continuous there: Continuity does not matter: f′(1)f'(1) fails to exist, which makes x=1x = 1 a critical point.
  • ff has a relative minimum at x=1x = 1, because f′(1)f'(1) does not exist: An undefined derivative makes a critical point, not an extremum. Values just left of 1 are smaller than f(1)f(1), values just right are larger.
  • Correct: x=1x = 1 is a critical point, but f(1)f(1) is not a relative extremum: f′(1)f'(1) does not exist, so it is a critical point; but ff rises on both sides of x=1x = 1, so there is no peak or valley.

3. What is the absolute minimum value of ff on [−2.5,4][-2.5, 4]?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

The lowest point of the graph is (−1,−2)(-1, -2), lower than both endpoint values f(−2.5)=0.25f(-2.5) = 0.25 and f(4)=0.5f(4) = 0.5. The absolute minimum value is −2-2.

  • Answer: -2

4. At which values of xx does ff have a relative extremum? Select each one.

  1. x=−1x = -1
  2. x=1x = 1
  3. x=2x = 2
  4. x=3x = 3
Show the answer

Relative minimum at x=−1x = -1 (value −2-2) and relative maximum at x=2x = 2 (value 2.52.5). The corner at x=1x = 1 is a critical point without an extremum.

  • Correct: x=−1x = -1: A valley: ff falls, then rises. Relative minimum.
  • x=1x = 1: The graph rises through the corner; no peak or valley.
  • Correct: x=2x = 2: A peak: ff rises, then falls. Relative maximum.
  • x=3x = 3: At x=3x = 3 the graph is falling; nothing special happens.

5. What is the absolute maximum of ff on [−2.5,4][-2.5, 4], and where does it occur?

  1. 2.52.5, at x=2x = 2
  2. 22, at the corner x=1x = 1
  3. 0.50.5, at the right endpoint x=4x = 4
  4. No maximum exists, since ff is not differentiable at x=1x = 1
Show the answer

ff is continuous on [−2.5,4][-2.5, 4], so the Extreme Value Theorem guarantees an absolute maximum. It is f(2)=2.5f(2) = 2.5, the highest point of the graph.

  • Correct: 2.52.5, at x=2x = 2: (2,2.5)(2, 2.5) is the highest point of the whole graph, above both endpoint values.
  • 22, at the corner x=1x = 1: f(1)=2f(1) = 2 is less than f(2)=2.5f(2) = 2.5.
  • 0.50.5, at the right endpoint x=4x = 4: An endpoint can hold the maximum, but f(4)=0.5f(4) = 0.5 is not the largest value.
  • No maximum exists, since ff is not differentiable at x=1x = 1: The Extreme Value Theorem needs continuity, not differentiability. ff is continuous on the closed interval.

6. Where does f(x)=(x−3)2/3f(x) = (x - 3)^{2/3} have a critical point?

  1. At x=0x = 0, where f′(x)=0f'(x) = 0
  2. At x=3x = 3, where f′(x)=0f'(x) = 0
  3. At x=3x = 3, where f′f' does not exist
  4. No critical point, since f′(x)=0f'(x) = 0 has no solution
Show the answer

f′(x)=23x−33f'(x) = \frac{2}{3\sqrt[3]{x - 3}} is never 0 and is undefined at x=3x = 3, which is in the domain of ff. So x=3x = 3 is the one critical point (a cusp).

  • At x=0x = 0, where f′(x)=0f'(x) = 0: f′(0)f'(0)=23(−3)−1/3≠0= \frac{2}{3}(-3)^{-1/3} \ne 0.
  • At x=3x = 3, where f′(x)=0f'(x) = 0: f′(3)f'(3) is undefined, not zero: the graph has a cusp.
  • Correct: At x=3x = 3, where f′f' does not exist: f′(x)f'(x)=23(x−3)−1/3= \frac{2}{3}(x - 3)^{-1/3} is undefined at x=3x = 3, an interior point of the domain.
  • No critical point, since f′(x)=0f'(x) = 0 has no solution: A critical point is also a point where f′f' does not exist.

Part 9 · Summary

Summary

A function continuous on a closed interval reaches an absolute maximum and minimum there (Extreme Value Theorem). Look for them at the endpoints and at critical points, interior points where f′f' is 0 or does not exist. A critical point is a candidate, not a guarantee of an extremum.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections