Back to AI Flashcard MakerMathematics /AP Calculus AB: 8.2.2 Maximum and Minimum

AP Calculus AB: 8.2.2 Maximum and Minimum

Mathematics13 CardsCreated 3 months ago

This flashcard set explains how to identify local (relative) and absolute (global) maxima and minima of functions. It covers the Extreme Value Theorem and its conditions, highlights when it fails, and teaches the Closed Interval Method to find absolute extrema by evaluating critical points and endpoints on closed intervals.

Maximum and Minimum

  • Identify relative, or local, maxima and minima and absolute, or global, maxima and minima.

  • Understand the Extreme Value Theorem and identify situations in which the Extreme Value Theorem fails to hold.

  • Apply the Closed Interval Method to find the absolute maxima and minima of a continuous function over a closed interval.

Tap or swipe ↕ to flip
Swipe ←→Navigate
1/13

Key Terms

Term
Definition

Maximum and Minimum

  • Identify relative, or local, maxima and minima and absolute, or global, maxima and minima.

  • Understand the Extreme Value Theo...

note 1

  • A function f has a relative, or local, maximum at c if f(c)
    is greater than or equal to f(x) for all x in a neighborhood of c.

  • <...

note 2

- The Extreme Value Theorem states that a continuous

function defined on a closed interval always attains an

absolute maximum and an ab...

note 3

  • Closed Interval Method for Finding Absolute Extrema on [a, b]:

  • Evaluate f at each critical point.

  • Evaluate th...

Find the absolute maximum and absolute minimum values of f (x) = x^ 2 − 2x + 1 on the interval [0, 3]

Absolute maximum value: 4

Absolute minimum value: 0

Find the absolute maximum and absolute minimum values of f (x) = e ^x − x on the interval [−1, 1].

Absolute maximum value: e − 1

Absolute minimum value: 1

Related Flashcard Decks

Study Tips

  • Press F to enter focus mode for distraction-free studying
  • Review cards regularly to improve retention
  • Try to recall the answer before flipping the card
  • Share this deck with friends to study together
TermDefinition

Maximum and Minimum

  • Identify relative, or local, maxima and minima and absolute, or global, maxima and minima.

  • Understand the Extreme Value Theorem and identify situations in which the Extreme Value Theorem fails to hold.

  • Apply the Closed Interval Method to find the absolute maxima and minima of a continuous function over a closed interval.

note 1

  • A function f has a relative, or local, maximum at c if f(c)
    is greater than or equal to f(x) for all x in a neighborhood of c.

  • A function f has a relative, or local minimum at c if f(c)
    is less than or equal to f(x) for all x in a neighborhood of c.

  • If a continuous function has a local maximum or a local
    minimum at a point c, then c is a critical point.

  • A function f has an absolute, or global,maximum at c if f(c) is greater than or equal to f(x) for all x in the domain of f. f(c) is the absolute maximum value.

  • A function f has an absolute, or global, minimum at c if f(c) is less than or equal to f(x) for all x in the domain of f. f(c) is the absolute minimum value.

  • Identifying the local and absolute maxima and minima of a continuous function by looking at its graph is straightforward. Just identify the critical points by seeing where the slope of the tangent to the graph is zero or undefined. Then for each critical point, determine whether the graph is at a high point (maximum), a low point (minimum), or neither.

  • For this graph, the critical point at x 2 is neither a maximum nor a minimum.

  • The very highest of the maximum points is the absolute
    maximum value, and the very lowest is the absolute minimum
    value.

note 2

- The Extreme Value Theorem states that a continuous

function defined on a closed interval always attains an

absolute maximum and an absolute minimum somewhere on the closed interval.

- Sometimes the absolute maximum or minimum will occur on the interior of the interval where there is a critical point, and sometimes it will occur at an endpoint (which may or may not be a critical point).

- Notice that there could be more than one absolute maximum or absolute minimum for a given continuous function on a closed interval. In this case, there are two absolute maxima.

- The graph on the left depicts a function that is discontinuous at x = 0. It has one critical point at x = 0, where the derivative does not exist. Notice that there is no line tangent to the curve at x = 0. It has no absolute maximum value, and no absolute minimum value on the closed interval [-1, 1]. The Extreme Value Theorem does not apply because the function is not continuous.

- The graph on the right depicts a function that is defined on an open interval. It has an absolute maximum value at x = 0, but it does not have an absolute minimum value. The Extreme Value Theorem does not apply because the function is not being considered over a closed interval.

note 3

  • Closed Interval Method for Finding Absolute Extrema on [a, b]:

  • Evaluate f at each critical point.

  • Evaluate the endpoints, f(a) and f(b).

  • The largest value is the absolute maximum value. The smallest value is the absolute minimum value.

Find the absolute maximum and absolute minimum values of f (x) = x^ 2 − 2x + 1 on the interval [0, 3]

Absolute maximum value: 4

Absolute minimum value: 0

Find the absolute maximum and absolute minimum values of f (x) = e ^x − x on the interval [−1, 1].

Absolute maximum value: e − 1

Absolute minimum value: 1

Use the graph to find the absolute and local maximum and minimum values of the function

Absolute maximum value: none
Absolute minimum value: −2
Local maximum values: 7
Local minimum values: −2, −1, and 2

Find the absolute maximum and absolute minimum values of f (x) = 4x^ −1 on the interval [−2, 1].

Absolute maximum value: none

Absolute minimum value: none

Find the absolute maximum and absolute minimum values of f (x) = −x^ 4 + 1 on the interval (−1, 1).

Absolute maximum value: 1

Absolute minimum value: none

Find the absolute maximum and absolute minimum values of f (x) = x ^3 − 6x ^2 + 9x − 3 on the interval [−1, 2].

Absolute maximum value: 1

Absolute minimum value: −19

Use the graph to find the absolute and local maximum and minimum values of the function.

Absolute maximum value: 9
Absolute minimum value: none
Local maximum values: 4 and 9
Local minimum value: −4

Find the absolute maximum and absolute minimum values of f (x) = 2|x|^1/2 on the interval [−1, 1].

Absolute maximum value: 2

Absolute minimum value: 0

Use the graph to find the absolute and local maximum and minimum values of the function.

Absolute maximum value: none
Absolute minimum value: none
Local maximum values: −1 and 5
Local minimum values: −8 and −6