4.1-Maximum-Minimum-Values
Objectives
Key Definitions:
Maximum, minimum, local maximum, local minimum.
Key Theorems:
Extreme Value Theorem.
Fermat’s Theorem.
Methodology:
The Closed Interval Method.
Definitions of Extremes
Global Maximum:
Value f(c) is an absolute maximum on D if f(c) ≥ f(x) for all x in D.
Global Minimum:
Value f(c) is an absolute minimum on D if f(c) ≤ f(x) for all x in D.
Extreme Values:
Maximum and minimum values (plural: maxima/minima/extrema).
Local Maximum:
f(c) is a local maximum if f(c) ≥ f(x) for x near c.
Local Minimum:
f(c) is a local minimum if f(c) ≤ f(x) for x near c.
Note: An extreme value is a y-value of the function.
Page 6: Further Clarifications
Extreme Value Definition:
Extreme values refer to the function values, not points (x, f(x)) or just x-values.
Neighborhood Definition:
"Near c" refers to values in an open interval around c.
Page 7: Example of Graph
Graph Analysis of y = f(x) on D = [−3, 4]:
Absolute Max:
f(−3) = 5.
Absolute Min:
f(−1) = 1.
Local Extrema:
f(−1) is a local minimum.
f(2) is another local minimum.
f(1) = 1 and f(3) = 4 are local minima.
f(4) is not a local extremum.
Page 8: Exercise 1
Graph Analysis: Determine extreme values from provided graphs.
Results:
Absolute Maximums: none; 6 examples provided.
Absolute Minimums: f(−1) = 0, others noted.
Local Maximums and Minimums also noted by graphs.
Page 9: Exercise 2
Find extreme values from the function f(x) = 3x^4 − 4x^3 − 12x^2 + 12 on [-2, 3].
Results:
Absolute Maximum: f(−2) = 44.
Absolute Minimum: f(2) = −20.
Local Max: f(0) = 12; Local Min: f(−1) = 7.
Page 10: Extreme Value Theorem
Statement: If f is continuous on a closed interval [a, b], then f has an absolute maximum and minimum on that interval.
Page 11: Non-Continuous Cases
If f is not continuous or the interval is open/half-open, absolute extrema may not exist.
Examples shown of inconclusive cases regarding extrema.
Page 12: Fermat’s Theorem
Statement: If f has a local extremum at x = c, then either f′(c) = 0 or f′(c) does not exist (implying horizontal tangent).
Page 13: Proof of Fermat's Theorem
A logical sequence proving that if a point is a local minimum, then its derivative equals zero or does not exist.
Page 14: Critical Numbers and Values
Definition: Critical number of f exists on interval D if f′(c) does not exist or f′(c) = 0.
Critical value is the value f(c).
Page 15: Example of Critical Numbers
Function: f(x) = √3 x(x−1).
Critical numbers found: x = 0 and x = 1/4 based on derivative analysis.
Page 16: Further Example of Critical Numbers
Function: f(x) = x^2/(x−1).
Critical numbers identified through derivative yield: x = 0 and x = 2.
Page 17: The Closed Interval Method
Objective: Find absolute extrema of f on [a, b]. Steps outlined:
Find critical numbers in (a, b).
Compute critical values.
Compute values at endpoints.
Compare values.
Page 18: Example of Closed Interval Method
Function: f(x) = 3x^4 − 4x^3 − 12x^2 + 12 on [-2, 3].
Results calculated and summarized: Abs Max = 44, Abs Min = −20.
Page 19: Exercise on Closed Interval Method
Function: f(x) = xe^(-x^2/3) on [−1, 5].
Steps outlined, critical values calculated leading to Abs Max = (3/2)e^(−1/2) and Abs Min = −e^(−1/3).
Page 20: Humor
Meme depicting the struggle with calculus concepts such as limits, derivatives, optimization, and integrals.