Use the Extreme Value Theorem to show that each function f in Exercises 49–54 has both a maximum and a minimum value on [a, b]. Then use a graphing utility to approximate values M and m in [a, b] at which f has a maximum and a minimum, respectively. You may assume that these functions are contin- uous everywhere. 49. f(x) = xª – 3x² – 2, [a, b] = [–2, 2] 50. f) — х1 — 3х? — 2, [а, b] — [0, 2] 51. f) %3D х* - Зx? - 2, [а, b] %—D [-1, 1] 52. f(x) = 3 – 2r² +x³, [a, b] = [-1, 2] 53. f(x) = 3 – 2x2 +x³, [a, b] = [0, 2] 54. f(x) = 3 – 2x?+x³, [a, b] = [–1, 1] %3D

College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter3: Functions
Section3.3: More On Functions; Piecewise-defined Functions
Problem 99E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...
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Use the Extreme Value Theorem to show that each function f
in Exercises 49–54 has both a maximum and a minimum value
on [a, b]. Then use a graphing utility to approximate values M
and m in [a, b] at which f has a maximum and a minimum,
respectively. You may assume that these functions are contin-
uous everywhere.
49. f(x) = xª – 3x² – 2, [a, b] = [–2, 2]
50. f) — х1 — 3х? — 2, [а, b] — [0, 2]
51. f) %3D х* - Зx? - 2, [а, b] %—D [-1, 1]
52. f(x) = 3 – 2r² +x³, [a, b] = [-1, 2]
53. f(x) = 3 – 2x2 +x³, [a, b] = [0, 2]
54. f(x) = 3 – 2x?+x³, [a, b] = [–1, 1]
%3D
Transcribed Image Text:Use the Extreme Value Theorem to show that each function f in Exercises 49–54 has both a maximum and a minimum value on [a, b]. Then use a graphing utility to approximate values M and m in [a, b] at which f has a maximum and a minimum, respectively. You may assume that these functions are contin- uous everywhere. 49. f(x) = xª – 3x² – 2, [a, b] = [–2, 2] 50. f) — х1 — 3х? — 2, [а, b] — [0, 2] 51. f) %3D х* - Зx? - 2, [а, b] %—D [-1, 1] 52. f(x) = 3 – 2r² +x³, [a, b] = [-1, 2] 53. f(x) = 3 – 2x2 +x³, [a, b] = [0, 2] 54. f(x) = 3 – 2x?+x³, [a, b] = [–1, 1] %3D
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