85. f(x) = &r* – 2x² + 5x – 1; [0, 1 ] 86. f(x) = x* + &r – x² + 2; [-1, 0] A 87. f(x) = 2x³ + 6x? – &x + 2; [-5, –4] 88. f(x) = 3x – 10x + 9; [-3, –2] %3D 89. f(x) = x – xrt + 7.x – 7x2 – 18r + 18; [1.4, 1.5] 90. f(x) = x – 3x* – 2r³ + 6x² + x + 2; [1.7, 1.8]

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section: Chapter Questions
Problem 5T
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In Problems 85–90, use the Intermediate Value Theorem to show that each function has a zero in the given interval. Approximate the zero
correct to two decimal places.

85. f(x) = &r* – 2x² + 5x – 1; [0, 1 ]
86. f(x) = x* + &r – x² + 2; [-1, 0]
A 87. f(x) = 2x³ + 6x? – &x + 2; [-5, –4]
88. f(x) = 3x – 10x + 9; [-3, –2]
%3D
89. f(x) = x – xrt + 7.x – 7x2 – 18r + 18; [1.4, 1.5]
90. f(x) = x – 3x* – 2r³ + 6x² + x + 2; [1.7, 1.8]
Transcribed Image Text:85. f(x) = &r* – 2x² + 5x – 1; [0, 1 ] 86. f(x) = x* + &r – x² + 2; [-1, 0] A 87. f(x) = 2x³ + 6x? – &x + 2; [-5, –4] 88. f(x) = 3x – 10x + 9; [-3, –2] %3D 89. f(x) = x – xrt + 7.x – 7x2 – 18r + 18; [1.4, 1.5] 90. f(x) = x – 3x* – 2r³ + 6x² + x + 2; [1.7, 1.8]
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