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For an equation like r = 0, a root exists at x 0. The Bisection Method cannot be adopted to solve this equation in spite of the root existing at.x 0 becatuse the function Ax)= x (A) is a polynomial (B) has repeated roots at x 0 (C) is always non-negative (D) has a slope equal to zero at x 0 OB OA C

For an equation like r = 0, a root exists at x 0. The Bisection Method cannot be adopted to solve this equation in spite of the root existing at.x 0 becatuse the function Ax)= x (A) is a polynomial (B) has repeated roots at x 0 (C) is always non-negative (D) has a slope equal to zero at x 0 OB OA C

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.5: Rational Functions
Problem 51E
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After five iterations of the Bisection Method, determine one real root of f (x) = -3x3 + 5x? + 14x – 16 on the interval (2, 3). 2.65625 O 2.78687 O 2.81911 O 2.58318
Transcribed Image Text:After five iterations of the Bisection Method, determine one real root of f (x) = -3x3 + 5x? + 14x – 16 on the interval (2, 3). 2.65625 O 2.78687 O 2.81911 O 2.58318
For an equation like r 0. a root exists at x 0. The Bisection Method cannot be adopted to solve this equation in spite of the root existing at.r 0 because the function (x)= r (A) is a polynomial (B) has repeated roots at x = 0 (C) is always non-negative (D) has a slope equal to zero at x = 0 В O A C OD
Transcribed Image Text:For an equation like r 0. a root exists at x 0. The Bisection Method cannot be adopted to solve this equation in spite of the root existing at.r 0 because the function (x)= r (A) is a polynomial (B) has repeated roots at x = 0 (C) is always non-negative (D) has a slope equal to zero at x = 0 В O A C OD
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