Find a solution of the given equation in the given interval using the Bisection Method with tolerance tol = 10-3. As a stopping condition use |f(xn)| < tol. (а) х5 — 29х — 23 — 0, хе x E [1,8]. (A) 2.4868690 (B) 2.4868692 (C) 2.4868694 (D) 2.4868696 (E) 2.4868698
Find a solution of the given equation in the given interval using the Bisection Method with tolerance tol = 10-3. As a stopping condition use |f(xn)| < tol. (а) х5 — 29х — 23 — 0, хе x E [1,8]. (A) 2.4868690 (B) 2.4868692 (C) 2.4868694 (D) 2.4868696 (E) 2.4868698
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section: Chapter Questions
Problem 15T
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![Find a solution of the given equation in the given interval using the Bisection Method
with tolerance tol = 10-3. As a stopping condition use |f(xn)|< tol.
(a) r5 – 29x – 23 = 0, x e [1,8].
||
(A) 2.4868690 (B) 2.4868692 (C) 2.4868694 (D) 2.4868696 (E) 2.4868698
(b) x5 – 23x – 7 = 0, x € [1,8].
(A) 2.26020241 (B) 2.26020243 (C) 2.26020245 (D) 2.26020247 (E) 2.26020249](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F95e3ef9e-8306-4f99-9f44-d7097574d202%2F753262d7-1f92-443f-ad37-51ff4f1ef144%2Foe9hqkn_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Find a solution of the given equation in the given interval using the Bisection Method
with tolerance tol = 10-3. As a stopping condition use |f(xn)|< tol.
(a) r5 – 29x – 23 = 0, x e [1,8].
||
(A) 2.4868690 (B) 2.4868692 (C) 2.4868694 (D) 2.4868696 (E) 2.4868698
(b) x5 – 23x – 7 = 0, x € [1,8].
(A) 2.26020241 (B) 2.26020243 (C) 2.26020245 (D) 2.26020247 (E) 2.26020249
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