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Given f (x) = 1 – 2 cos (72) that is defined over [1,6] with a step (h) = 1. Solve points (6, 7,8, and 9) based on divided difference interpolation. • The values of A fo and Aªf1are respectively: -16, -32 O-16, -8 O-16, -64 None P1(2.5) is: O2 None -3 O-7

Given f (x) = 1 – 2 cos (72) that is defined over [1,6] with a step (h) = 1. Solve points (6, 7,8, and 9) based on divided difference interpolation. • The values of A fo and Aªf1are respectively: -16, -32 O-16, -8 O-16, -64 None P1(2.5) is: O2 None -3 O-7

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter5: Inverse, Exponential, And Logarithmic Functions
Section: Chapter Questions
Problem 21T
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Given f (x) = 1 – 2 cos (7x) that is defined over [1,6] with a step (h) = 1. Solve points (6, 7,8, and 9) based on divided difference interpolation. • The values of A°fo and Aªf1are respectively: O-16, -32 O-16, -8 O-16, -64 None P1(2.5) is: O2 None O-3 O-7
Transcribed Image Text:Given f (x) = 1 – 2 cos (7x) that is defined over [1,6] with a step (h) = 1. Solve points (6, 7,8, and 9) based on divided difference interpolation. • The values of A°fo and Aªf1are respectively: O-16, -32 O-16, -8 O-16, -64 None P1(2.5) is: O2 None O-3 O-7
Starting from (x=2), P3(s) at s=5 is: None The absolute error |P2(4) - f(4)| is: O None O 22 O 12 O 16
Transcribed Image Text:Starting from (x=2), P3(s) at s=5 is: None The absolute error |P2(4) - f(4)| is: O None O 22 O 12 O 16
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