Question 15 1 pts TT Given f (x) = x+ – 23 -sin (12) defined over the interval [0, 6] where h=1. Use N.G.F. Interpolatio

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Question 15 1 pts TT Given f (x) = x+ – 23 -sin (12) defined over the interval [0, 6] where h=1. Use N.G.F. Interpolatio

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Question 15 1 Pts Tt Given F X X 23 Sin 12 Defined Over The Interval 0 6 Where H 1 Use N G F Interpolatio 1
Question 15 1 Pts Tt Given F X X 23 Sin 12 Defined Over The Interval 0 6 Where H 1 Use N G F Interpolatio 1 (37.38 KiB) Viewed 63 times
Question 15 1 Pts Tt Given F X X 23 Sin 12 Defined Over The Interval 0 6 Where H 1 Use N G F Interpolatio 2
Question 15 1 Pts Tt Given F X X 23 Sin 12 Defined Over The Interval 0 6 Where H 1 Use N G F Interpolatio 2 (19.06 KiB) Viewed 63 times
Question 15 1 Pts Tt Given F X X 23 Sin 12 Defined Over The Interval 0 6 Where H 1 Use N G F Interpolatio 3
Question 15 1 Pts Tt Given F X X 23 Sin 12 Defined Over The Interval 0 6 Where H 1 Use N G F Interpolatio 3 (12.74 KiB) Viewed 63 times
Question 15 1 Pts Tt Given F X X 23 Sin 12 Defined Over The Interval 0 6 Where H 1 Use N G F Interpolatio 4
Question 15 1 Pts Tt Given F X X 23 Sin 12 Defined Over The Interval 0 6 Where H 1 Use N G F Interpolatio 4 (25.65 KiB) Viewed 63 times
Question 15 1 Pts Tt Given F X X 23 Sin 12 Defined Over The Interval 0 6 Where H 1 Use N G F Interpolatio 5
Question 15 1 Pts Tt Given F X X 23 Sin 12 Defined Over The Interval 0 6 Where H 1 Use N G F Interpolatio 5 (21.22 KiB) Viewed 63 times
15 to 19 depend on question 15
Question 15 1 pts TT Given f (x) = x+ – 23 -sin (12) defined over the interval [0, 6] where h=1. Use N.G.F. Interpolation to solve questions (15 to 19). * The maximum order of the polynomial that we get from the data is:

Question 16 1 pts Starting from (x=2), d4P4(x)/dx4 at x=2 using central derivative is:

Question 17 2 pts dP4(x)/dx at x=0 is:

Question 18 2 pts Starting from x=2; The absolute error |dP2(x)/dx - df(x)/dxl at x=4 is: 27 025 O None 023

Question 19 1 pts Starting from x=1; d2P3(x)/dx? at X=1 is: -38 0-16 None 0-54
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