1. Linear Interpolation and Extrapolation

You have a set of data provided below:

x	f(x)
0.000	10.6000
1.0000	-5.7392	f(3.5) =	-15.2006
2.0000	-4.9876
3.0000	3.5875
4.0000	5.9518	f(6.2) =	-14.5306
5.0000	-5.1424
6.0000	-7.5320

a) Find the f(x) for x = 3.5 using the data above. Find the error.

f(3.5) =

4.7697

Error =

-19.9703

b) Find the f(x) for x = 6.2 using the data above. Find the error.

f(6.2) =

-8.0099

Error =

-6.5207

c) Find the f(x) for x = 3.5 using the new data now given. Find the error.

x	f(x)
3.000	3.5875
3.2000	-6.8922	f(3.5) =	-15.2006
3.4000	-14.4274
3.6000	-13.6225
3.8000	-4.8575	f(6.2) =	-14.5306
4.0000	5.9518

f(3.5) =

-14.0250

Error =

-1.1756

%Program to Solve Problem 1 for Linear Curve Fitting Homework

%version 2007-10-10 D.W. Donovan

clear all;

f3p5E = -15.2006;

f6p2E = -14.5306;

f3p0 = 3.5875;

f4p0 = 5.9518;

f5p0 = -5.1424;

f6p0 = -7.5320;

f3p4 = -14.4274;

f3p6 = -13.6225;

f3p51 = f4p0-((f4p0-f3p0)/(4-3))*(4-3.5)

e3p51 = f3p5E-f3p51

f6p2 = f6p0+((f6p0-f5p0)/(6-5))*(6.2-6)

e6p2 = f6p2E-f6p2

f3p52 = f3p6-((f3p6-f3p4)/(3.6-3.4))*(3.6-3.5)

e3p52 = f3p5E-f3p52

%{

f3p51 = 4.7697

e3p51 = -19.9703

f6p2 = -8.0099

e6p2 = -6.5207

f3p52 = -14.0250

e3p52 = -1.1756

%}

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This page last updated on October 9, 2015