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How can I find it?

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Accepted Answer

Walter Roberson
Walter Roberson on 22 Jun 2021
There is not just one answer: it depends upon what model function you use
x = [-0.4, 0, 0.4, 0.8, 1.2]
x = 1×5
-0.4000 0 0.4000 0.8000 1.2000
fx = [-0.204, -0.07, -0.006, 0.442, 1.658]
fx = 1×5
-0.2040 -0.0700 -0.0060 0.4420 1.6580
fx03 = interp1(x, fx, -0.3)
fx03 = -0.1705
fx03s = interp1(x, fx, -0.3, 'spline')
fx03s = -0.1382
p3 = polyfit(x, fx, 3);
fx03p3 = polyval(p3, -0.3)
fx03p3 = -0.1427
The cubic fit looks pretty good.
  1 Comment
Furkan Ozkan
Furkan Ozkan on 22 Jun 2021
thanks!

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More Answers (1)

Chunru
Chunru on 22 Jun 2021
You can use interp1 to interpolate. 'doc interp1' for more details.
x = [-0.4 0 0.4 0.8 1.2];
f = [-0.204 -0.07 -0.006 0.442 1.658];
y = interp1(x, f, -0.3)
y = -0.1705

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