How to find x values from y value in "fit" function?

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Kanan Yagublu
Kanan Yagublu el 10 de Mayo de 2022
Comentada: Kanan Yagublu el 11 de Mayo de 2022
Hi, I have some data and with this data I create a fit function:
f = fit(x,y,'smoothingspline');
plot(f,x,y);
I need to find X values in two points like in the picture.
I tried to use :
z = x(y==y_i);
But what I get is:
0×1 empty double column vector
How can I solve this problem?
Thanks in advance

Respuesta aceptada

David Hill
David Hill el 10 de Mayo de 2022
yval=-62;
eqn=@(x)feval(f,x)-yval
X1=fzero(eqn,4.8);
X2=fzero(eqn,5.3);
  2 comentarios
Kanan Yagublu
Kanan Yagublu el 11 de Mayo de 2022
Thank you for your answer, but how did you get 4.8 and 5.3? Are they approximation from the graph?
Because when I change the data set the code doesn't work
Kanan Yagublu
Kanan Yagublu el 11 de Mayo de 2022
@David Hill and how can I modify it so it will not depend on 4.8 and 5.3 ?

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Steven Lord
Steven Lord el 10 de Mayo de 2022
Set up a sample polynomial fit.
x = randn(10, 1);
y = (x-1).*(x+1); % polynomial is y = x^2-1 = (x-1)*(x+1)
p = fit(x, y, 'poly2')
p =
Linear model Poly2: p(x) = p1*x^2 + p2*x + p3 Coefficients (with 95% confidence bounds): p1 = 1 (1, 1) p2 = 6.259e-17 (-2.697e-16, 3.949e-16) p3 = -1 (-1, -1)
Find the points where p takes on the value y = 3.
plusOneSolution = fzero(@(x) p(x)-3, 1)
plusOneSolution = 2
minusOneSolution = fzero(@(x) p(x)-3, -1)
minusOneSolution = -2
Check that evaluating the fit at those two points gives us the value y = 3.
check = p([plusOneSolution, minusOneSolution])
check = 2×1
3.0000 3.0000

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