MATLAB: How to find inflection point, PLEASE HELP!

helpinflectioninflectionpointMATLABpoint

Hi!I have no experience in Matlab, but I was given the assignment to find point of inflection in this graph. What I have is: x = [ 7.0 7.2 7.4 7.6 8.4 8.8 9.2 9.6 10.0]; y = [ 0.692 0.719 0.723 0.732 0.719 0.712 1.407 1.714 1.99];
I would REALLY appreciate if someone could help me because I don't know what to do and how to find it. Thank you all!

Best Answer

Inflection points are defined where the curve changes direction, and the derivative is equal to zero.
See if this does what you want:
x = [ 7.0   7.2   7.4    7.6    8.4    8.8   9.2    9.6  10.0];
y = [ 0.692 0.719  0.723   0.732  0.719   0.712    1.407  1.714  1.99];
dydx = gradient(y) ./ gradient(x);                                      % Derivative Of Unevenly-Sampled Data
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);                    % Returns Approximate Zero-Crossing Indices Of Argument Vector
zxidx = zci(dydx);                                                      % Approximate Indices Where ‘dydx=0’
for k1 = 1:numel(zxidx)                                                 % Loop Finds ‘x’ & ‘y’ For ‘dydx=0’

    inflptx(k1) = interp1(dydx(zxidx(k1):zxidx(k1)+1), x(zxidx(k1):zxidx(k1)+1), 0, 'linear');
    inflpty(k1) = interp1(x(zxidx(k1):zxidx(k1)+1), y(zxidx(k1):zxidx(k1)+1), inflptx(k1), 'linear');
end
figure(1)
plot(x, y)
hold on
plot(x, dydx)
plot(inflptx, inflpty, 'pg', 'MarkerFaceColor','g')
hold off
grid
legend('Data', 'Derivative', 'Inflection Points')
inflpts = regexp(sprintf('(%5.3f,%5.3f)\n', [inflptx; inflpty]), '\n', 'split');
text(inflptx, inflpty, inflpts(1:end-1),'FontSize',8, 'HorizontalAlignment','center', 'VerticalAlignment','bottom')
The ‘inflptx’ and ‘inflpty’ vectors (derived from a linear interpolation of the derivative) are the coordinates of the inflection points. They are approximate because your data are discrete. The code does not find an inflection point where what is apparently a spline interpolation might create one, because that is not in your original data.
EDIT — (8 Jun 2020 at 12:21)
While the original code works for the original data here, if the identified zero-crossings of the derivative are close to the ends of the vector, it will crash with indices that exceed the number of vector elements.
This version of the loop fixes that: 
for k1 = 1:numel(zxidx)                                                 % Loop Finds ‘x’ & ‘y’ For ‘dydx=0’
    ixrng = max(zxidx(k1)-2,1):min(zxidx(k1)+2,numel(x));
    inflptx(k1) = interp1(dydx(ixrng), x(ixrng), 0, 'linear');
    inflpty(k1) = interp1(x(ixrng), y(ixrng), inflptx(k1), 'linear');
end
The ‘inflpts’ assignment also needs to be modified: 
inflpts = sprintfc('(%5.3f, %5.3f)', [inflptx; inflpty].');
text(inflptx, inflpty, inflpts,'FontSize',8, 'HorizontalAlignment','center', 'VerticalAlignment','bottom')
.
Related Question