Here we look for a root of 
that is close to 1.
that is close to 1. f = @(x) x + cos(10*x);
interval = [0.5,1.5];
fplot(f,interval)
set(gca,'ygrid','on'), axis(axis) % ignore this line
title('Objective function') % ignore this line
xlabel('x'), ylabel('y') % ignore this line
r = fzero(f,1)
We choose three values to get the iteration started.
x = [0.8 1.2 1]';
y = f(x);
hold on, plot(x,y,'.')
title('Three initial points') % ignore this line
If we were using "forward" interpolation, we would ask for the polynomial interpolant of y as a function of x. But that parabola has no real roots.
c = polyfit(x,y,2); % coefficients of interpolant
q = @(x) polyval(c,x);
fplot(q,interval,'--')
title('Parabola model') % ignore this line
To do inverse interpolation, we swap the roles of x and y in the interpolation.
cla, fplot(f,interval), plot(x,y,'.') % ignore this line
c = polyfit(y,x,2); % coefficients of interpolating polynomial
q = @(y) polyval(c,y);
fplot(q,@(y) y,ylim,'--') % plot x=q(y), y=y
title('Sideways parabola') % ignore this line
We seek the value of x that makes y zero. This means evaluating q at zero.
x = [x; q(0)];
y = [y; f(x(end))]
We repeat the process a few more times.
for k = 4:8
c = polyfit(y(k-2:k),x(k-2:k),2);
x(k+1) = polyval(c,0);
y(k+1) = f(x(k+1));
end
Here is the sequence of errors.
format short e
err = x - r
The error seems to be superlinear, but subquadratic.
logerr = log(abs(err));
ratios = logerr(2:end) ./ logerr(1:end-1)