Is there a way to find the exact roots of a fitted function?

I used the following code to produce a fit function for some data:

f = fittype(@(a, b, c, d, x) a + b.*abs(sin(c.*x + d)));
guess = [1,1, 7, 2];
[fitobject] = fit(freq, amp, f, 'StartPoint', guess);

Its quite a messy function with 4 parameters but the shape looks correct when plotting:

Is there a way in which I can solve for the zero points of the model function which is contained in [fitobject]?

I know how to use fzero on a function if you type it out, but is there any easy way I can find the zero points? Ideally I would also like to find the uncertainty in the roots as well but I could do that manually.

Thanks

Answer 1

[Star Strider]

It would appear to be relatively straightforward.

Since the function is 0 at:

c*x + d = 0

and ‘a’ appears to be 0 or close to it, solving for ‘x’ at 0:

x = -d/c

Since it appears that the function is periodic such that x is 0 every 0.4 units, and since the sine function crosses 0 every π radians, it would appear to be periodic at radians. You could expect or observe zeros at: 0.175, 0.575, 0.975, 1.375, 1.7750, ...

At least that’s how I analyse it, seeing only the plot image.