Download my SLM toolbox. Well, not because you need to use the SLM engine itself, but because you want to use slmsolve. The nice thing is, it can work on the pp form produced by csapi.
Spl = csapi (1:10,rand(1,10) - .5);
Spl
Spl =
struct with fields:
form: 'pp'
breaks: [1 2 3 4 5 6 7 8 9 10]
coefs: [9×4 double]
pieces: 9
order: 4
dim: 1
x0 = slmsolve(Spl,0)
x0 =
2.37436802433669 3.52837291901417 5.18859998252568 7.8567313904662
So there were 4 zero crossings for that particular randomly generated curve.
If you need to know which of those roots are cases where the curve went from negative to positive, then just evaluate the derivative of the curve at those locations.
fnval(fnder(Spl,1),x0) > 0
ans =
1×4 logical array
0 1 0 1
