Without knowing the relevant context, it is difficult to say.
The first syntax:
fzero(f,x0)
implies that ‘f’ is a function of one variable, and already exists in the workspace as a function handle.
The second syntax:
fsolve(@(x)f,x0)
actually throws this error when I run it (in R2018b):
Error using fsolve (line 281)
FSOLVE requires all values returned by functions to be of data type double.
The correct way to use that syntax is to express ‘f’ as a function to be evaluated:
fsolve(@(x)f(x),x0)
and this works correctly.
