Can be derived from the previous function.

①x & lt； =- 1

y = f(x+ 1)+ 1 =(x+ 1)+ 1+ 1 = x+3

At this point, let y=0 be available and x =-3.

So y has a zero at this time and x=-3.

②- 1 & lt； x & lt=0

y = f(x+ 1)+ 1 = log(2x+2)+ 1

Let y=0 be available and x=-0.95, which is within (-1, 0).

So at this point y has a zero point x=-0.95.

③0 & lt； x & lt=0.5

y=f(log(2x))+ 1=log(2x)+2

Let y=0 be available, and x = 1 0 (-2)/2 =1/200 = 0.005, which is within (0,1).

So at this point y has a zero point x=0.005.

④x & gt； 0.5 hours

y = f(log2x)+ 1 = log(2 log(2x))+ 1

To sum up, y*** has four zeros.