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.