A/-2-5=0 A=-10;a=0;a=10
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Taking this example into account, we can see that setting the first value equal to 1, we obtain that 
and 
. Using this information, we find that 
. It shows that when x is positive, the succussive terms are increasing.
Referring to that finding, if we set initial value less than zero, which means that we are solving 0.5x+1<0 and taking a number in the interval of the solution, which means x ∈ (- ∞, -20). Setting x=-19, we find that
and x=-40. In the next iteration, 
. In the next iteration, 
. By this way, we find that even if the initial value is less than zero, value of the successive iterations is increasing.
Using the function
and taking the initial value equal to 4, we find that 
and x=-2. In the next iteration, 
. If we continue the iterations we'll see that they are decreasing.
Setting the initial value equal to 2, we find that
and x=0. The next iteration is 
. In this case, the interations are also decreasing.
If we set the initial value equal to 1, we find that
and x=1. In the next iteration, 
and the iterations are decreasing.
Referring to that finding, if we set initial value less than zero, which means that we are solving 0.5x+1<0 and taking a number in the interval of the solution, which means x ∈ (- ∞, -20). Setting x=-19, we find that
Using the function
Setting the initial value equal to 2, we find that
If we set the initial value equal to 1, we find that