Question

Example 2:

Answer 1

Taking this example into account, we can see that setting the first value equal to 1, we obtain that $F(x)=0.5x+1=4$ and $x=6$. Using this information, we find that $F(x+1)=0.5(x+1)+1=0.5(6+1)+1=4.5$. 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 $F(x)=0.5x+1=-19$ and x=-40. In the next iteration, $F(x+1)=0.5(x+1)+1=0.5(1-40)+1=-18.5$. In the next iteration, $F(x+2)=0.5(x+2)+1=0.5(2-40)+1=-18$. 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 $g(x)=-x+2$ and taking the initial value equal to 4, we find that $g(x)=-x+2=4$ and x=-2. In the next iteration, $g(x+1)=-(x+1)+2=-(-2+1)+2=3$. If we continue the iterations we'll see that they are decreasing.
Setting the initial value equal to 2, we find that $g(x)=-x+2=2$ and x=0. The next iteration is $g(x+1)=-(x+1)+2=1$. In this case, the interations are also decreasing. 
If we set the initial value equal to 1, we find that $g(x)=-x+2=1$ and x=1. In the next iteration, $g(x+1)=-(x+1)+2=0$ and the iterations are decreasing.