Answer:
[tex]S=\left \{ 6 \right \}[/tex]
Step-by-step explanation:
The Newton's Method to approximate the zeros is useful, despite the fact that it might not work. The key point here is find tangent lines, and applying that formula again and again till it gets closer to the roots.
It fits for any polynomial equation even though it is much more used, to higher degree polynomial.
*For the function given, f(x)=x-2x+6, since the root is an integer, 6, in few trials we'll find the root, all of them bigger than 0.001 so let's replace th
1) Check graph below.
2) Taking a closer look, to f(x)=-x+6 we can see that the root is between x=5 and x=7, we already know is -6.
Let's try x=7
Call it [tex]x_{1}[/tex]
Plugging into the formula:
[tex]X_{n+1}= x_{n}-\frac{f(x)}{f'(x)} \\x_{2}=7-\frac{-x+6}{-1}\\ x_{2}=7-\frac{-7+6}{-1}\\x_{2}=6\\[/tex]
[tex]x_{3}=6-\frac{-6+6}{-1}\Rightarrow x_{3}= 6[/tex]
In this function, the root is an integer, all the approximations were integer numbers and larger than 0.001.
On the other hand, if the function had irrational roots then we could easily show results lesser than 0.001.
