Looking at the function $f(x) = -\frac{1}{2}x+3$
I want to show that all real numbers are in the stable set of its fixed point using the mean value theorem.
I know that the fixed point is equal to 2
I did an orbit diagram to see that all points are indeed heading toward that fixed point.
The mean value theorem is that there exists a c in the domain of the function such that $f(b)-f(a) = f'(c)(b-a)$
I am not really sure how to incorporate the mean value theorem.
For a real number to be in the stable set. When iterated by the function it should approach 2.
IS my goal to show that the interval shrinks until only 2 can belong ?
As you have noted, $x_0=2$ is the only fixed point under $f$. Let $x \in \mathbb{R}\setminus \{2\}$ be given. Then, by the Mean Value Theorem, $$|f(x) - 2| = |f(x) - f(2)| = \frac{1}{2} |x-2|. $$ By induction, we have that $$|f^n (x) - 2| = \frac{1}{2^n} |x-2| \longrightarrow 0$$ as $n \to \infty$.
For this particular problem, direct computation is simple enough. For a more general approach, you would probably want to aim towards the Banach fixed-point theorem. Note that the proof of the Banach fixed point theorem has a similar flavor as above. What matters isn't so much that our function $f$ satisfies the Mean Value Theorem, but rather that $f$ is Lipschitz with a constant $L<1$ (which can be shown by means of the Mean Value Theorem).