- #1

lemonthree

- 51

- 0

Justify your answer.

a.$ g(x) = e^{\frac{-x}{2}}$

b.$ g(x) = 3x - 1$

Let me attempt for part a first. For fixed points, g(p) = p. I believe it is a yes, because it fulfils the conditions of having a convergence in a fix point iteration.

From the graph of $ e^{\frac{-x}{2}}$, I know that g(x) is continuous from $\left [ 0,1 \right ]$. So By Intermediate Value Theorem, I know that there exists a fixed point on $\left [ 0,1 \right ]$. Furthermore, $g'(x) = -0.5e^{\frac{-x}{2}}$, so$ \left | g'(x) \right | \leq 1$. Therefore, fixed point converges and there is a unique fixed point.

Am I right to be saying this? What else should I include to justify my answer, or is the above not helpful at all?