# A question that combines Darboux's theorem, Mean Value Theorem, and possibly Intermediate Value Theorem.

I encounter this question in an exam paper of my uni, and to my amazement it somehow combines three important theorem in elementary analysis into a single question. My problem is with the a part of the question.

Statement of the question

Let $$a,b$$ $$\in \mathbb{R}$$ with $$a. And let $$f:\mathbb{R} \to \mathbb{R}$$ be a differentiable function on $$\mathbb{R}$$. Now suppose that $$f(a)<0, f(b)>0$$ and the derivative $$f'$$ is strictly decreasing. Prove that for the following sequence:

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} ; x_1 = a$$ that $$1. a\leq x_n $$2. f(x_n)<0$$ $$3. f'(x_n)>0$$

I think I already succeed in proving that first and the third part of the question by induction. The problem is:

How can the second part be proved?

Here is my attempt for the remaining two parts:

First note that $$f'(a)>0$$, and for $$n=1$$ all the statements are correct. Then assume that they are true for $$n=k$$, so $$f(x_k)<0$$ and $$f'(x_k)>0$$. Then assume to the contrary that $$x_{k+1} >b$$, we have $$x_k - \frac{f(x_k)}{f'(x_k)} >b$$ $$- \frac{f(x_k)}{b-x_k} > f'(x_k)$$ This implies: $$\frac{f(b)}{b-x_k}- \frac{f(x_k)}{b-x_k} > f'(x_k)$$ Then by the Mean Value Theorem, there exists a point $$c\in (x_k,b)$$ such that $$f'(c)>f'(x_k)$$ but $$c>x_k$$, this is a contradiction. Thus the first statement is proved. For the third statement, assume to the contrary that $$f'(x_{k+1}) \leq 0$$, then by Darboux's theorem we have a point $$c \in (x_k, x_{k+1})$$ such that $$f'(c)=0$$, this combined with the fact that the derivatives are decreasing implies that $$f(b)<0$$, a contradiction again.