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 $a<b$. 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 $$1. a\leq x_n<b$$ $$2. f(x_n)<0$$ $$3. f'(x_n)>0$$

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.


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