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.


Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.