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












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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
$$1. aleq x_n<b$$
$$2. f(x_n)<0$$
$$3. f'(x_n)>0$$



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 $cin (x_k,b)$ 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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    $begingroup$


    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
    $$1. aleq x_n<b$$
    $$2. f(x_n)<0$$
    $$3. f'(x_n)>0$$



    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 $cin (x_k,b)$ 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.










    share|cite|improve this question









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      $begingroup$


      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
      $$1. aleq x_n<b$$
      $$2. f(x_n)<0$$
      $$3. f'(x_n)>0$$



      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 $cin (x_k,b)$ 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.










      share|cite|improve this question









      $endgroup$




      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
      $$1. aleq x_n<b$$
      $$2. f(x_n)<0$$
      $$3. f'(x_n)>0$$



      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 $cin (x_k,b)$ 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.







      real-analysis sequences-and-series derivatives






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      asked Dec 6 '18 at 16:25









      hephaeshephaes

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