Prove there are $xi$, $eta$ with $f'(xi)f'(eta)=1$












4












$begingroup$


Let $f:[0,1]tomathbb{R}$ be continuous with $f(0)=0$, $f(1)=1$ and $f$ is differentiable on $(0,1)$. Show that there are distinct $xi,etain(0,1)$ so that $f'(xi)f'(eta)=1$.



I think this requires mean value theorem. But this does not help since if apply the theorem to $f$ we can only get $f'(c)=1$ for some $c$.






calculus derivatives







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$endgroup$












  • $begingroup$
    The result is very general and for any given positive integer $n$ it is possible to find $n$ distinct points $x_{1}, x_{2},dots, x_{n}$ all in $(0,1)$ such that $prod_{i = 1}^{n}f'(x_{i}) = 1$. This is already proved in MSE but unable to find the link.
    $endgroup$
    – Paramanand Singh
    Jan 12 '16 at 3:54










  • $begingroup$
    @ParamanandSingh: That's How to prove there exist distinct $a_{i}$ such $f'(a_{1})f'(a_{2})f'(a_{3})cdots f'(a_{n})=1$.
    $endgroup$
    – Martin R
    Jan 24 at 12:14
















4












$begingroup$


Let $f:[0,1]tomathbb{R}$ be continuous with $f(0)=0$, $f(1)=1$ and $f$ is differentiable on $(0,1)$. Show that there are distinct $xi,etain(0,1)$ so that $f'(xi)f'(eta)=1$.



I think this requires mean value theorem. But this does not help since if apply the theorem to $f$ we can only get $f'(c)=1$ for some $c$.






calculus derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    The result is very general and for any given positive integer $n$ it is possible to find $n$ distinct points $x_{1}, x_{2},dots, x_{n}$ all in $(0,1)$ such that $prod_{i = 1}^{n}f'(x_{i}) = 1$. This is already proved in MSE but unable to find the link.
    $endgroup$
    – Paramanand Singh
    Jan 12 '16 at 3:54










  • $begingroup$
    @ParamanandSingh: That's How to prove there exist distinct $a_{i}$ such $f'(a_{1})f'(a_{2})f'(a_{3})cdots f'(a_{n})=1$.
    $endgroup$
    – Martin R
    Jan 24 at 12:14














4












4








4


3



$begingroup$


Let $f:[0,1]tomathbb{R}$ be continuous with $f(0)=0$, $f(1)=1$ and $f$ is differentiable on $(0,1)$. Show that there are distinct $xi,etain(0,1)$ so that $f'(xi)f'(eta)=1$.



I think this requires mean value theorem. But this does not help since if apply the theorem to $f$ we can only get $f'(c)=1$ for some $c$.






calculus derivatives







share|cite|improve this question











$endgroup$




Let $f:[0,1]tomathbb{R}$ be continuous with $f(0)=0$, $f(1)=1$ and $f$ is differentiable on $(0,1)$. Show that there are distinct $xi,etain(0,1)$ so that $f'(xi)f'(eta)=1$.



I think this requires mean value theorem. But this does not help since if apply the theorem to $f$ we can only get $f'(c)=1$ for some $c$.






calculus derivatives




calculus derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 11 '16 at 12:16









Johannes Z

19517




19517










asked Nov 25 '13 at 15:56









RickyRicky

211




211












  • $begingroup$
    The result is very general and for any given positive integer $n$ it is possible to find $n$ distinct points $x_{1}, x_{2},dots, x_{n}$ all in $(0,1)$ such that $prod_{i = 1}^{n}f'(x_{i}) = 1$. This is already proved in MSE but unable to find the link.
    $endgroup$
    – Paramanand Singh
    Jan 12 '16 at 3:54










  • $begingroup$
    @ParamanandSingh: That's How to prove there exist distinct $a_{i}$ such $f'(a_{1})f'(a_{2})f'(a_{3})cdots f'(a_{n})=1$.
    $endgroup$
    – Martin R
    Jan 24 at 12:14


















  • $begingroup$
    The result is very general and for any given positive integer $n$ it is possible to find $n$ distinct points $x_{1}, x_{2},dots, x_{n}$ all in $(0,1)$ such that $prod_{i = 1}^{n}f'(x_{i}) = 1$. This is already proved in MSE but unable to find the link.
    $endgroup$
    – Paramanand Singh
    Jan 12 '16 at 3:54










  • $begingroup$
    @ParamanandSingh: That's How to prove there exist distinct $a_{i}$ such $f'(a_{1})f'(a_{2})f'(a_{3})cdots f'(a_{n})=1$.
    $endgroup$
    – Martin R
    Jan 24 at 12:14
















$begingroup$
The result is very general and for any given positive integer $n$ it is possible to find $n$ distinct points $x_{1}, x_{2},dots, x_{n}$ all in $(0,1)$ such that $prod_{i = 1}^{n}f'(x_{i}) = 1$. This is already proved in MSE but unable to find the link.
$endgroup$
– Paramanand Singh
Jan 12 '16 at 3:54




$begingroup$
The result is very general and for any given positive integer $n$ it is possible to find $n$ distinct points $x_{1}, x_{2},dots, x_{n}$ all in $(0,1)$ such that $prod_{i = 1}^{n}f'(x_{i}) = 1$. This is already proved in MSE but unable to find the link.
$endgroup$
– Paramanand Singh
Jan 12 '16 at 3:54












$begingroup$
@ParamanandSingh: That's How to prove there exist distinct $a_{i}$ such $f'(a_{1})f'(a_{2})f'(a_{3})cdots f'(a_{n})=1$.
$endgroup$
– Martin R
Jan 24 at 12:14




$begingroup$
@ParamanandSingh: That's How to prove there exist distinct $a_{i}$ such $f'(a_{1})f'(a_{2})f'(a_{3})cdots f'(a_{n})=1$.
$endgroup$
– Martin R
Jan 24 at 12:14










2 Answers
2






active

oldest

votes


















5












$begingroup$

Here is one way to do this:



Let $g(x) = f(x)^2 - x^2$. Applying Rolle's theorem to $g(x)$, we obtain a value $eta in (0,1)$ such that $2f(eta)f^prime(eta) - 2eta = 0$, or $f(eta) = frac{eta}{f^prime(eta)}$. Now apply the mean value theorem to $f$ on the domain $[0, eta]$ to obtain a value $xi in (0,eta)$ such that $f^prime(xi) = frac{f(eta)}{eta}$. Then $displaystyle f^prime(xi)f^prime(eta) = frac{f(eta)}{eta}f^prime(eta) = frac{eta}{eta f^prime(eta)}f^prime(eta) = 1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sir what is the motivation behind considering the function $f(x)^2-x^2$?
    $endgroup$
    – TheBox
    Jan 12 '16 at 5:41



















0












$begingroup$

Clearly the equation $f(x) = x$ has at least two roots in $[0, 1]$ (namely $x = 0, x = 1$). If there was another root in $(0, 1)$ (say $c$) then we know that there will be $xi in (0, c), eta in (c, 1)$ such that $f'(xi) = 1 = f'(eta)$ and we are done.



Hence let's assume that $x = 0, 1$ are the only roots of $f(x) = x$ in $[0, 1]$. Thus there are two possibilities: either $f(x) > x$ for all $x in (0, 1)$ or $f(x) < x$ for all $x in (0, 1)$.



We take the case when $f(x) > x$ for all $x in (0, 1)$. Consider the point $d in (0, 1)$ where $f'(d) = 1$ (its existence is guaranteed by Mean Value Theorem). Clearly $f(d) > d$ and hence there is a point $p in (0, d)$ with $f'(p) = f(d)/d > 1$ and another point $q in (d, 1)$ with $f'(q) = (1 - f(d))/(1 - d) < 1$. Thus we have found two points $p, q in (0, 1)$ with $B = f'(p) > 1, A = f'(q) < 1$.



Now the proof is almost obvious. By Darboux theorem derivatives possess the intermediate value property and hence $f'(x)$ takes all values between $A$ and $1$ and all values between $1$ and $B$. And clearly we can choose two such values of $f'(x)$ whose product is $1$.



More generally we can choose as many pairs of values of $x$ as we want such that the product of derivative $f'$ at these points (corresponding to one chosen pair) is $1$. This proves the general result that for any $n$ we can find distinct points $x_{i}$ such that $prod f'(x_{i}) = 1 $.






share|cite|improve this answer









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    2 Answers
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    2 Answers
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    active

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    5












    $begingroup$

    Here is one way to do this:



    Let $g(x) = f(x)^2 - x^2$. Applying Rolle's theorem to $g(x)$, we obtain a value $eta in (0,1)$ such that $2f(eta)f^prime(eta) - 2eta = 0$, or $f(eta) = frac{eta}{f^prime(eta)}$. Now apply the mean value theorem to $f$ on the domain $[0, eta]$ to obtain a value $xi in (0,eta)$ such that $f^prime(xi) = frac{f(eta)}{eta}$. Then $displaystyle f^prime(xi)f^prime(eta) = frac{f(eta)}{eta}f^prime(eta) = frac{eta}{eta f^prime(eta)}f^prime(eta) = 1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sir what is the motivation behind considering the function $f(x)^2-x^2$?
      $endgroup$
      – TheBox
      Jan 12 '16 at 5:41
















    5












    $begingroup$

    Here is one way to do this:



    Let $g(x) = f(x)^2 - x^2$. Applying Rolle's theorem to $g(x)$, we obtain a value $eta in (0,1)$ such that $2f(eta)f^prime(eta) - 2eta = 0$, or $f(eta) = frac{eta}{f^prime(eta)}$. Now apply the mean value theorem to $f$ on the domain $[0, eta]$ to obtain a value $xi in (0,eta)$ such that $f^prime(xi) = frac{f(eta)}{eta}$. Then $displaystyle f^prime(xi)f^prime(eta) = frac{f(eta)}{eta}f^prime(eta) = frac{eta}{eta f^prime(eta)}f^prime(eta) = 1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Sir what is the motivation behind considering the function $f(x)^2-x^2$?
      $endgroup$
      – TheBox
      Jan 12 '16 at 5:41














    5












    5








    5





    $begingroup$

    Here is one way to do this:



    Let $g(x) = f(x)^2 - x^2$. Applying Rolle's theorem to $g(x)$, we obtain a value $eta in (0,1)$ such that $2f(eta)f^prime(eta) - 2eta = 0$, or $f(eta) = frac{eta}{f^prime(eta)}$. Now apply the mean value theorem to $f$ on the domain $[0, eta]$ to obtain a value $xi in (0,eta)$ such that $f^prime(xi) = frac{f(eta)}{eta}$. Then $displaystyle f^prime(xi)f^prime(eta) = frac{f(eta)}{eta}f^prime(eta) = frac{eta}{eta f^prime(eta)}f^prime(eta) = 1$.






    share|cite|improve this answer









    $endgroup$



    Here is one way to do this:



    Let $g(x) = f(x)^2 - x^2$. Applying Rolle's theorem to $g(x)$, we obtain a value $eta in (0,1)$ such that $2f(eta)f^prime(eta) - 2eta = 0$, or $f(eta) = frac{eta}{f^prime(eta)}$. Now apply the mean value theorem to $f$ on the domain $[0, eta]$ to obtain a value $xi in (0,eta)$ such that $f^prime(xi) = frac{f(eta)}{eta}$. Then $displaystyle f^prime(xi)f^prime(eta) = frac{f(eta)}{eta}f^prime(eta) = frac{eta}{eta f^prime(eta)}f^prime(eta) = 1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 25 '13 at 16:21









    universalsetuniversalset

    7,3811330




    7,3811330












    • $begingroup$
      Sir what is the motivation behind considering the function $f(x)^2-x^2$?
      $endgroup$
      – TheBox
      Jan 12 '16 at 5:41


















    • $begingroup$
      Sir what is the motivation behind considering the function $f(x)^2-x^2$?
      $endgroup$
      – TheBox
      Jan 12 '16 at 5:41
















    $begingroup$
    Sir what is the motivation behind considering the function $f(x)^2-x^2$?
    $endgroup$
    – TheBox
    Jan 12 '16 at 5:41




    $begingroup$
    Sir what is the motivation behind considering the function $f(x)^2-x^2$?
    $endgroup$
    – TheBox
    Jan 12 '16 at 5:41











    0












    $begingroup$

    Clearly the equation $f(x) = x$ has at least two roots in $[0, 1]$ (namely $x = 0, x = 1$). If there was another root in $(0, 1)$ (say $c$) then we know that there will be $xi in (0, c), eta in (c, 1)$ such that $f'(xi) = 1 = f'(eta)$ and we are done.



    Hence let's assume that $x = 0, 1$ are the only roots of $f(x) = x$ in $[0, 1]$. Thus there are two possibilities: either $f(x) > x$ for all $x in (0, 1)$ or $f(x) < x$ for all $x in (0, 1)$.



    We take the case when $f(x) > x$ for all $x in (0, 1)$. Consider the point $d in (0, 1)$ where $f'(d) = 1$ (its existence is guaranteed by Mean Value Theorem). Clearly $f(d) > d$ and hence there is a point $p in (0, d)$ with $f'(p) = f(d)/d > 1$ and another point $q in (d, 1)$ with $f'(q) = (1 - f(d))/(1 - d) < 1$. Thus we have found two points $p, q in (0, 1)$ with $B = f'(p) > 1, A = f'(q) < 1$.



    Now the proof is almost obvious. By Darboux theorem derivatives possess the intermediate value property and hence $f'(x)$ takes all values between $A$ and $1$ and all values between $1$ and $B$. And clearly we can choose two such values of $f'(x)$ whose product is $1$.



    More generally we can choose as many pairs of values of $x$ as we want such that the product of derivative $f'$ at these points (corresponding to one chosen pair) is $1$. This proves the general result that for any $n$ we can find distinct points $x_{i}$ such that $prod f'(x_{i}) = 1 $.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Clearly the equation $f(x) = x$ has at least two roots in $[0, 1]$ (namely $x = 0, x = 1$). If there was another root in $(0, 1)$ (say $c$) then we know that there will be $xi in (0, c), eta in (c, 1)$ such that $f'(xi) = 1 = f'(eta)$ and we are done.



      Hence let's assume that $x = 0, 1$ are the only roots of $f(x) = x$ in $[0, 1]$. Thus there are two possibilities: either $f(x) > x$ for all $x in (0, 1)$ or $f(x) < x$ for all $x in (0, 1)$.



      We take the case when $f(x) > x$ for all $x in (0, 1)$. Consider the point $d in (0, 1)$ where $f'(d) = 1$ (its existence is guaranteed by Mean Value Theorem). Clearly $f(d) > d$ and hence there is a point $p in (0, d)$ with $f'(p) = f(d)/d > 1$ and another point $q in (d, 1)$ with $f'(q) = (1 - f(d))/(1 - d) < 1$. Thus we have found two points $p, q in (0, 1)$ with $B = f'(p) > 1, A = f'(q) < 1$.



      Now the proof is almost obvious. By Darboux theorem derivatives possess the intermediate value property and hence $f'(x)$ takes all values between $A$ and $1$ and all values between $1$ and $B$. And clearly we can choose two such values of $f'(x)$ whose product is $1$.



      More generally we can choose as many pairs of values of $x$ as we want such that the product of derivative $f'$ at these points (corresponding to one chosen pair) is $1$. This proves the general result that for any $n$ we can find distinct points $x_{i}$ such that $prod f'(x_{i}) = 1 $.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Clearly the equation $f(x) = x$ has at least two roots in $[0, 1]$ (namely $x = 0, x = 1$). If there was another root in $(0, 1)$ (say $c$) then we know that there will be $xi in (0, c), eta in (c, 1)$ such that $f'(xi) = 1 = f'(eta)$ and we are done.



        Hence let's assume that $x = 0, 1$ are the only roots of $f(x) = x$ in $[0, 1]$. Thus there are two possibilities: either $f(x) > x$ for all $x in (0, 1)$ or $f(x) < x$ for all $x in (0, 1)$.



        We take the case when $f(x) > x$ for all $x in (0, 1)$. Consider the point $d in (0, 1)$ where $f'(d) = 1$ (its existence is guaranteed by Mean Value Theorem). Clearly $f(d) > d$ and hence there is a point $p in (0, d)$ with $f'(p) = f(d)/d > 1$ and another point $q in (d, 1)$ with $f'(q) = (1 - f(d))/(1 - d) < 1$. Thus we have found two points $p, q in (0, 1)$ with $B = f'(p) > 1, A = f'(q) < 1$.



        Now the proof is almost obvious. By Darboux theorem derivatives possess the intermediate value property and hence $f'(x)$ takes all values between $A$ and $1$ and all values between $1$ and $B$. And clearly we can choose two such values of $f'(x)$ whose product is $1$.



        More generally we can choose as many pairs of values of $x$ as we want such that the product of derivative $f'$ at these points (corresponding to one chosen pair) is $1$. This proves the general result that for any $n$ we can find distinct points $x_{i}$ such that $prod f'(x_{i}) = 1 $.






        share|cite|improve this answer









        $endgroup$



        Clearly the equation $f(x) = x$ has at least two roots in $[0, 1]$ (namely $x = 0, x = 1$). If there was another root in $(0, 1)$ (say $c$) then we know that there will be $xi in (0, c), eta in (c, 1)$ such that $f'(xi) = 1 = f'(eta)$ and we are done.



        Hence let's assume that $x = 0, 1$ are the only roots of $f(x) = x$ in $[0, 1]$. Thus there are two possibilities: either $f(x) > x$ for all $x in (0, 1)$ or $f(x) < x$ for all $x in (0, 1)$.



        We take the case when $f(x) > x$ for all $x in (0, 1)$. Consider the point $d in (0, 1)$ where $f'(d) = 1$ (its existence is guaranteed by Mean Value Theorem). Clearly $f(d) > d$ and hence there is a point $p in (0, d)$ with $f'(p) = f(d)/d > 1$ and another point $q in (d, 1)$ with $f'(q) = (1 - f(d))/(1 - d) < 1$. Thus we have found two points $p, q in (0, 1)$ with $B = f'(p) > 1, A = f'(q) < 1$.



        Now the proof is almost obvious. By Darboux theorem derivatives possess the intermediate value property and hence $f'(x)$ takes all values between $A$ and $1$ and all values between $1$ and $B$. And clearly we can choose two such values of $f'(x)$ whose product is $1$.



        More generally we can choose as many pairs of values of $x$ as we want such that the product of derivative $f'$ at these points (corresponding to one chosen pair) is $1$. This proves the general result that for any $n$ we can find distinct points $x_{i}$ such that $prod f'(x_{i}) = 1 $.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 '16 at 4:12









        Paramanand SinghParamanand Singh

        50.5k556168




        50.5k556168






























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