A non constant function satisfying the condition of Rolle's theorem cannot be monotonic.

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I'm reading a book of elementary calculus there a found the statement "A non constant function satisfying the condition of Rolle's theorem cannot be monotonic."



But i'm finding myself unable to prove it.I need help in proving this




calculus real-analysis means




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  • 3




    Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
    – Rhys Steele
    Aug 6 at 21:49






  • 1




    As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
    – AdamW
    Aug 6 at 21:57














up vote
0
down vote

favorite












I'm reading a book of elementary calculus there a found the statement "A non constant function satisfying the condition of Rolle's theorem cannot be monotonic."



But i'm finding myself unable to prove it.I need help in proving this




calculus real-analysis means




share|cite|improve this question















  • 3




    Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
    – Rhys Steele
    Aug 6 at 21:49






  • 1




    As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
    – AdamW
    Aug 6 at 21:57












up vote
0
down vote

favorite









up vote
0
down vote

favorite











I'm reading a book of elementary calculus there a found the statement "A non constant function satisfying the condition of Rolle's theorem cannot be monotonic."



But i'm finding myself unable to prove it.I need help in proving this




calculus real-analysis means




share|cite|improve this question











I'm reading a book of elementary calculus there a found the statement "A non constant function satisfying the condition of Rolle's theorem cannot be monotonic."



But i'm finding myself unable to prove it.I need help in proving this





calculus real-analysis means





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 6 at 21:43









PK Styles

1,394624




1,394624







  • 3




    Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
    – Rhys Steele
    Aug 6 at 21:49






  • 1




    As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
    – AdamW
    Aug 6 at 21:57












  • 3




    Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
    – Rhys Steele
    Aug 6 at 21:49






  • 1




    As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
    – AdamW
    Aug 6 at 21:57







3




3




Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
– Rhys Steele
Aug 6 at 21:49




Well $f(a) = f(b)$. I'm reminded of the phrase "What goes up must come down".
– Rhys Steele
Aug 6 at 21:49




1




1




As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
– AdamW
Aug 6 at 21:57




As written, it seems technically false. The function may be constant on the interval $[a,b]$ of Rolle's theorem but monotonically strictly increasing outside of that. However, if we restrict to being non-constant on $[a,b]$, the proof of Rolle's theorem on wiki shows that the derivative must change sign in the given region, which is sufficient.
– AdamW
Aug 6 at 21:57










2 Answers
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We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.



We assuming that $f$ is not constant hence there exists $x_0in(a,b)$ such that $f(x_0)ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).



So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.






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  • What does $f(a)<f(x_0)>f(b)$ mean?
    – Nosrati
    Aug 7 at 7:10

















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0
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Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)ge f(x)$. Now choose the ends of the interval, so you get $f(b)ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)ne f(a)$. Since we use monotonically increasing function, it means that $f(c)gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)ge f(c) gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.






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    2 Answers
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    2 Answers
    2






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    active

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    up vote
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    We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.



    We assuming that $f$ is not constant hence there exists $x_0in(a,b)$ such that $f(x_0)ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).



    So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.






    share|cite|improve this answer























    • What does $f(a)<f(x_0)>f(b)$ mean?
      – Nosrati
      Aug 7 at 7:10














    up vote
    1
    down vote













    We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.



    We assuming that $f$ is not constant hence there exists $x_0in(a,b)$ such that $f(x_0)ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).



    So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.






    share|cite|improve this answer























    • What does $f(a)<f(x_0)>f(b)$ mean?
      – Nosrati
      Aug 7 at 7:10












    up vote
    1
    down vote










    up vote
    1
    down vote









    We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.



    We assuming that $f$ is not constant hence there exists $x_0in(a,b)$ such that $f(x_0)ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).



    So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.






    share|cite|improve this answer















    We have $f$ continuous on $[a,b]$ and differentiable on $(a,b)$ with $f(a)=f(b)$.



    We assuming that $f$ is not constant hence there exists $x_0in(a,b)$ such that $f(x_0)ne f(a)$(I will assume that $f(x_0)> f(a)$, but it is all the same if it is $<$).



    So we have $a<x_0<b$ but $f(a)<f(x_0)>f(b)$, i.e. $f(a)<f(x_0)$ and $f(x_0)>f(b)$.







    share|cite|improve this answer















    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 7 at 10:54


























    answered Aug 6 at 22:01









    Holo

    4,2662629




    4,2662629











    • What does $f(a)<f(x_0)>f(b)$ mean?
      – Nosrati
      Aug 7 at 7:10
















    • What does $f(a)<f(x_0)>f(b)$ mean?
      – Nosrati
      Aug 7 at 7:10















    What does $f(a)<f(x_0)>f(b)$ mean?
    – Nosrati
    Aug 7 at 7:10




    What does $f(a)<f(x_0)>f(b)$ mean?
    – Nosrati
    Aug 7 at 7:10










    up vote
    0
    down vote













    Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)ge f(x)$. Now choose the ends of the interval, so you get $f(b)ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)ne f(a)$. Since we use monotonically increasing function, it means that $f(c)gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)ge f(c) gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)ge f(x)$. Now choose the ends of the interval, so you get $f(b)ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)ne f(a)$. Since we use monotonically increasing function, it means that $f(c)gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)ge f(c) gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)ge f(x)$. Now choose the ends of the interval, so you get $f(b)ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)ne f(a)$. Since we use monotonically increasing function, it means that $f(c)gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)ge f(c) gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.






        share|cite|improve this answer













        Without loss of generality, let's use a monotonically increasing function. It means that for any $y>x$ you have $f(y)ge f(x)$. Now choose the ends of the interval, so you get $f(b)ge f(a)$. Since we want Rolles' theorem conditions, $f(b)=f(a)$. Obviously this is true for the constant function (you can always call a constant function monotonically increasing). Now let's assume that the function is not constant, so that means that there exists at least one $c$ in $(a,b)$ where $f(c)ne f(a)$. Since we use monotonically increasing function, it means that $f(c)gt f(a)$ (strict inequality). Now apply monotonic definition between $b$ and $c$ and you get $f(b)ge f(c) gt f(a)$, or $f(b)>f(a)$. This does not obey the Rolles' theorem definition.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Aug 6 at 21:59









        Andrei

        7,5852822




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