Given second and first derivatives at 2 points, prove that some point in between them has third derivative...












3














Let $f$ be a function that is $C^3$ on an open interval containing $[0,1]$ - that is, the third derivative $f'''$ exists and is continuous on an open interval containing $[0,1]$. Assume that $f(0) = f'(0) = f''(0) = 0$ and that $f'(1) = f''(1) = 0$. If $f(1) = 1$, prove that there is some $cinleft(0,1right)$ such that $f'''(c) geq 24$.



Not sure how to approach this problem, been stuck on it for a while. Any help would be nice.



I tried using the Mean Value Theorem, and got that at some point $c_1inleft[0,1right], f'(c_1) = frac{f(1)-f(0)}{1-0}=1$ and there is some point $c_2 in [c_1,1]$ such that $f''(c_2) = frac{f'(1)-f'(c_1)}{1-c_1}=frac{-1}{1-c_1}<-1$ and there is some point $c_3in[c_2,1]$ such that $f'''(c_3) = frac{f''(1)-f''(c_2)}{1-c_2}>1$. I'm not sure how to extend this to $24$ though, or even if this method will work.



Is there a way to use Taylor Polynomials perhaps?










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  • What have you tried?
    – saulspatz
    Nov 24 at 20:45










  • I've tried using the mean value theorem but couldn't come up with anything.
    – You Zhou
    Nov 24 at 20:55










  • I don't know how to do it off the top of my head, but I suspect that Taylor polynomials come into it. I advise you to edit the question to explain what you have tried. The question may well be closed if you don't give any more context than you have. Many people will vote to close without reading the comments, so please edit the question body.
    – saulspatz
    Nov 24 at 21:04






  • 2




    The function $f(x)=6x^2-8x^3+3x^4$ satisfies the conditions. Note that $f'''(x)=-48+72x$, so there is no $c in (0,1)$ where $f$ has large enough third derivative, but $f'''(1)=24$. That is to say, I think there's a typo somewhere and $c$ should be allowed to be an endpoint.
    – Micah
    Nov 24 at 22:20






  • 3




    @Micah If $g(x)=6 x^2 - 8 x^3 + 3 x^4$ then $g''(x)=12 - 48x + 36x^2$ and $g''(0)=12neq0,$ so this does not satisfy all the conditions.
    – David K
    Nov 25 at 20:12
















3














Let $f$ be a function that is $C^3$ on an open interval containing $[0,1]$ - that is, the third derivative $f'''$ exists and is continuous on an open interval containing $[0,1]$. Assume that $f(0) = f'(0) = f''(0) = 0$ and that $f'(1) = f''(1) = 0$. If $f(1) = 1$, prove that there is some $cinleft(0,1right)$ such that $f'''(c) geq 24$.



Not sure how to approach this problem, been stuck on it for a while. Any help would be nice.



I tried using the Mean Value Theorem, and got that at some point $c_1inleft[0,1right], f'(c_1) = frac{f(1)-f(0)}{1-0}=1$ and there is some point $c_2 in [c_1,1]$ such that $f''(c_2) = frac{f'(1)-f'(c_1)}{1-c_1}=frac{-1}{1-c_1}<-1$ and there is some point $c_3in[c_2,1]$ such that $f'''(c_3) = frac{f''(1)-f''(c_2)}{1-c_2}>1$. I'm not sure how to extend this to $24$ though, or even if this method will work.



Is there a way to use Taylor Polynomials perhaps?










share|cite|improve this question
























  • What have you tried?
    – saulspatz
    Nov 24 at 20:45










  • I've tried using the mean value theorem but couldn't come up with anything.
    – You Zhou
    Nov 24 at 20:55










  • I don't know how to do it off the top of my head, but I suspect that Taylor polynomials come into it. I advise you to edit the question to explain what you have tried. The question may well be closed if you don't give any more context than you have. Many people will vote to close without reading the comments, so please edit the question body.
    – saulspatz
    Nov 24 at 21:04






  • 2




    The function $f(x)=6x^2-8x^3+3x^4$ satisfies the conditions. Note that $f'''(x)=-48+72x$, so there is no $c in (0,1)$ where $f$ has large enough third derivative, but $f'''(1)=24$. That is to say, I think there's a typo somewhere and $c$ should be allowed to be an endpoint.
    – Micah
    Nov 24 at 22:20






  • 3




    @Micah If $g(x)=6 x^2 - 8 x^3 + 3 x^4$ then $g''(x)=12 - 48x + 36x^2$ and $g''(0)=12neq0,$ so this does not satisfy all the conditions.
    – David K
    Nov 25 at 20:12














3












3








3


2





Let $f$ be a function that is $C^3$ on an open interval containing $[0,1]$ - that is, the third derivative $f'''$ exists and is continuous on an open interval containing $[0,1]$. Assume that $f(0) = f'(0) = f''(0) = 0$ and that $f'(1) = f''(1) = 0$. If $f(1) = 1$, prove that there is some $cinleft(0,1right)$ such that $f'''(c) geq 24$.



Not sure how to approach this problem, been stuck on it for a while. Any help would be nice.



I tried using the Mean Value Theorem, and got that at some point $c_1inleft[0,1right], f'(c_1) = frac{f(1)-f(0)}{1-0}=1$ and there is some point $c_2 in [c_1,1]$ such that $f''(c_2) = frac{f'(1)-f'(c_1)}{1-c_1}=frac{-1}{1-c_1}<-1$ and there is some point $c_3in[c_2,1]$ such that $f'''(c_3) = frac{f''(1)-f''(c_2)}{1-c_2}>1$. I'm not sure how to extend this to $24$ though, or even if this method will work.



Is there a way to use Taylor Polynomials perhaps?










share|cite|improve this question















Let $f$ be a function that is $C^3$ on an open interval containing $[0,1]$ - that is, the third derivative $f'''$ exists and is continuous on an open interval containing $[0,1]$. Assume that $f(0) = f'(0) = f''(0) = 0$ and that $f'(1) = f''(1) = 0$. If $f(1) = 1$, prove that there is some $cinleft(0,1right)$ such that $f'''(c) geq 24$.



Not sure how to approach this problem, been stuck on it for a while. Any help would be nice.



I tried using the Mean Value Theorem, and got that at some point $c_1inleft[0,1right], f'(c_1) = frac{f(1)-f(0)}{1-0}=1$ and there is some point $c_2 in [c_1,1]$ such that $f''(c_2) = frac{f'(1)-f'(c_1)}{1-c_1}=frac{-1}{1-c_1}<-1$ and there is some point $c_3in[c_2,1]$ such that $f'''(c_3) = frac{f''(1)-f''(c_2)}{1-c_2}>1$. I'm not sure how to extend this to $24$ though, or even if this method will work.



Is there a way to use Taylor Polynomials perhaps?







derivatives






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edited Nov 25 at 20:37

























asked Nov 24 at 20:40









You Zhou

406




406












  • What have you tried?
    – saulspatz
    Nov 24 at 20:45










  • I've tried using the mean value theorem but couldn't come up with anything.
    – You Zhou
    Nov 24 at 20:55










  • I don't know how to do it off the top of my head, but I suspect that Taylor polynomials come into it. I advise you to edit the question to explain what you have tried. The question may well be closed if you don't give any more context than you have. Many people will vote to close without reading the comments, so please edit the question body.
    – saulspatz
    Nov 24 at 21:04






  • 2




    The function $f(x)=6x^2-8x^3+3x^4$ satisfies the conditions. Note that $f'''(x)=-48+72x$, so there is no $c in (0,1)$ where $f$ has large enough third derivative, but $f'''(1)=24$. That is to say, I think there's a typo somewhere and $c$ should be allowed to be an endpoint.
    – Micah
    Nov 24 at 22:20






  • 3




    @Micah If $g(x)=6 x^2 - 8 x^3 + 3 x^4$ then $g''(x)=12 - 48x + 36x^2$ and $g''(0)=12neq0,$ so this does not satisfy all the conditions.
    – David K
    Nov 25 at 20:12


















  • What have you tried?
    – saulspatz
    Nov 24 at 20:45










  • I've tried using the mean value theorem but couldn't come up with anything.
    – You Zhou
    Nov 24 at 20:55










  • I don't know how to do it off the top of my head, but I suspect that Taylor polynomials come into it. I advise you to edit the question to explain what you have tried. The question may well be closed if you don't give any more context than you have. Many people will vote to close without reading the comments, so please edit the question body.
    – saulspatz
    Nov 24 at 21:04






  • 2




    The function $f(x)=6x^2-8x^3+3x^4$ satisfies the conditions. Note that $f'''(x)=-48+72x$, so there is no $c in (0,1)$ where $f$ has large enough third derivative, but $f'''(1)=24$. That is to say, I think there's a typo somewhere and $c$ should be allowed to be an endpoint.
    – Micah
    Nov 24 at 22:20






  • 3




    @Micah If $g(x)=6 x^2 - 8 x^3 + 3 x^4$ then $g''(x)=12 - 48x + 36x^2$ and $g''(0)=12neq0,$ so this does not satisfy all the conditions.
    – David K
    Nov 25 at 20:12
















What have you tried?
– saulspatz
Nov 24 at 20:45




What have you tried?
– saulspatz
Nov 24 at 20:45












I've tried using the mean value theorem but couldn't come up with anything.
– You Zhou
Nov 24 at 20:55




I've tried using the mean value theorem but couldn't come up with anything.
– You Zhou
Nov 24 at 20:55












I don't know how to do it off the top of my head, but I suspect that Taylor polynomials come into it. I advise you to edit the question to explain what you have tried. The question may well be closed if you don't give any more context than you have. Many people will vote to close without reading the comments, so please edit the question body.
– saulspatz
Nov 24 at 21:04




I don't know how to do it off the top of my head, but I suspect that Taylor polynomials come into it. I advise you to edit the question to explain what you have tried. The question may well be closed if you don't give any more context than you have. Many people will vote to close without reading the comments, so please edit the question body.
– saulspatz
Nov 24 at 21:04




2




2




The function $f(x)=6x^2-8x^3+3x^4$ satisfies the conditions. Note that $f'''(x)=-48+72x$, so there is no $c in (0,1)$ where $f$ has large enough third derivative, but $f'''(1)=24$. That is to say, I think there's a typo somewhere and $c$ should be allowed to be an endpoint.
– Micah
Nov 24 at 22:20




The function $f(x)=6x^2-8x^3+3x^4$ satisfies the conditions. Note that $f'''(x)=-48+72x$, so there is no $c in (0,1)$ where $f$ has large enough third derivative, but $f'''(1)=24$. That is to say, I think there's a typo somewhere and $c$ should be allowed to be an endpoint.
– Micah
Nov 24 at 22:20




3




3




@Micah If $g(x)=6 x^2 - 8 x^3 + 3 x^4$ then $g''(x)=12 - 48x + 36x^2$ and $g''(0)=12neq0,$ so this does not satisfy all the conditions.
– David K
Nov 25 at 20:12




@Micah If $g(x)=6 x^2 - 8 x^3 + 3 x^4$ then $g''(x)=12 - 48x + 36x^2$ and $g''(0)=12neq0,$ so this does not satisfy all the conditions.
– David K
Nov 25 at 20:12










2 Answers
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I think the idea is that if $f'''(x) < 24$ when $0 < x < 1,$ then
the conditions $f'(0)=f''(0)=0$ imply that $f'(x) < 12x^2$
when $0 < x leq frac12,$
the conditions $f'(1)=f''(1)=0$ imply that $f'(x) < 12(x-1)^2$
when $frac12 leq x < 1,$
and together these imply that $f(1) - f(0) < 1.$



Note that in order for $f'(x)$ to get close to these limits when $x approxfrac12,$
we need $f''(x)$ to change quickly from something near $12$ to something near $-12,$
which requires $f'''(x)$ to be much less than $-24.$
If the conclusion were of the form $lvert f'''(x)rvert geq L$
then I think we could set $L = 32,$ though this seems a bit harder to prove.






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  • How on Earth this answer did not receive any upvote yet? +1
    – Oldboy
    Nov 26 at 21:48



















0














Here's a proof.



We will prove it by contradiction. Let us assume that $f(x)$ satisfies the conditions, and that $f'''(x) < 24$ for $0 < x < 1$.



First, we show that we can assume that the function is symmetric around the point $(frac{1}{2},frac{1}{2})$.



We first note that if a function $f$ satisfies the conditions (including $f'''(x) < 24$), then $g(x) = 1 - f(1-x)$ also does.
Why is this true? Because when you reflect horizontally — taking $f(x)$ to $f(1-x)$ — you negate all the odd derivatives. And when you reflect vertically — taking $f(1-x)$ to $1-f(1-x)$ — you negate all the derivatives. So $g'''(x) = f'''(1-x) < 24$ as well.



Now, consider the function $h(x) = frac{f(x) + g(x)}{2}$. This is symmetric around the point $(frac{1}{2},frac{1}{2})$, and also satisfies the conditions. So $h(x)$ is also a counterexample, and $h(frac{1}{2})=frac{1}{2}$.



Consider the function $p(x) = 4x^3, 0 leq x leq frac{1}{2}$.



This has $p(frac{1}{2})=frac{1}{2}$, the third derivative $p'''$ is 24 for $0 leq x leq frac{1}{2}$, and $p(0)=p'(0) = p''(0)$. So if $h'''(x) < 24$ for $0 leq x leq frac{1}{2}$, then $h(x) < p(x)$ for $0 < x leq frac{1}{2}$. This contradicts $h(frac{1}{2})=frac{1}{2}$.






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    2 Answers
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    2 Answers
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    active

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    I think the idea is that if $f'''(x) < 24$ when $0 < x < 1,$ then
    the conditions $f'(0)=f''(0)=0$ imply that $f'(x) < 12x^2$
    when $0 < x leq frac12,$
    the conditions $f'(1)=f''(1)=0$ imply that $f'(x) < 12(x-1)^2$
    when $frac12 leq x < 1,$
    and together these imply that $f(1) - f(0) < 1.$



    Note that in order for $f'(x)$ to get close to these limits when $x approxfrac12,$
    we need $f''(x)$ to change quickly from something near $12$ to something near $-12,$
    which requires $f'''(x)$ to be much less than $-24.$
    If the conclusion were of the form $lvert f'''(x)rvert geq L$
    then I think we could set $L = 32,$ though this seems a bit harder to prove.






    share|cite|improve this answer





















    • How on Earth this answer did not receive any upvote yet? +1
      – Oldboy
      Nov 26 at 21:48
















    0














    I think the idea is that if $f'''(x) < 24$ when $0 < x < 1,$ then
    the conditions $f'(0)=f''(0)=0$ imply that $f'(x) < 12x^2$
    when $0 < x leq frac12,$
    the conditions $f'(1)=f''(1)=0$ imply that $f'(x) < 12(x-1)^2$
    when $frac12 leq x < 1,$
    and together these imply that $f(1) - f(0) < 1.$



    Note that in order for $f'(x)$ to get close to these limits when $x approxfrac12,$
    we need $f''(x)$ to change quickly from something near $12$ to something near $-12,$
    which requires $f'''(x)$ to be much less than $-24.$
    If the conclusion were of the form $lvert f'''(x)rvert geq L$
    then I think we could set $L = 32,$ though this seems a bit harder to prove.






    share|cite|improve this answer





















    • How on Earth this answer did not receive any upvote yet? +1
      – Oldboy
      Nov 26 at 21:48














    0












    0








    0






    I think the idea is that if $f'''(x) < 24$ when $0 < x < 1,$ then
    the conditions $f'(0)=f''(0)=0$ imply that $f'(x) < 12x^2$
    when $0 < x leq frac12,$
    the conditions $f'(1)=f''(1)=0$ imply that $f'(x) < 12(x-1)^2$
    when $frac12 leq x < 1,$
    and together these imply that $f(1) - f(0) < 1.$



    Note that in order for $f'(x)$ to get close to these limits when $x approxfrac12,$
    we need $f''(x)$ to change quickly from something near $12$ to something near $-12,$
    which requires $f'''(x)$ to be much less than $-24.$
    If the conclusion were of the form $lvert f'''(x)rvert geq L$
    then I think we could set $L = 32,$ though this seems a bit harder to prove.






    share|cite|improve this answer












    I think the idea is that if $f'''(x) < 24$ when $0 < x < 1,$ then
    the conditions $f'(0)=f''(0)=0$ imply that $f'(x) < 12x^2$
    when $0 < x leq frac12,$
    the conditions $f'(1)=f''(1)=0$ imply that $f'(x) < 12(x-1)^2$
    when $frac12 leq x < 1,$
    and together these imply that $f(1) - f(0) < 1.$



    Note that in order for $f'(x)$ to get close to these limits when $x approxfrac12,$
    we need $f''(x)$ to change quickly from something near $12$ to something near $-12,$
    which requires $f'''(x)$ to be much less than $-24.$
    If the conclusion were of the form $lvert f'''(x)rvert geq L$
    then I think we could set $L = 32,$ though this seems a bit harder to prove.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 26 at 12:56









    David K

    52.6k340115




    52.6k340115












    • How on Earth this answer did not receive any upvote yet? +1
      – Oldboy
      Nov 26 at 21:48


















    • How on Earth this answer did not receive any upvote yet? +1
      – Oldboy
      Nov 26 at 21:48
















    How on Earth this answer did not receive any upvote yet? +1
    – Oldboy
    Nov 26 at 21:48




    How on Earth this answer did not receive any upvote yet? +1
    – Oldboy
    Nov 26 at 21:48











    0














    Here's a proof.



    We will prove it by contradiction. Let us assume that $f(x)$ satisfies the conditions, and that $f'''(x) < 24$ for $0 < x < 1$.



    First, we show that we can assume that the function is symmetric around the point $(frac{1}{2},frac{1}{2})$.



    We first note that if a function $f$ satisfies the conditions (including $f'''(x) < 24$), then $g(x) = 1 - f(1-x)$ also does.
    Why is this true? Because when you reflect horizontally — taking $f(x)$ to $f(1-x)$ — you negate all the odd derivatives. And when you reflect vertically — taking $f(1-x)$ to $1-f(1-x)$ — you negate all the derivatives. So $g'''(x) = f'''(1-x) < 24$ as well.



    Now, consider the function $h(x) = frac{f(x) + g(x)}{2}$. This is symmetric around the point $(frac{1}{2},frac{1}{2})$, and also satisfies the conditions. So $h(x)$ is also a counterexample, and $h(frac{1}{2})=frac{1}{2}$.



    Consider the function $p(x) = 4x^3, 0 leq x leq frac{1}{2}$.



    This has $p(frac{1}{2})=frac{1}{2}$, the third derivative $p'''$ is 24 for $0 leq x leq frac{1}{2}$, and $p(0)=p'(0) = p''(0)$. So if $h'''(x) < 24$ for $0 leq x leq frac{1}{2}$, then $h(x) < p(x)$ for $0 < x leq frac{1}{2}$. This contradicts $h(frac{1}{2})=frac{1}{2}$.






    share|cite|improve this answer




























      0














      Here's a proof.



      We will prove it by contradiction. Let us assume that $f(x)$ satisfies the conditions, and that $f'''(x) < 24$ for $0 < x < 1$.



      First, we show that we can assume that the function is symmetric around the point $(frac{1}{2},frac{1}{2})$.



      We first note that if a function $f$ satisfies the conditions (including $f'''(x) < 24$), then $g(x) = 1 - f(1-x)$ also does.
      Why is this true? Because when you reflect horizontally — taking $f(x)$ to $f(1-x)$ — you negate all the odd derivatives. And when you reflect vertically — taking $f(1-x)$ to $1-f(1-x)$ — you negate all the derivatives. So $g'''(x) = f'''(1-x) < 24$ as well.



      Now, consider the function $h(x) = frac{f(x) + g(x)}{2}$. This is symmetric around the point $(frac{1}{2},frac{1}{2})$, and also satisfies the conditions. So $h(x)$ is also a counterexample, and $h(frac{1}{2})=frac{1}{2}$.



      Consider the function $p(x) = 4x^3, 0 leq x leq frac{1}{2}$.



      This has $p(frac{1}{2})=frac{1}{2}$, the third derivative $p'''$ is 24 for $0 leq x leq frac{1}{2}$, and $p(0)=p'(0) = p''(0)$. So if $h'''(x) < 24$ for $0 leq x leq frac{1}{2}$, then $h(x) < p(x)$ for $0 < x leq frac{1}{2}$. This contradicts $h(frac{1}{2})=frac{1}{2}$.






      share|cite|improve this answer


























        0












        0








        0






        Here's a proof.



        We will prove it by contradiction. Let us assume that $f(x)$ satisfies the conditions, and that $f'''(x) < 24$ for $0 < x < 1$.



        First, we show that we can assume that the function is symmetric around the point $(frac{1}{2},frac{1}{2})$.



        We first note that if a function $f$ satisfies the conditions (including $f'''(x) < 24$), then $g(x) = 1 - f(1-x)$ also does.
        Why is this true? Because when you reflect horizontally — taking $f(x)$ to $f(1-x)$ — you negate all the odd derivatives. And when you reflect vertically — taking $f(1-x)$ to $1-f(1-x)$ — you negate all the derivatives. So $g'''(x) = f'''(1-x) < 24$ as well.



        Now, consider the function $h(x) = frac{f(x) + g(x)}{2}$. This is symmetric around the point $(frac{1}{2},frac{1}{2})$, and also satisfies the conditions. So $h(x)$ is also a counterexample, and $h(frac{1}{2})=frac{1}{2}$.



        Consider the function $p(x) = 4x^3, 0 leq x leq frac{1}{2}$.



        This has $p(frac{1}{2})=frac{1}{2}$, the third derivative $p'''$ is 24 for $0 leq x leq frac{1}{2}$, and $p(0)=p'(0) = p''(0)$. So if $h'''(x) < 24$ for $0 leq x leq frac{1}{2}$, then $h(x) < p(x)$ for $0 < x leq frac{1}{2}$. This contradicts $h(frac{1}{2})=frac{1}{2}$.






        share|cite|improve this answer














        Here's a proof.



        We will prove it by contradiction. Let us assume that $f(x)$ satisfies the conditions, and that $f'''(x) < 24$ for $0 < x < 1$.



        First, we show that we can assume that the function is symmetric around the point $(frac{1}{2},frac{1}{2})$.



        We first note that if a function $f$ satisfies the conditions (including $f'''(x) < 24$), then $g(x) = 1 - f(1-x)$ also does.
        Why is this true? Because when you reflect horizontally — taking $f(x)$ to $f(1-x)$ — you negate all the odd derivatives. And when you reflect vertically — taking $f(1-x)$ to $1-f(1-x)$ — you negate all the derivatives. So $g'''(x) = f'''(1-x) < 24$ as well.



        Now, consider the function $h(x) = frac{f(x) + g(x)}{2}$. This is symmetric around the point $(frac{1}{2},frac{1}{2})$, and also satisfies the conditions. So $h(x)$ is also a counterexample, and $h(frac{1}{2})=frac{1}{2}$.



        Consider the function $p(x) = 4x^3, 0 leq x leq frac{1}{2}$.



        This has $p(frac{1}{2})=frac{1}{2}$, the third derivative $p'''$ is 24 for $0 leq x leq frac{1}{2}$, and $p(0)=p'(0) = p''(0)$. So if $h'''(x) < 24$ for $0 leq x leq frac{1}{2}$, then $h(x) < p(x)$ for $0 < x leq frac{1}{2}$. This contradicts $h(frac{1}{2})=frac{1}{2}$.







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        edited Dec 3 at 3:07

























        answered Nov 29 at 2:07









        Peter Shor

        2,6001622




        2,6001622






























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