Spivak, A Twice Differentiable Function












0












$begingroup$


I am currently working on Spivak. I am in need of help or a hint on this question, because every which way I try this question whether it be with Cauchy MVT or just plain MVT, I get to some promising point and end up coming back out with nothing. I've been trying at this problem since last Friday.



The problem:




Prove that if $f$ is a twice differentiable function with $f(0)=0$ and
$f(1)=1$ and $f'(0) = f'(1) = 0$, then $|f''(x)|geq4$ for some $x in (0, 1).$
In more picturesque terms: A particle which travels a unit distance in
a unit time, and starts and ends with velocity $0$, has at some time an
acceleration $>4$. Hint: Prove that either $f''(x)geq4$ for some $xin(0,1)$,
or else $f''(x)leq4$ for some $xin(0,1)$.




I really don't know how else to approach this. I've tried arbitrary intervals say $(0,x)$, intervals of $(0,1/2)$ and $(1/2,1)$ via Cauchy MVT and 'regular' MVT, I've tried his "hint", I've tried breaking it into cases where $f'(1/2)<1$, $f'(1/2)>1$ and $f'(1/2)=1$ with no real luck.



I'm sure there's something I'm not seeing that I should be seeing, but I figure after a week I should probably ask around, say, not at school. Any help is much appreciated.






calculus derivatives inequality







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I am currently working on Spivak. I am in need of help or a hint on this question, because every which way I try this question whether it be with Cauchy MVT or just plain MVT, I get to some promising point and end up coming back out with nothing. I've been trying at this problem since last Friday.



    The problem:




    Prove that if $f$ is a twice differentiable function with $f(0)=0$ and
    $f(1)=1$ and $f'(0) = f'(1) = 0$, then $|f''(x)|geq4$ for some $x in (0, 1).$
    In more picturesque terms: A particle which travels a unit distance in
    a unit time, and starts and ends with velocity $0$, has at some time an
    acceleration $>4$. Hint: Prove that either $f''(x)geq4$ for some $xin(0,1)$,
    or else $f''(x)leq4$ for some $xin(0,1)$.




    I really don't know how else to approach this. I've tried arbitrary intervals say $(0,x)$, intervals of $(0,1/2)$ and $(1/2,1)$ via Cauchy MVT and 'regular' MVT, I've tried his "hint", I've tried breaking it into cases where $f'(1/2)<1$, $f'(1/2)>1$ and $f'(1/2)=1$ with no real luck.



    I'm sure there's something I'm not seeing that I should be seeing, but I figure after a week I should probably ask around, say, not at school. Any help is much appreciated.






    calculus derivatives inequality







    share|cite|improve this question











    $endgroup$















      0












      0








      0


      1



      $begingroup$


      I am currently working on Spivak. I am in need of help or a hint on this question, because every which way I try this question whether it be with Cauchy MVT or just plain MVT, I get to some promising point and end up coming back out with nothing. I've been trying at this problem since last Friday.



      The problem:




      Prove that if $f$ is a twice differentiable function with $f(0)=0$ and
      $f(1)=1$ and $f'(0) = f'(1) = 0$, then $|f''(x)|geq4$ for some $x in (0, 1).$
      In more picturesque terms: A particle which travels a unit distance in
      a unit time, and starts and ends with velocity $0$, has at some time an
      acceleration $>4$. Hint: Prove that either $f''(x)geq4$ for some $xin(0,1)$,
      or else $f''(x)leq4$ for some $xin(0,1)$.




      I really don't know how else to approach this. I've tried arbitrary intervals say $(0,x)$, intervals of $(0,1/2)$ and $(1/2,1)$ via Cauchy MVT and 'regular' MVT, I've tried his "hint", I've tried breaking it into cases where $f'(1/2)<1$, $f'(1/2)>1$ and $f'(1/2)=1$ with no real luck.



      I'm sure there's something I'm not seeing that I should be seeing, but I figure after a week I should probably ask around, say, not at school. Any help is much appreciated.






      calculus derivatives inequality







      share|cite|improve this question











      $endgroup$




      I am currently working on Spivak. I am in need of help or a hint on this question, because every which way I try this question whether it be with Cauchy MVT or just plain MVT, I get to some promising point and end up coming back out with nothing. I've been trying at this problem since last Friday.



      The problem:




      Prove that if $f$ is a twice differentiable function with $f(0)=0$ and
      $f(1)=1$ and $f'(0) = f'(1) = 0$, then $|f''(x)|geq4$ for some $x in (0, 1).$
      In more picturesque terms: A particle which travels a unit distance in
      a unit time, and starts and ends with velocity $0$, has at some time an
      acceleration $>4$. Hint: Prove that either $f''(x)geq4$ for some $xin(0,1)$,
      or else $f''(x)leq4$ for some $xin(0,1)$.




      I really don't know how else to approach this. I've tried arbitrary intervals say $(0,x)$, intervals of $(0,1/2)$ and $(1/2,1)$ via Cauchy MVT and 'regular' MVT, I've tried his "hint", I've tried breaking it into cases where $f'(1/2)<1$, $f'(1/2)>1$ and $f'(1/2)=1$ with no real luck.



      I'm sure there's something I'm not seeing that I should be seeing, but I figure after a week I should probably ask around, say, not at school. Any help is much appreciated.






      calculus derivatives inequality




      calculus derivatives inequality






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 10:36









      egreg

      181k1485203




      181k1485203










      asked Dec 8 '18 at 7:48









      kyle campbellkyle campbell

      725




      725






















          1 Answer
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          $begingroup$

          If $bigl(forall xin[0,1]bigr):bigllvert f''(x)bigrrvert<4$, then there a $x_0inleft[0,frac12right]$ such thatbegin{align}fleft(frac12right)&=f(0)+frac12f'(0)+left(frac12right)^2frac{f''(x_0)}2\&=left(frac12right)^2frac{f''(x_0)}2\&=frac{f''(x_0)}8\&<frac48\&=frac12end{align}and there is a $x_1inleft[frac12,1right]$ such thatbegin{align}fleft(frac12right)&=f(1)+left(frac12-1right)f'(1)+left(frac12-1right)^2frac{f''(x_1)}2\&= 1+frac{f''(x_1)}8\&>1-frac48\&=frac12.end{align}






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Hi Jose! Thank you. Are you using Taylor's remainder theorem here? Is there any way to express this only using Cauchy MVT &/or MVT? I believe that this theorem isn't introduced for another 200 or so pages in Spivak. Very slick though.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 8:26










          • $begingroup$
            You apply the MVT to $f'$: if $f''(x)<4$ for all $xin[0,1]$, then$$frac{f'(x)-f'(0)}x<4$$and therefore $f'(x)<4x$. So, again by MVT$$frac{f(x)-f(0)}x<4x$$and so $f(x)<4x^2$. In particular, $fleft(frac12right)<frac12$.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 8:37












          • $begingroup$
            Right, so that's where I was at originally, but how are you getting $f(1/2)<1/2$? I don't see how it follow that if $f(x)<4x^2$ then $f(1/2)<1/2$? The lowest bound I see for $f(1/2)$ if $f(x)<4x^2$ is $f(1/2)<4cdot(1/2)^2=1$.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 8:41










          • $begingroup$
            Sorry, it doesn't follow from what I wrote. My mistake. But then I went to see the answer book and… it has the same answer (with more details)! Perhaps that Spivak made an error here.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 9:00






          • 1




            $begingroup$
            Indeed, I'm still not sure how to proceed with the question. I enjoy the Taylor method, but I figure since only MVT has been introduced I suspect you should be able to prove it with only those tools introduced thusfar.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 9:38











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          2












          $begingroup$

          If $bigl(forall xin[0,1]bigr):bigllvert f''(x)bigrrvert<4$, then there a $x_0inleft[0,frac12right]$ such thatbegin{align}fleft(frac12right)&=f(0)+frac12f'(0)+left(frac12right)^2frac{f''(x_0)}2\&=left(frac12right)^2frac{f''(x_0)}2\&=frac{f''(x_0)}8\&<frac48\&=frac12end{align}and there is a $x_1inleft[frac12,1right]$ such thatbegin{align}fleft(frac12right)&=f(1)+left(frac12-1right)f'(1)+left(frac12-1right)^2frac{f''(x_1)}2\&= 1+frac{f''(x_1)}8\&>1-frac48\&=frac12.end{align}






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Hi Jose! Thank you. Are you using Taylor's remainder theorem here? Is there any way to express this only using Cauchy MVT &/or MVT? I believe that this theorem isn't introduced for another 200 or so pages in Spivak. Very slick though.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 8:26










          • $begingroup$
            You apply the MVT to $f'$: if $f''(x)<4$ for all $xin[0,1]$, then$$frac{f'(x)-f'(0)}x<4$$and therefore $f'(x)<4x$. So, again by MVT$$frac{f(x)-f(0)}x<4x$$and so $f(x)<4x^2$. In particular, $fleft(frac12right)<frac12$.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 8:37












          • $begingroup$
            Right, so that's where I was at originally, but how are you getting $f(1/2)<1/2$? I don't see how it follow that if $f(x)<4x^2$ then $f(1/2)<1/2$? The lowest bound I see for $f(1/2)$ if $f(x)<4x^2$ is $f(1/2)<4cdot(1/2)^2=1$.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 8:41










          • $begingroup$
            Sorry, it doesn't follow from what I wrote. My mistake. But then I went to see the answer book and… it has the same answer (with more details)! Perhaps that Spivak made an error here.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 9:00






          • 1




            $begingroup$
            Indeed, I'm still not sure how to proceed with the question. I enjoy the Taylor method, but I figure since only MVT has been introduced I suspect you should be able to prove it with only those tools introduced thusfar.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 9:38
















          2












          $begingroup$

          If $bigl(forall xin[0,1]bigr):bigllvert f''(x)bigrrvert<4$, then there a $x_0inleft[0,frac12right]$ such thatbegin{align}fleft(frac12right)&=f(0)+frac12f'(0)+left(frac12right)^2frac{f''(x_0)}2\&=left(frac12right)^2frac{f''(x_0)}2\&=frac{f''(x_0)}8\&<frac48\&=frac12end{align}and there is a $x_1inleft[frac12,1right]$ such thatbegin{align}fleft(frac12right)&=f(1)+left(frac12-1right)f'(1)+left(frac12-1right)^2frac{f''(x_1)}2\&= 1+frac{f''(x_1)}8\&>1-frac48\&=frac12.end{align}






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            Hi Jose! Thank you. Are you using Taylor's remainder theorem here? Is there any way to express this only using Cauchy MVT &/or MVT? I believe that this theorem isn't introduced for another 200 or so pages in Spivak. Very slick though.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 8:26










          • $begingroup$
            You apply the MVT to $f'$: if $f''(x)<4$ for all $xin[0,1]$, then$$frac{f'(x)-f'(0)}x<4$$and therefore $f'(x)<4x$. So, again by MVT$$frac{f(x)-f(0)}x<4x$$and so $f(x)<4x^2$. In particular, $fleft(frac12right)<frac12$.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 8:37












          • $begingroup$
            Right, so that's where I was at originally, but how are you getting $f(1/2)<1/2$? I don't see how it follow that if $f(x)<4x^2$ then $f(1/2)<1/2$? The lowest bound I see for $f(1/2)$ if $f(x)<4x^2$ is $f(1/2)<4cdot(1/2)^2=1$.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 8:41










          • $begingroup$
            Sorry, it doesn't follow from what I wrote. My mistake. But then I went to see the answer book and… it has the same answer (with more details)! Perhaps that Spivak made an error here.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 9:00






          • 1




            $begingroup$
            Indeed, I'm still not sure how to proceed with the question. I enjoy the Taylor method, but I figure since only MVT has been introduced I suspect you should be able to prove it with only those tools introduced thusfar.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 9:38














          2












          2








          2





          $begingroup$

          If $bigl(forall xin[0,1]bigr):bigllvert f''(x)bigrrvert<4$, then there a $x_0inleft[0,frac12right]$ such thatbegin{align}fleft(frac12right)&=f(0)+frac12f'(0)+left(frac12right)^2frac{f''(x_0)}2\&=left(frac12right)^2frac{f''(x_0)}2\&=frac{f''(x_0)}8\&<frac48\&=frac12end{align}and there is a $x_1inleft[frac12,1right]$ such thatbegin{align}fleft(frac12right)&=f(1)+left(frac12-1right)f'(1)+left(frac12-1right)^2frac{f''(x_1)}2\&= 1+frac{f''(x_1)}8\&>1-frac48\&=frac12.end{align}






          share|cite|improve this answer









          $endgroup$



          If $bigl(forall xin[0,1]bigr):bigllvert f''(x)bigrrvert<4$, then there a $x_0inleft[0,frac12right]$ such thatbegin{align}fleft(frac12right)&=f(0)+frac12f'(0)+left(frac12right)^2frac{f''(x_0)}2\&=left(frac12right)^2frac{f''(x_0)}2\&=frac{f''(x_0)}8\&<frac48\&=frac12end{align}and there is a $x_1inleft[frac12,1right]$ such thatbegin{align}fleft(frac12right)&=f(1)+left(frac12-1right)f'(1)+left(frac12-1right)^2frac{f''(x_1)}2\&= 1+frac{f''(x_1)}8\&>1-frac48\&=frac12.end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 8:18









          José Carlos SantosJosé Carlos Santos

          160k22127232




          160k22127232








          • 1




            $begingroup$
            Hi Jose! Thank you. Are you using Taylor's remainder theorem here? Is there any way to express this only using Cauchy MVT &/or MVT? I believe that this theorem isn't introduced for another 200 or so pages in Spivak. Very slick though.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 8:26










          • $begingroup$
            You apply the MVT to $f'$: if $f''(x)<4$ for all $xin[0,1]$, then$$frac{f'(x)-f'(0)}x<4$$and therefore $f'(x)<4x$. So, again by MVT$$frac{f(x)-f(0)}x<4x$$and so $f(x)<4x^2$. In particular, $fleft(frac12right)<frac12$.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 8:37












          • $begingroup$
            Right, so that's where I was at originally, but how are you getting $f(1/2)<1/2$? I don't see how it follow that if $f(x)<4x^2$ then $f(1/2)<1/2$? The lowest bound I see for $f(1/2)$ if $f(x)<4x^2$ is $f(1/2)<4cdot(1/2)^2=1$.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 8:41










          • $begingroup$
            Sorry, it doesn't follow from what I wrote. My mistake. But then I went to see the answer book and… it has the same answer (with more details)! Perhaps that Spivak made an error here.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 9:00






          • 1




            $begingroup$
            Indeed, I'm still not sure how to proceed with the question. I enjoy the Taylor method, but I figure since only MVT has been introduced I suspect you should be able to prove it with only those tools introduced thusfar.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 9:38














          • 1




            $begingroup$
            Hi Jose! Thank you. Are you using Taylor's remainder theorem here? Is there any way to express this only using Cauchy MVT &/or MVT? I believe that this theorem isn't introduced for another 200 or so pages in Spivak. Very slick though.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 8:26










          • $begingroup$
            You apply the MVT to $f'$: if $f''(x)<4$ for all $xin[0,1]$, then$$frac{f'(x)-f'(0)}x<4$$and therefore $f'(x)<4x$. So, again by MVT$$frac{f(x)-f(0)}x<4x$$and so $f(x)<4x^2$. In particular, $fleft(frac12right)<frac12$.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 8:37












          • $begingroup$
            Right, so that's where I was at originally, but how are you getting $f(1/2)<1/2$? I don't see how it follow that if $f(x)<4x^2$ then $f(1/2)<1/2$? The lowest bound I see for $f(1/2)$ if $f(x)<4x^2$ is $f(1/2)<4cdot(1/2)^2=1$.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 8:41










          • $begingroup$
            Sorry, it doesn't follow from what I wrote. My mistake. But then I went to see the answer book and… it has the same answer (with more details)! Perhaps that Spivak made an error here.
            $endgroup$
            – José Carlos Santos
            Dec 8 '18 at 9:00






          • 1




            $begingroup$
            Indeed, I'm still not sure how to proceed with the question. I enjoy the Taylor method, but I figure since only MVT has been introduced I suspect you should be able to prove it with only those tools introduced thusfar.
            $endgroup$
            – kyle campbell
            Dec 8 '18 at 9:38








          1




          1




          $begingroup$
          Hi Jose! Thank you. Are you using Taylor's remainder theorem here? Is there any way to express this only using Cauchy MVT &/or MVT? I believe that this theorem isn't introduced for another 200 or so pages in Spivak. Very slick though.
          $endgroup$
          – kyle campbell
          Dec 8 '18 at 8:26




          $begingroup$
          Hi Jose! Thank you. Are you using Taylor's remainder theorem here? Is there any way to express this only using Cauchy MVT &/or MVT? I believe that this theorem isn't introduced for another 200 or so pages in Spivak. Very slick though.
          $endgroup$
          – kyle campbell
          Dec 8 '18 at 8:26












          $begingroup$
          You apply the MVT to $f'$: if $f''(x)<4$ for all $xin[0,1]$, then$$frac{f'(x)-f'(0)}x<4$$and therefore $f'(x)<4x$. So, again by MVT$$frac{f(x)-f(0)}x<4x$$and so $f(x)<4x^2$. In particular, $fleft(frac12right)<frac12$.
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 8:37






          $begingroup$
          You apply the MVT to $f'$: if $f''(x)<4$ for all $xin[0,1]$, then$$frac{f'(x)-f'(0)}x<4$$and therefore $f'(x)<4x$. So, again by MVT$$frac{f(x)-f(0)}x<4x$$and so $f(x)<4x^2$. In particular, $fleft(frac12right)<frac12$.
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 8:37














          $begingroup$
          Right, so that's where I was at originally, but how are you getting $f(1/2)<1/2$? I don't see how it follow that if $f(x)<4x^2$ then $f(1/2)<1/2$? The lowest bound I see for $f(1/2)$ if $f(x)<4x^2$ is $f(1/2)<4cdot(1/2)^2=1$.
          $endgroup$
          – kyle campbell
          Dec 8 '18 at 8:41




          $begingroup$
          Right, so that's where I was at originally, but how are you getting $f(1/2)<1/2$? I don't see how it follow that if $f(x)<4x^2$ then $f(1/2)<1/2$? The lowest bound I see for $f(1/2)$ if $f(x)<4x^2$ is $f(1/2)<4cdot(1/2)^2=1$.
          $endgroup$
          – kyle campbell
          Dec 8 '18 at 8:41












          $begingroup$
          Sorry, it doesn't follow from what I wrote. My mistake. But then I went to see the answer book and… it has the same answer (with more details)! Perhaps that Spivak made an error here.
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 9:00




          $begingroup$
          Sorry, it doesn't follow from what I wrote. My mistake. But then I went to see the answer book and… it has the same answer (with more details)! Perhaps that Spivak made an error here.
          $endgroup$
          – José Carlos Santos
          Dec 8 '18 at 9:00




          1




          1




          $begingroup$
          Indeed, I'm still not sure how to proceed with the question. I enjoy the Taylor method, but I figure since only MVT has been introduced I suspect you should be able to prove it with only those tools introduced thusfar.
          $endgroup$
          – kyle campbell
          Dec 8 '18 at 9:38




          $begingroup$
          Indeed, I'm still not sure how to proceed with the question. I enjoy the Taylor method, but I figure since only MVT has been introduced I suspect you should be able to prove it with only those tools introduced thusfar.
          $endgroup$
          – kyle campbell
          Dec 8 '18 at 9:38


















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