Theorem 5.8 Baby Rudin. Some questions












4














This is with reference to the page number 107 of Baby Rudin.




Theorem 5.8 Let $f$ be defined on $[a,b]$.If a point $xin(a,b)$ is a
local maximum of function $f$, and if $f'(x)$ exist, then $f'(x)=0$.




I have some questions. Thanks in advance for reading and helping out.




  1. What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?


  2. The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function function with local maximum/minimum at some point say $p$ but derivative undefined at the point.



Edits:



We can take $f(x)=x^2$ on $[-1,1]$. $f'(0)=0$ but $0$ is not local maximum. I think this works for (1).










share|cite|improve this question





























    4














    This is with reference to the page number 107 of Baby Rudin.




    Theorem 5.8 Let $f$ be defined on $[a,b]$.If a point $xin(a,b)$ is a
    local maximum of function $f$, and if $f'(x)$ exist, then $f'(x)=0$.




    I have some questions. Thanks in advance for reading and helping out.




    1. What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?


    2. The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function function with local maximum/minimum at some point say $p$ but derivative undefined at the point.



    Edits:



    We can take $f(x)=x^2$ on $[-1,1]$. $f'(0)=0$ but $0$ is not local maximum. I think this works for (1).










    share|cite|improve this question



























      4












      4








      4







      This is with reference to the page number 107 of Baby Rudin.




      Theorem 5.8 Let $f$ be defined on $[a,b]$.If a point $xin(a,b)$ is a
      local maximum of function $f$, and if $f'(x)$ exist, then $f'(x)=0$.




      I have some questions. Thanks in advance for reading and helping out.




      1. What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?


      2. The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function function with local maximum/minimum at some point say $p$ but derivative undefined at the point.



      Edits:



      We can take $f(x)=x^2$ on $[-1,1]$. $f'(0)=0$ but $0$ is not local maximum. I think this works for (1).










      share|cite|improve this question















      This is with reference to the page number 107 of Baby Rudin.




      Theorem 5.8 Let $f$ be defined on $[a,b]$.If a point $xin(a,b)$ is a
      local maximum of function $f$, and if $f'(x)$ exist, then $f'(x)=0$.




      I have some questions. Thanks in advance for reading and helping out.




      1. What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?


      2. The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function function with local maximum/minimum at some point say $p$ but derivative undefined at the point.



      Edits:



      We can take $f(x)=x^2$ on $[-1,1]$. $f'(0)=0$ but $0$ is not local maximum. I think this works for (1).







      real-analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited yesterday

























      asked yesterday









      StammeringMathematician

      2,2481322




      2,2481322






















          4 Answers
          4






          active

          oldest

          votes


















          9














          Answer to question 1



          The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.



          Answer to question 2



          Take $g(x)=vert x vert$, again look at zero.






          share|cite|improve this answer





























            5














            If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.



            For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.






            share|cite|improve this answer





























              2
















              1. What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?




              No. It could be a minimum, or it could be neither. Such examples:




              • $f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.


              • $f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.



              (I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)



              This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.







              1. The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?




              Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.






              share|cite|improve this answer





























                2














                Let me try:



                1) $f'(x)=0; $



                Example:$ f(x)=x^3$, at $x= 0$, inflection point, not an extremum



                2) $f(x)= |x|$ , has a local minimum at $x=0.$



                $f(x)= -|x|$ has a local maximum at $x=0$



                (f'(0) does not exist)(Why?)






                share|cite|improve this answer

















                • 1




                  $f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
                  – StammeringMathematician
                  yesterday










                • Stammering.Yes,exactly!!Greetings.
                  – Peter Szilas
                  yesterday











                Your Answer





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                4 Answers
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                4 Answers
                4






                active

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                active

                oldest

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                active

                oldest

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                9














                Answer to question 1



                The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.



                Answer to question 2



                Take $g(x)=vert x vert$, again look at zero.






                share|cite|improve this answer


























                  9














                  Answer to question 1



                  The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.



                  Answer to question 2



                  Take $g(x)=vert x vert$, again look at zero.






                  share|cite|improve this answer
























                    9












                    9








                    9






                    Answer to question 1



                    The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.



                    Answer to question 2



                    Take $g(x)=vert x vert$, again look at zero.






                    share|cite|improve this answer












                    Answer to question 1



                    The derivative of $f(x)=x^3$ vanishes at $0$ but $f$ doesn’t have a minimum nor at maximum at $0$.



                    Answer to question 2



                    Take $g(x)=vert x vert$, again look at zero.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered yesterday









                    mathcounterexamples.net

                    25.2k21953




                    25.2k21953























                        5














                        If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.



                        For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.






                        share|cite|improve this answer


























                          5














                          If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.



                          For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.






                          share|cite|improve this answer
























                            5












                            5








                            5






                            If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.



                            For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.






                            share|cite|improve this answer












                            If $f'(x)=0$, then $x$ is said to be a stationary point. It can be something like $x=0$ in $f(x)=x^3$ too, for example.



                            For (2), we can even have a function with a global maximum at $p$ but without a derivative in this point. See, for example, $f(x)=-|x|$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered yesterday









                            Gabriel Ribeiro

                            1,408422




                            1,408422























                                2
















                                1. What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?




                                No. It could be a minimum, or it could be neither. Such examples:




                                • $f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.


                                • $f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.



                                (I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)



                                This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.







                                1. The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?




                                Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.






                                share|cite|improve this answer


























                                  2
















                                  1. What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?




                                  No. It could be a minimum, or it could be neither. Such examples:




                                  • $f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.


                                  • $f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.



                                  (I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)



                                  This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.







                                  1. The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?




                                  Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.






                                  share|cite|improve this answer
























                                    2












                                    2








                                    2








                                    1. What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?




                                    No. It could be a minimum, or it could be neither. Such examples:




                                    • $f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.


                                    • $f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.



                                    (I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)



                                    This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.







                                    1. The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?




                                    Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.






                                    share|cite|improve this answer














                                    1. What about the converse of this theorem? Suppose $f'(x)=0$ at some point, can we always conclude that it is local maximum?




                                    No. It could be a minimum, or it could be neither. Such examples:




                                    • $f(x) = x^2$. Here, $f'(x) = 2x$ thus implying $f'(x) = 0$ for $x = 0$. It is immediately clear on looking at the graph that this is a local minimum for some intervals of the function.


                                    • $f(x) = x^3$. Here, $f'(x) = 3x^2$. Thus, $f'(x) = 0$ only if $x=0$. On looking at the graph, it is clearly not a local extremum in either respect, up to choice of the interval.



                                    (I say "up to choice of the interval" because $x=0$ yields a local minimum on the choice of interval $[0,1]$ for both functions, and on $[-1,0]$ for example the latter has a maximum instead. If we generally state intervals $[a,b]$ with $a<0<b$, though, they respectively have a minimum and neither respectively at $x=0$.)



                                    This touches on the nature of "critical points" you might have learned in your introductory calculus courses: points where $f'(x) = 0$ or is undefined are candidates for local extrema.







                                    1. The theorem add a condition that $f'(x)$ should exist. This raised a question, Can we have a function with local maximum/minimum at some point say $p$ but derivative undefined at the point?




                                    Yes. Consider $f(x) = |x|$ on the interval $[a,b]$ where $a<0<b$. The function clearly has a minimum at $x=0$, but the derivative is undefined at that point.







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered yesterday









                                    Eevee Trainer

                                    4,9971634




                                    4,9971634























                                        2














                                        Let me try:



                                        1) $f'(x)=0; $



                                        Example:$ f(x)=x^3$, at $x= 0$, inflection point, not an extremum



                                        2) $f(x)= |x|$ , has a local minimum at $x=0.$



                                        $f(x)= -|x|$ has a local maximum at $x=0$



                                        (f'(0) does not exist)(Why?)






                                        share|cite|improve this answer

















                                        • 1




                                          $f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
                                          – StammeringMathematician
                                          yesterday










                                        • Stammering.Yes,exactly!!Greetings.
                                          – Peter Szilas
                                          yesterday
















                                        2














                                        Let me try:



                                        1) $f'(x)=0; $



                                        Example:$ f(x)=x^3$, at $x= 0$, inflection point, not an extremum



                                        2) $f(x)= |x|$ , has a local minimum at $x=0.$



                                        $f(x)= -|x|$ has a local maximum at $x=0$



                                        (f'(0) does not exist)(Why?)






                                        share|cite|improve this answer

















                                        • 1




                                          $f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
                                          – StammeringMathematician
                                          yesterday










                                        • Stammering.Yes,exactly!!Greetings.
                                          – Peter Szilas
                                          yesterday














                                        2












                                        2








                                        2






                                        Let me try:



                                        1) $f'(x)=0; $



                                        Example:$ f(x)=x^3$, at $x= 0$, inflection point, not an extremum



                                        2) $f(x)= |x|$ , has a local minimum at $x=0.$



                                        $f(x)= -|x|$ has a local maximum at $x=0$



                                        (f'(0) does not exist)(Why?)






                                        share|cite|improve this answer












                                        Let me try:



                                        1) $f'(x)=0; $



                                        Example:$ f(x)=x^3$, at $x= 0$, inflection point, not an extremum



                                        2) $f(x)= |x|$ , has a local minimum at $x=0.$



                                        $f(x)= -|x|$ has a local maximum at $x=0$



                                        (f'(0) does not exist)(Why?)







                                        share|cite|improve this answer












                                        share|cite|improve this answer



                                        share|cite|improve this answer










                                        answered yesterday









                                        Peter Szilas

                                        10.7k2720




                                        10.7k2720








                                        • 1




                                          $f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
                                          – StammeringMathematician
                                          yesterday










                                        • Stammering.Yes,exactly!!Greetings.
                                          – Peter Szilas
                                          yesterday














                                        • 1




                                          $f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
                                          – StammeringMathematician
                                          yesterday










                                        • Stammering.Yes,exactly!!Greetings.
                                          – Peter Szilas
                                          yesterday








                                        1




                                        1




                                        $f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
                                        – StammeringMathematician
                                        yesterday




                                        $f'(0)$ does not exist because RHD and LHD are not equal. If we take difference coefficient $f(x)-f(0)/(x-0) $ then depending upon from where we approach $0$ we get two different limits. Thanks a lot.
                                        – StammeringMathematician
                                        yesterday












                                        Stammering.Yes,exactly!!Greetings.
                                        – Peter Szilas
                                        yesterday




                                        Stammering.Yes,exactly!!Greetings.
                                        – Peter Szilas
                                        yesterday


















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