Suppose that $f(x)$ is differentiable at any point and $f(0) = 0$ and $|f′(x)|leq 1$ . Prove that $|f(x)|...












-2














Suppose that:





  • $f(x)$ is differentiable at any point

  • $f(0) = 0$

  • $|f′(x)| ≤ 1$


How to prove $|f(x)| leq |x|$ for any $x$?






calculus







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put on hold as off-topic by Nosrati, Davide Giraudo, amWhy, mrtaurho, John Wayland Bales yesterday


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    -2














    Suppose that:





    • $f(x)$ is differentiable at any point

    • $f(0) = 0$

    • $|f′(x)| ≤ 1$


    How to prove $|f(x)| leq |x|$ for any $x$?






    calculus







    share|cite|improve this question









    New contributor




    Johnathan1994 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.











    put on hold as off-topic by Nosrati, Davide Giraudo, amWhy, mrtaurho, John Wayland Bales yesterday


    This question appears to be off-topic. The users who voted to close gave this specific reason:


    • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Davide Giraudo, amWhy, mrtaurho, John Wayland Bales

    If this question can be reworded to fit the rules in the help center, please edit the question.
















      -2












      -2








      -2







      Suppose that:





      • $f(x)$ is differentiable at any point

      • $f(0) = 0$

      • $|f′(x)| ≤ 1$


      How to prove $|f(x)| leq |x|$ for any $x$?






      calculus







      share|cite|improve this question









      New contributor




      Johnathan1994 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Suppose that:





      • $f(x)$ is differentiable at any point

      • $f(0) = 0$

      • $|f′(x)| ≤ 1$


      How to prove $|f(x)| leq |x|$ for any $x$?






      calculus




      calculus






      share|cite|improve this question









      New contributor




      Johnathan1994 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question









      New contributor




      Johnathan1994 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question








      edited 2 days ago









      amWhy

      192k28224439




      192k28224439






      New contributor




      Johnathan1994 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 2 days ago









      Johnathan1994

      1




      1




      New contributor




      Johnathan1994 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      New contributor





      Johnathan1994 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      Johnathan1994 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.




      put on hold as off-topic by Nosrati, Davide Giraudo, amWhy, mrtaurho, John Wayland Bales yesterday


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Davide Giraudo, amWhy, mrtaurho, John Wayland Bales

      If this question can be reworded to fit the rules in the help center, please edit the question.




      put on hold as off-topic by Nosrati, Davide Giraudo, amWhy, mrtaurho, John Wayland Bales yesterday


      This question appears to be off-topic. The users who voted to close gave this specific reason:


      • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Nosrati, Davide Giraudo, amWhy, mrtaurho, John Wayland Bales

      If this question can be reworded to fit the rules in the help center, please edit the question.






















          1 Answer
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          Hint: By mean value theorem, there exists $xiinBbb R $ such that $$f'(xi)=frac{f (x)-f (0)}{x-0}=frac{f (x)}{x}$$. Now use the fact that $|f'(xi)|leq 1$.






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            2 days ago










          • Please read this before asking next question.
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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          Hint: By mean value theorem, there exists $xiinBbb R $ such that $$f'(xi)=frac{f (x)-f (0)}{x-0}=frac{f (x)}{x}$$. Now use the fact that $|f'(xi)|leq 1$.






          share|cite|improve this answer

















          • 1




            Thanks a lot !!!
            – Johnathan1994
            2 days ago










          • Please read this before asking next question.
            – Thomas Shelby
            yesterday
















          1














          Hint: By mean value theorem, there exists $xiinBbb R $ such that $$f'(xi)=frac{f (x)-f (0)}{x-0}=frac{f (x)}{x}$$. Now use the fact that $|f'(xi)|leq 1$.






          share|cite|improve this answer

















          • 1




            Thanks a lot !!!
            – Johnathan1994
            2 days ago










          • Please read this before asking next question.
            – Thomas Shelby
            yesterday














          1












          1








          1






          Hint: By mean value theorem, there exists $xiinBbb R $ such that $$f'(xi)=frac{f (x)-f (0)}{x-0}=frac{f (x)}{x}$$. Now use the fact that $|f'(xi)|leq 1$.






          share|cite|improve this answer












          Hint: By mean value theorem, there exists $xiinBbb R $ such that $$f'(xi)=frac{f (x)-f (0)}{x-0}=frac{f (x)}{x}$$. Now use the fact that $|f'(xi)|leq 1$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Thomas Shelby

          1,680216




          1,680216








          • 1




            Thanks a lot !!!
            – Johnathan1994
            2 days ago










          • Please read this before asking next question.
            – Thomas Shelby
            yesterday














          • 1




            Thanks a lot !!!
            – Johnathan1994
            2 days ago










          • Please read this before asking next question.
            – Thomas Shelby
            yesterday








          1




          1




          Thanks a lot !!!
          – Johnathan1994
          2 days ago




          Thanks a lot !!!
          – Johnathan1994
          2 days ago












          Please read this before asking next question.
          – Thomas Shelby
          yesterday




          Please read this before asking next question.
          – Thomas Shelby
          yesterday



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