If $f(x)≤x$ , then $f′(x)≤1$?












5












$begingroup$


I'm studying Calculus and having a trouble solving this question.



1) If $f(x)leq x$, then $f′(x)leq 1$ for all $x$?



2) What if $f(0)=0$, $f′(x)$ exists for all $x$?



I could easily find the counter example for 1) (Therefore it is false)



But I'm not sure about 2)



If $f(0)=0$ and $f′(x)$ exists for all $x$ &
$f(x)leq x$ , then $f′(x)leq 1$ for all $x$?



Please leave a comment if you don't mind :)






calculus derivatives







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$endgroup$












  • $begingroup$
    The question is unclear: is the exercise ``consider a differentiable function $f$ such that $f(0)=0$ and $f(x)leq x$ for all $x$. Show that $f'(x)leq 1$ for all $x$.''?
    $endgroup$
    – b00n heT
    Mar 17 at 11:46








  • 1




    $begingroup$
    I'm not sure that the new edits improved the question. I preferred the original version of the question, which gave a little more context.
    $endgroup$
    – littleO
    Mar 17 at 11:50








  • 1




    $begingroup$
    Don't remove relevant information from your question!
    $endgroup$
    – user21820
    Mar 17 at 12:39
















5












$begingroup$


I'm studying Calculus and having a trouble solving this question.



1) If $f(x)leq x$, then $f′(x)leq 1$ for all $x$?



2) What if $f(0)=0$, $f′(x)$ exists for all $x$?



I could easily find the counter example for 1) (Therefore it is false)



But I'm not sure about 2)



If $f(0)=0$ and $f′(x)$ exists for all $x$ &
$f(x)leq x$ , then $f′(x)leq 1$ for all $x$?



Please leave a comment if you don't mind :)






calculus derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    The question is unclear: is the exercise ``consider a differentiable function $f$ such that $f(0)=0$ and $f(x)leq x$ for all $x$. Show that $f'(x)leq 1$ for all $x$.''?
    $endgroup$
    – b00n heT
    Mar 17 at 11:46








  • 1




    $begingroup$
    I'm not sure that the new edits improved the question. I preferred the original version of the question, which gave a little more context.
    $endgroup$
    – littleO
    Mar 17 at 11:50








  • 1




    $begingroup$
    Don't remove relevant information from your question!
    $endgroup$
    – user21820
    Mar 17 at 12:39














5












5








5


2



$begingroup$


I'm studying Calculus and having a trouble solving this question.



1) If $f(x)leq x$, then $f′(x)leq 1$ for all $x$?



2) What if $f(0)=0$, $f′(x)$ exists for all $x$?



I could easily find the counter example for 1) (Therefore it is false)



But I'm not sure about 2)



If $f(0)=0$ and $f′(x)$ exists for all $x$ &
$f(x)leq x$ , then $f′(x)leq 1$ for all $x$?



Please leave a comment if you don't mind :)






calculus derivatives







share|cite|improve this question











$endgroup$




I'm studying Calculus and having a trouble solving this question.



1) If $f(x)leq x$, then $f′(x)leq 1$ for all $x$?



2) What if $f(0)=0$, $f′(x)$ exists for all $x$?



I could easily find the counter example for 1) (Therefore it is false)



But I'm not sure about 2)



If $f(0)=0$ and $f′(x)$ exists for all $x$ &
$f(x)leq x$ , then $f′(x)leq 1$ for all $x$?



Please leave a comment if you don't mind :)






calculus derivatives




calculus derivatives






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 12:39









user21820

39.8k544158




39.8k544158










asked Mar 17 at 10:43









Mighty QWERTYMighty QWERTY

325




325












  • $begingroup$
    The question is unclear: is the exercise ``consider a differentiable function $f$ such that $f(0)=0$ and $f(x)leq x$ for all $x$. Show that $f'(x)leq 1$ for all $x$.''?
    $endgroup$
    – b00n heT
    Mar 17 at 11:46








  • 1




    $begingroup$
    I'm not sure that the new edits improved the question. I preferred the original version of the question, which gave a little more context.
    $endgroup$
    – littleO
    Mar 17 at 11:50








  • 1




    $begingroup$
    Don't remove relevant information from your question!
    $endgroup$
    – user21820
    Mar 17 at 12:39


















  • $begingroup$
    The question is unclear: is the exercise ``consider a differentiable function $f$ such that $f(0)=0$ and $f(x)leq x$ for all $x$. Show that $f'(x)leq 1$ for all $x$.''?
    $endgroup$
    – b00n heT
    Mar 17 at 11:46








  • 1




    $begingroup$
    I'm not sure that the new edits improved the question. I preferred the original version of the question, which gave a little more context.
    $endgroup$
    – littleO
    Mar 17 at 11:50








  • 1




    $begingroup$
    Don't remove relevant information from your question!
    $endgroup$
    – user21820
    Mar 17 at 12:39
















$begingroup$
The question is unclear: is the exercise ``consider a differentiable function $f$ such that $f(0)=0$ and $f(x)leq x$ for all $x$. Show that $f'(x)leq 1$ for all $x$.''?
$endgroup$
– b00n heT
Mar 17 at 11:46






$begingroup$
The question is unclear: is the exercise ``consider a differentiable function $f$ such that $f(0)=0$ and $f(x)leq x$ for all $x$. Show that $f'(x)leq 1$ for all $x$.''?
$endgroup$
– b00n heT
Mar 17 at 11:46






1




1




$begingroup$
I'm not sure that the new edits improved the question. I preferred the original version of the question, which gave a little more context.
$endgroup$
– littleO
Mar 17 at 11:50






$begingroup$
I'm not sure that the new edits improved the question. I preferred the original version of the question, which gave a little more context.
$endgroup$
– littleO
Mar 17 at 11:50






1




1




$begingroup$
Don't remove relevant information from your question!
$endgroup$
– user21820
Mar 17 at 12:39




$begingroup$
Don't remove relevant information from your question!
$endgroup$
– user21820
Mar 17 at 12:39










3 Answers
3






active

oldest

votes


















4












$begingroup$

Hint: Consider $$f(x)=x-A sin^2 x $$
for large $A$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Draw the line $y=x$, and then draw any kind of squiggly function you want that stays below or touches the line. In particular, the function $f(x)=x-e^{-x}$ has $f'(x)gt1$ for all $x$, while $f(x)=x-{1over2}x^2$ satisfies $f(0)=0$ but $f'(x)gt1$ for $xlt0$.



    Remark: The original version of the OP's question had two parts, with the condition $f(0)=0$ being added in the second part. The function $f(x)=x-e^{-x}$, of course, does not satisfy that condition.






    share|cite|improve this answer











    $endgroup$





















      -2












      $begingroup$

      Let for example $f(x)=xsin x$, then $f(0)=0$ and for any real $x$ $f(x)leq x$. Then the derivative will look like this: $f'(x)=sin x+xcos x$ and so will exist for all $x$. Taking $x=2pi n$, where $n$ is positive integer we get $f'(x)=2pi n$, which evidently has no upper limitation.






      share|cite|improve this answer











      $endgroup$













      • $begingroup$
        $f(x)leq x$ fails for $x$ negative.
        $endgroup$
        – Wojowu
        Mar 17 at 11:47











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4












      $begingroup$

      Hint: Consider $$f(x)=x-A sin^2 x $$
      for large $A$.






      share|cite|improve this answer









      $endgroup$


















        4












        $begingroup$

        Hint: Consider $$f(x)=x-A sin^2 x $$
        for large $A$.






        share|cite|improve this answer









        $endgroup$
















          4












          4








          4





          $begingroup$

          Hint: Consider $$f(x)=x-A sin^2 x $$
          for large $A$.






          share|cite|improve this answer









          $endgroup$



          Hint: Consider $$f(x)=x-A sin^2 x $$
          for large $A$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Mar 17 at 10:51









          user1337user1337

          16.8k43592




          16.8k43592























              4












              $begingroup$

              Draw the line $y=x$, and then draw any kind of squiggly function you want that stays below or touches the line. In particular, the function $f(x)=x-e^{-x}$ has $f'(x)gt1$ for all $x$, while $f(x)=x-{1over2}x^2$ satisfies $f(0)=0$ but $f'(x)gt1$ for $xlt0$.



              Remark: The original version of the OP's question had two parts, with the condition $f(0)=0$ being added in the second part. The function $f(x)=x-e^{-x}$, of course, does not satisfy that condition.






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                Draw the line $y=x$, and then draw any kind of squiggly function you want that stays below or touches the line. In particular, the function $f(x)=x-e^{-x}$ has $f'(x)gt1$ for all $x$, while $f(x)=x-{1over2}x^2$ satisfies $f(0)=0$ but $f'(x)gt1$ for $xlt0$.



                Remark: The original version of the OP's question had two parts, with the condition $f(0)=0$ being added in the second part. The function $f(x)=x-e^{-x}$, of course, does not satisfy that condition.






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  Draw the line $y=x$, and then draw any kind of squiggly function you want that stays below or touches the line. In particular, the function $f(x)=x-e^{-x}$ has $f'(x)gt1$ for all $x$, while $f(x)=x-{1over2}x^2$ satisfies $f(0)=0$ but $f'(x)gt1$ for $xlt0$.



                  Remark: The original version of the OP's question had two parts, with the condition $f(0)=0$ being added in the second part. The function $f(x)=x-e^{-x}$, of course, does not satisfy that condition.






                  share|cite|improve this answer











                  $endgroup$



                  Draw the line $y=x$, and then draw any kind of squiggly function you want that stays below or touches the line. In particular, the function $f(x)=x-e^{-x}$ has $f'(x)gt1$ for all $x$, while $f(x)=x-{1over2}x^2$ satisfies $f(0)=0$ but $f'(x)gt1$ for $xlt0$.



                  Remark: The original version of the OP's question had two parts, with the condition $f(0)=0$ being added in the second part. The function $f(x)=x-e^{-x}$, of course, does not satisfy that condition.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Mar 17 at 11:56

























                  answered Mar 17 at 11:24









                  Barry CipraBarry Cipra

                  60.5k655128




                  60.5k655128























                      -2












                      $begingroup$

                      Let for example $f(x)=xsin x$, then $f(0)=0$ and for any real $x$ $f(x)leq x$. Then the derivative will look like this: $f'(x)=sin x+xcos x$ and so will exist for all $x$. Taking $x=2pi n$, where $n$ is positive integer we get $f'(x)=2pi n$, which evidently has no upper limitation.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        $f(x)leq x$ fails for $x$ negative.
                        $endgroup$
                        – Wojowu
                        Mar 17 at 11:47
















                      -2












                      $begingroup$

                      Let for example $f(x)=xsin x$, then $f(0)=0$ and for any real $x$ $f(x)leq x$. Then the derivative will look like this: $f'(x)=sin x+xcos x$ and so will exist for all $x$. Taking $x=2pi n$, where $n$ is positive integer we get $f'(x)=2pi n$, which evidently has no upper limitation.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        $f(x)leq x$ fails for $x$ negative.
                        $endgroup$
                        – Wojowu
                        Mar 17 at 11:47














                      -2












                      -2








                      -2





                      $begingroup$

                      Let for example $f(x)=xsin x$, then $f(0)=0$ and for any real $x$ $f(x)leq x$. Then the derivative will look like this: $f'(x)=sin x+xcos x$ and so will exist for all $x$. Taking $x=2pi n$, where $n$ is positive integer we get $f'(x)=2pi n$, which evidently has no upper limitation.






                      share|cite|improve this answer











                      $endgroup$



                      Let for example $f(x)=xsin x$, then $f(0)=0$ and for any real $x$ $f(x)leq x$. Then the derivative will look like this: $f'(x)=sin x+xcos x$ and so will exist for all $x$. Taking $x=2pi n$, where $n$ is positive integer we get $f'(x)=2pi n$, which evidently has no upper limitation.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Mar 17 at 11:39









                      Max

                      9211319




                      9211319










                      answered Mar 17 at 11:19









                      Alex KovalevskyAlex Kovalevsky

                      11




                      11












                      • $begingroup$
                        $f(x)leq x$ fails for $x$ negative.
                        $endgroup$
                        – Wojowu
                        Mar 17 at 11:47


















                      • $begingroup$
                        $f(x)leq x$ fails for $x$ negative.
                        $endgroup$
                        – Wojowu
                        Mar 17 at 11:47
















                      $begingroup$
                      $f(x)leq x$ fails for $x$ negative.
                      $endgroup$
                      – Wojowu
                      Mar 17 at 11:47




                      $begingroup$
                      $f(x)leq x$ fails for $x$ negative.
                      $endgroup$
                      – Wojowu
                      Mar 17 at 11:47


















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