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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  • $begingroup$
    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
    14 hours ago








  • 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
    14 hours ago












  • $begingroup$
    Don't remove relevant information from your question!
    $endgroup$
    – user21820
    13 hours ago
















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









New contributor




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







$endgroup$












  • $begingroup$
    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
    14 hours ago








  • 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
    14 hours ago












  • $begingroup$
    Don't remove relevant information from your question!
    $endgroup$
    – user21820
    13 hours ago














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









New contributor




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







$endgroup$




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






share|cite|improve this question









New contributor




Mighty QWERTY 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




Mighty QWERTY 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 13 hours ago









user21820

39.4k543155




39.4k543155






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asked 15 hours ago









Mighty QWERTYMighty QWERTY

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  • $begingroup$
    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
    14 hours ago








  • 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
    14 hours ago












  • $begingroup$
    Don't remove relevant information from your question!
    $endgroup$
    – user21820
    13 hours ago


















  • $begingroup$
    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
    14 hours ago








  • 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
    14 hours ago












  • $begingroup$
    Don't remove relevant information from your question!
    $endgroup$
    – user21820
    13 hours ago
















$begingroup$
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
14 hours ago






$begingroup$
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
14 hours ago






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
14 hours ago






$begingroup$
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
14 hours ago














$begingroup$
Don't remove relevant information from your question!
$endgroup$
– user21820
13 hours ago




$begingroup$
Don't remove relevant information from your question!
$endgroup$
– user21820
13 hours ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

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












    $begingroup$

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






    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










      New contributor




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






      $endgroup$













      • $begingroup$
        $f(x)leq x$ fails for $x$ negative.
        $endgroup$
        – Wojowu
        14 hours ago











      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      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











      $endgroup$


















        3












        $begingroup$

        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











        $endgroup$
















          3












          3








          3





          $begingroup$

          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 14 hours ago

























          answered 14 hours ago









          Barry CipraBarry Cipra

          60.3k654126




          60.3k654126























              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 15 hours ago









                  user1337user1337

                  16.8k43592




                  16.8k43592























                      -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










                      New contributor




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






                      $endgroup$













                      • $begingroup$
                        $f(x)leq x$ fails for $x$ negative.
                        $endgroup$
                        – Wojowu
                        14 hours ago
















                      -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










                      New contributor




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






                      $endgroup$













                      • $begingroup$
                        $f(x)leq x$ fails for $x$ negative.
                        $endgroup$
                        – Wojowu
                        14 hours ago














                      -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










                      New contributor




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






                      $endgroup$



                      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










                      New contributor




                      Alex Kovalevsky 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 answer



                      share|cite|improve this answer








                      edited 14 hours ago









                      Max

                      881318




                      881318






                      New contributor




                      Alex Kovalevsky is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                      answered 15 hours ago









                      Alex KovalevskyAlex Kovalevsky

                      11




                      11




                      New contributor




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                      New contributor





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












                      • $begingroup$
                        $f(x)leq x$ fails for $x$ negative.
                        $endgroup$
                        – Wojowu
                        14 hours ago


















                      • $begingroup$
                        $f(x)leq x$ fails for $x$ negative.
                        $endgroup$
                        – Wojowu
                        14 hours ago
















                      $begingroup$
                      $f(x)leq x$ fails for $x$ negative.
                      $endgroup$
                      – Wojowu
                      14 hours ago




                      $begingroup$
                      $f(x)leq x$ fails for $x$ negative.
                      $endgroup$
                      – Wojowu
                      14 hours ago










                      Mighty QWERTY is a new contributor. Be nice, and check out our Code of Conduct.










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                      南乌拉尔铁路局

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                      南乌拉尔铁路局 Южно-Уральская железная дорога 成立 1934年1月4日 国家   蘇聯（1934年—1991年）   俄羅斯（1991年—） 总部 车里雅宾斯克 隶属 俄罗斯铁路股份公司 路局代码 80 铁路里程 4,934.9公里 奖项 网址 http://yuzd.rzd.ru/ 查 论 编 南乌拉尔铁路局 （俄语： Южно-Уральская железная дорога ， 羅馬化： Yuzhno-Uralskaya zheleznaya doroga ，简称 ЮУЖД ）是俄罗斯铁路股份公司属下的铁路管理局之一，总部位于车里雅宾斯克，负责管辖俄罗斯联邦乌拉尔南部地区的铁路系统，范围包括奥伦堡州、车里雅宾斯克州、库尔干州的大部分铁路，以及萨马拉州、斯维尔德洛夫斯克州、巴什科尔托斯坦共和国的少部分铁路，并有部分铁路延伸至或穿过哈萨克斯坦境内。南乌拉尔铁路局的营业里程为4,934.9公里，与伏尔加铁路局、古比雪夫铁路局、斯维尔德洛夫斯克铁路局、西西伯利亚铁路局相邻，并与南方的哈萨克斯坦铁路连接。 外部链接 （俄文） Южно-Уральская железная дорога — филиал ОАО РЖД （俄文） 南乌拉尔铁路局管辖路线图（一） （俄文） 南乌拉尔铁路局管辖路线图（二） 查 论 编 俄罗斯铁路（集团）股份公司 十月铁路局 · 莫斯科铁路局 · 北高加索铁路局 · 高尔基铁路局 · 古比雪夫铁路局 · 北方铁路局 · 伏尔加铁路局 · 东南铁路局 · 斯维尔德洛夫斯克铁路局 · 南乌拉尔铁路局 · 西西伯利亚铁路局 · 克拉斯诺亚尔斯克铁路局 · 东西伯利亚铁路局 · 后贝加尔铁路局 · 远东铁路局 · 加里宁格勒铁路局 This page is only for reference, If you need detailed information, please check here
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