$ex leq e^x$ inequality using derivatives












0












$begingroup$


So I was going through a sum for



Prove $ex leq e^x$ , $forall x in mathbb{R} $



I took $g(x) = e^x - ex$



Then $g'(x)= e^x - e$



I understood the case when $x>1$ function is strictly increasing i.e $g'(x) >0$ then $e^x>ex$ but what about when $x leq 1$






calculus







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$endgroup$








  • 3




    $begingroup$
    Then the function is decreasing. That's all you need. Think about it for a moment.
    $endgroup$
    – saulspatz
    Nov 24 '18 at 16:32










  • $begingroup$
    Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
    $endgroup$
    – Andrei
    Nov 24 '18 at 16:35












  • $begingroup$
    I'm sorry I'm still getting a bit confused
    $endgroup$
    – Sumukh Sai
    Nov 24 '18 at 16:45
















0












$begingroup$


So I was going through a sum for



Prove $ex leq e^x$ , $forall x in mathbb{R} $



I took $g(x) = e^x - ex$



Then $g'(x)= e^x - e$



I understood the case when $x>1$ function is strictly increasing i.e $g'(x) >0$ then $e^x>ex$ but what about when $x leq 1$






calculus







share|cite|improve this question









$endgroup$








  • 3




    $begingroup$
    Then the function is decreasing. That's all you need. Think about it for a moment.
    $endgroup$
    – saulspatz
    Nov 24 '18 at 16:32










  • $begingroup$
    Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
    $endgroup$
    – Andrei
    Nov 24 '18 at 16:35












  • $begingroup$
    I'm sorry I'm still getting a bit confused
    $endgroup$
    – Sumukh Sai
    Nov 24 '18 at 16:45














0












0








0





$begingroup$


So I was going through a sum for



Prove $ex leq e^x$ , $forall x in mathbb{R} $



I took $g(x) = e^x - ex$



Then $g'(x)= e^x - e$



I understood the case when $x>1$ function is strictly increasing i.e $g'(x) >0$ then $e^x>ex$ but what about when $x leq 1$






calculus







share|cite|improve this question









$endgroup$




So I was going through a sum for



Prove $ex leq e^x$ , $forall x in mathbb{R} $



I took $g(x) = e^x - ex$



Then $g'(x)= e^x - e$



I understood the case when $x>1$ function is strictly increasing i.e $g'(x) >0$ then $e^x>ex$ but what about when $x leq 1$






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 '18 at 16:29









Sumukh SaiSumukh Sai

206




206








  • 3




    $begingroup$
    Then the function is decreasing. That's all you need. Think about it for a moment.
    $endgroup$
    – saulspatz
    Nov 24 '18 at 16:32










  • $begingroup$
    Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
    $endgroup$
    – Andrei
    Nov 24 '18 at 16:35












  • $begingroup$
    I'm sorry I'm still getting a bit confused
    $endgroup$
    – Sumukh Sai
    Nov 24 '18 at 16:45














  • 3




    $begingroup$
    Then the function is decreasing. That's all you need. Think about it for a moment.
    $endgroup$
    – saulspatz
    Nov 24 '18 at 16:32










  • $begingroup$
    Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
    $endgroup$
    – Andrei
    Nov 24 '18 at 16:35












  • $begingroup$
    I'm sorry I'm still getting a bit confused
    $endgroup$
    – Sumukh Sai
    Nov 24 '18 at 16:45








3




3




$begingroup$
Then the function is decreasing. That's all you need. Think about it for a moment.
$endgroup$
– saulspatz
Nov 24 '18 at 16:32




$begingroup$
Then the function is decreasing. That's all you need. Think about it for a moment.
$endgroup$
– saulspatz
Nov 24 '18 at 16:32












$begingroup$
Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
$endgroup$
– Andrei
Nov 24 '18 at 16:35






$begingroup$
Just re-wording what @saulspatz says: what does it mean for a function to have only one minima or maxima? Write it as inequality
$endgroup$
– Andrei
Nov 24 '18 at 16:35














$begingroup$
I'm sorry I'm still getting a bit confused
$endgroup$
– Sumukh Sai
Nov 24 '18 at 16:45




$begingroup$
I'm sorry I'm still getting a bit confused
$endgroup$
– Sumukh Sai
Nov 24 '18 at 16:45










2 Answers
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$begingroup$

it is easy to see $x=1$ is the absolute minimum, that is, $ f(1)leq f(x)$ for all $xin R$ so that $0leq ex-e^x$.Thus $exleq e^x$ for all $xin R.$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    We have that



    $$g(x)=e^x-ex implies g'(x)=e^x-e=0quad x=1$$



    moreover $g''(x)=e^x > 0$, therefore $x=1$ is a point of minimum for $g(x)$.






    share|cite|improve this answer











    $endgroup$













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      2 Answers
      2






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      2 Answers
      2






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      1












      $begingroup$

      it is easy to see $x=1$ is the absolute minimum, that is, $ f(1)leq f(x)$ for all $xin R$ so that $0leq ex-e^x$.Thus $exleq e^x$ for all $xin R.$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        it is easy to see $x=1$ is the absolute minimum, that is, $ f(1)leq f(x)$ for all $xin R$ so that $0leq ex-e^x$.Thus $exleq e^x$ for all $xin R.$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          it is easy to see $x=1$ is the absolute minimum, that is, $ f(1)leq f(x)$ for all $xin R$ so that $0leq ex-e^x$.Thus $exleq e^x$ for all $xin R.$






          share|cite|improve this answer









          $endgroup$



          it is easy to see $x=1$ is the absolute minimum, that is, $ f(1)leq f(x)$ for all $xin R$ so that $0leq ex-e^x$.Thus $exleq e^x$ for all $xin R.$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 24 '18 at 17:07









          gb2017gb2017

          944




          944























              0












              $begingroup$

              We have that



              $$g(x)=e^x-ex implies g'(x)=e^x-e=0quad x=1$$



              moreover $g''(x)=e^x > 0$, therefore $x=1$ is a point of minimum for $g(x)$.






              share|cite|improve this answer











              $endgroup$


















                0












                $begingroup$

                We have that



                $$g(x)=e^x-ex implies g'(x)=e^x-e=0quad x=1$$



                moreover $g''(x)=e^x > 0$, therefore $x=1$ is a point of minimum for $g(x)$.






                share|cite|improve this answer











                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  We have that



                  $$g(x)=e^x-ex implies g'(x)=e^x-e=0quad x=1$$



                  moreover $g''(x)=e^x > 0$, therefore $x=1$ is a point of minimum for $g(x)$.






                  share|cite|improve this answer











                  $endgroup$



                  We have that



                  $$g(x)=e^x-ex implies g'(x)=e^x-e=0quad x=1$$



                  moreover $g''(x)=e^x > 0$, therefore $x=1$ is a point of minimum for $g(x)$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 24 '18 at 17:25

























                  answered Nov 24 '18 at 16:57









                  gimusigimusi

                  1




                  1






























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