Prove $ (b-1)e^{-b} < (a-1)e^{-a} $ for every $ 2 < a < b $












1












$begingroup$


Prove: for every $ 2 < a < b $ : $ (b-1)e^{-b} < (a-1)e^{-a} $



My first try was to try and build a function and show it is going down:



$ f(x) = (b-1)e^{-b} - (a-1)e^{-a} $



but I don't know how to handle such function. Is there any easier way?






calculus







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$endgroup$












  • $begingroup$
    Where is the variable $x$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 6 at 16:33
















1












$begingroup$


Prove: for every $ 2 < a < b $ : $ (b-1)e^{-b} < (a-1)e^{-a} $



My first try was to try and build a function and show it is going down:



$ f(x) = (b-1)e^{-b} - (a-1)e^{-a} $



but I don't know how to handle such function. Is there any easier way?






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    Where is the variable $x$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 6 at 16:33














1












1








1


1



$begingroup$


Prove: for every $ 2 < a < b $ : $ (b-1)e^{-b} < (a-1)e^{-a} $



My first try was to try and build a function and show it is going down:



$ f(x) = (b-1)e^{-b} - (a-1)e^{-a} $



but I don't know how to handle such function. Is there any easier way?






calculus







share|cite|improve this question









$endgroup$




Prove: for every $ 2 < a < b $ : $ (b-1)e^{-b} < (a-1)e^{-a} $



My first try was to try and build a function and show it is going down:



$ f(x) = (b-1)e^{-b} - (a-1)e^{-a} $



but I don't know how to handle such function. Is there any easier way?






calculus




calculus






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Jan 6 at 16:30









bm1125bm1125

69116




69116












  • $begingroup$
    Where is the variable $x$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 6 at 16:33


















  • $begingroup$
    Where is the variable $x$?
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 6 at 16:33
















$begingroup$
Where is the variable $x$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:33




$begingroup$
Where is the variable $x$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 6 at 16:33










2 Answers
2






active

oldest

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2












$begingroup$

Take
$$f(x) = (x-1)e^{-x}$$
then
$$f'(x) = e^{-x}-e^{-x}(x-1)=e^{-x}(2-x)<0$$
when $x>2$ and so $f$ is decreasing when $x>2.$ Thus for $2<a<b$ we have that
$$f(a)>f(b)implies (b-1)e^{-b} < (a-1)e^{-a}.$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint/Guide



    Consider the function
    $$
    f(x)=frac{x-1}{e^x}.
    $$
    By computing its derivative, show that $f$ is decreasing for $x>2$ from which the claim will follow.






    share|cite|improve this answer









    $endgroup$














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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

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      active

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      active

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      2












      $begingroup$

      Take
      $$f(x) = (x-1)e^{-x}$$
      then
      $$f'(x) = e^{-x}-e^{-x}(x-1)=e^{-x}(2-x)<0$$
      when $x>2$ and so $f$ is decreasing when $x>2.$ Thus for $2<a<b$ we have that
      $$f(a)>f(b)implies (b-1)e^{-b} < (a-1)e^{-a}.$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Take
        $$f(x) = (x-1)e^{-x}$$
        then
        $$f'(x) = e^{-x}-e^{-x}(x-1)=e^{-x}(2-x)<0$$
        when $x>2$ and so $f$ is decreasing when $x>2.$ Thus for $2<a<b$ we have that
        $$f(a)>f(b)implies (b-1)e^{-b} < (a-1)e^{-a}.$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Take
          $$f(x) = (x-1)e^{-x}$$
          then
          $$f'(x) = e^{-x}-e^{-x}(x-1)=e^{-x}(2-x)<0$$
          when $x>2$ and so $f$ is decreasing when $x>2.$ Thus for $2<a<b$ we have that
          $$f(a)>f(b)implies (b-1)e^{-b} < (a-1)e^{-a}.$$






          share|cite|improve this answer









          $endgroup$



          Take
          $$f(x) = (x-1)e^{-x}$$
          then
          $$f'(x) = e^{-x}-e^{-x}(x-1)=e^{-x}(2-x)<0$$
          when $x>2$ and so $f$ is decreasing when $x>2.$ Thus for $2<a<b$ we have that
          $$f(a)>f(b)implies (b-1)e^{-b} < (a-1)e^{-a}.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 6 at 16:37









          model_checkermodel_checker

          4,45521931




          4,45521931























              1












              $begingroup$

              Hint/Guide



              Consider the function
              $$
              f(x)=frac{x-1}{e^x}.
              $$
              By computing its derivative, show that $f$ is decreasing for $x>2$ from which the claim will follow.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Hint/Guide



                Consider the function
                $$
                f(x)=frac{x-1}{e^x}.
                $$
                By computing its derivative, show that $f$ is decreasing for $x>2$ from which the claim will follow.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Hint/Guide



                  Consider the function
                  $$
                  f(x)=frac{x-1}{e^x}.
                  $$
                  By computing its derivative, show that $f$ is decreasing for $x>2$ from which the claim will follow.






                  share|cite|improve this answer









                  $endgroup$



                  Hint/Guide



                  Consider the function
                  $$
                  f(x)=frac{x-1}{e^x}.
                  $$
                  By computing its derivative, show that $f$ is decreasing for $x>2$ from which the claim will follow.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 16:34









                  Foobaz JohnFoobaz John

                  22.9k41552




                  22.9k41552






























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