How can I show the monotincity of $frac{x}{1+|x|}$ without differential calculus?












0












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I am not able to work out this exercise because I am not sure of the behavior of Absolute value for monotony.






calculus functions continuity







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  • $begingroup$
    Discuss by case $x geqslant 0$ and $x <0$. The monotonicity could be shown by directly check the definition.
    $endgroup$
    – xbh
    Dec 3 '18 at 13:17
















0












$begingroup$


I am not able to work out this exercise because I am not sure of the behavior of Absolute value for monotony.






calculus functions continuity







share|cite|improve this question











$endgroup$












  • $begingroup$
    Discuss by case $x geqslant 0$ and $x <0$. The monotonicity could be shown by directly check the definition.
    $endgroup$
    – xbh
    Dec 3 '18 at 13:17














0












0








0





$begingroup$


I am not able to work out this exercise because I am not sure of the behavior of Absolute value for monotony.






calculus functions continuity







share|cite|improve this question











$endgroup$




I am not able to work out this exercise because I am not sure of the behavior of Absolute value for monotony.






calculus functions continuity




calculus functions continuity






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 23:51









greedoid

39.2k114797




39.2k114797










asked Dec 3 '18 at 13:11









Andreas LetoAndreas Leto

214




214












  • $begingroup$
    Discuss by case $x geqslant 0$ and $x <0$. The monotonicity could be shown by directly check the definition.
    $endgroup$
    – xbh
    Dec 3 '18 at 13:17


















  • $begingroup$
    Discuss by case $x geqslant 0$ and $x <0$. The monotonicity could be shown by directly check the definition.
    $endgroup$
    – xbh
    Dec 3 '18 at 13:17
















$begingroup$
Discuss by case $x geqslant 0$ and $x <0$. The monotonicity could be shown by directly check the definition.
$endgroup$
– xbh
Dec 3 '18 at 13:17




$begingroup$
Discuss by case $x geqslant 0$ and $x <0$. The monotonicity could be shown by directly check the definition.
$endgroup$
– xbh
Dec 3 '18 at 13:17










5 Answers
5






active

oldest

votes


















8












$begingroup$

Let's prove (first) it is injective: $$f(a) = f(b)implies {aover 1+|a|} = {bover 1+|b|}$$
so $a$ and $b$ must have equaly sign. Say both are positive, then $${aover 1+a} = {bover 1+b} implies a=b$$
and the same if both are negative. So $f$ is injective and it is continuous so it must be monotonic.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That's my favourite answer, +1.
    $endgroup$
    – Michael Hoppe
    Dec 3 '18 at 15:56



















3












$begingroup$

Note that this is an odd function, so increasing for positive $x$ implies increasing for negative $x$.



Then you only have to think about $frac{x}{1+x}=1-frac{1}{1+x}$ for positive $x$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Hint:




    1. A function $f$ is monotonous if, for $x_1<x_2$, the expression $f(x_1)-f(x_2)$ always has the same sign.

    2. Separate three cases when calculating $f(x_1)-f(x_2)$: one when $x_1,x_2$ are both non-negative, one where they are both negative, and one where one is non-negative and one is negative






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Hint:




      • If $x >0$, then we have $$frac{x}{1+|x|}=frac{1}{frac1x+1}$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        Since $f(1)>f(0)$ we can guess that $f(x)$ is strictly increasing, then we need to show that $forall a>0$



        $$frac{x+a}{1+|x+a|}>frac{x}{1+|x|} iff (1+|x|)(x+a)>x(1+|x+a|) $$$$iff a+a|x|+x|x|-x|x+a|>0$$



        then consider the cases





        • $xge 0 implies |x|=x $ $$implies a+a|x|+x|x|-x|x+a|=a+ax+x^2-x^2-ax=a>0$$


        • $x <0implies |x|=-x$ $$implies a+a|x|+x|x|-x|x+a|=a-ax-x^2+x^2+ax=a>0$$






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          Why would $f(x+1)>f(x)$ imply that $f$ is strictly increasing?
          $endgroup$
          – Shubham Johri
          Dec 3 '18 at 13:31










        • $begingroup$
          @ShubhamJohri Yes sorry, I don't know why I've used one, I fix that . Thanks
          $endgroup$
          – gimusi
          Dec 3 '18 at 13:32











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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        8












        $begingroup$

        Let's prove (first) it is injective: $$f(a) = f(b)implies {aover 1+|a|} = {bover 1+|b|}$$
        so $a$ and $b$ must have equaly sign. Say both are positive, then $${aover 1+a} = {bover 1+b} implies a=b$$
        and the same if both are negative. So $f$ is injective and it is continuous so it must be monotonic.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          That's my favourite answer, +1.
          $endgroup$
          – Michael Hoppe
          Dec 3 '18 at 15:56
















        8












        $begingroup$

        Let's prove (first) it is injective: $$f(a) = f(b)implies {aover 1+|a|} = {bover 1+|b|}$$
        so $a$ and $b$ must have equaly sign. Say both are positive, then $${aover 1+a} = {bover 1+b} implies a=b$$
        and the same if both are negative. So $f$ is injective and it is continuous so it must be monotonic.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          That's my favourite answer, +1.
          $endgroup$
          – Michael Hoppe
          Dec 3 '18 at 15:56














        8












        8








        8





        $begingroup$

        Let's prove (first) it is injective: $$f(a) = f(b)implies {aover 1+|a|} = {bover 1+|b|}$$
        so $a$ and $b$ must have equaly sign. Say both are positive, then $${aover 1+a} = {bover 1+b} implies a=b$$
        and the same if both are negative. So $f$ is injective and it is continuous so it must be monotonic.






        share|cite|improve this answer











        $endgroup$



        Let's prove (first) it is injective: $$f(a) = f(b)implies {aover 1+|a|} = {bover 1+|b|}$$
        so $a$ and $b$ must have equaly sign. Say both are positive, then $${aover 1+a} = {bover 1+b} implies a=b$$
        and the same if both are negative. So $f$ is injective and it is continuous so it must be monotonic.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 13:38

























        answered Dec 3 '18 at 13:22









        greedoidgreedoid

        39.2k114797




        39.2k114797












        • $begingroup$
          That's my favourite answer, +1.
          $endgroup$
          – Michael Hoppe
          Dec 3 '18 at 15:56


















        • $begingroup$
          That's my favourite answer, +1.
          $endgroup$
          – Michael Hoppe
          Dec 3 '18 at 15:56
















        $begingroup$
        That's my favourite answer, +1.
        $endgroup$
        – Michael Hoppe
        Dec 3 '18 at 15:56




        $begingroup$
        That's my favourite answer, +1.
        $endgroup$
        – Michael Hoppe
        Dec 3 '18 at 15:56











        3












        $begingroup$

        Note that this is an odd function, so increasing for positive $x$ implies increasing for negative $x$.



        Then you only have to think about $frac{x}{1+x}=1-frac{1}{1+x}$ for positive $x$.






        share|cite|improve this answer









        $endgroup$


















          3












          $begingroup$

          Note that this is an odd function, so increasing for positive $x$ implies increasing for negative $x$.



          Then you only have to think about $frac{x}{1+x}=1-frac{1}{1+x}$ for positive $x$.






          share|cite|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            Note that this is an odd function, so increasing for positive $x$ implies increasing for negative $x$.



            Then you only have to think about $frac{x}{1+x}=1-frac{1}{1+x}$ for positive $x$.






            share|cite|improve this answer









            $endgroup$



            Note that this is an odd function, so increasing for positive $x$ implies increasing for negative $x$.



            Then you only have to think about $frac{x}{1+x}=1-frac{1}{1+x}$ for positive $x$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 3 '18 at 13:17









            paw88789paw88789

            29.1k12349




            29.1k12349























                2












                $begingroup$

                Hint:




                1. A function $f$ is monotonous if, for $x_1<x_2$, the expression $f(x_1)-f(x_2)$ always has the same sign.

                2. Separate three cases when calculating $f(x_1)-f(x_2)$: one when $x_1,x_2$ are both non-negative, one where they are both negative, and one where one is non-negative and one is negative






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  Hint:




                  1. A function $f$ is monotonous if, for $x_1<x_2$, the expression $f(x_1)-f(x_2)$ always has the same sign.

                  2. Separate three cases when calculating $f(x_1)-f(x_2)$: one when $x_1,x_2$ are both non-negative, one where they are both negative, and one where one is non-negative and one is negative






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    Hint:




                    1. A function $f$ is monotonous if, for $x_1<x_2$, the expression $f(x_1)-f(x_2)$ always has the same sign.

                    2. Separate three cases when calculating $f(x_1)-f(x_2)$: one when $x_1,x_2$ are both non-negative, one where they are both negative, and one where one is non-negative and one is negative






                    share|cite|improve this answer









                    $endgroup$



                    Hint:




                    1. A function $f$ is monotonous if, for $x_1<x_2$, the expression $f(x_1)-f(x_2)$ always has the same sign.

                    2. Separate three cases when calculating $f(x_1)-f(x_2)$: one when $x_1,x_2$ are both non-negative, one where they are both negative, and one where one is non-negative and one is negative







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 3 '18 at 13:17









                    5xum5xum

                    90.2k394161




                    90.2k394161























                        1












                        $begingroup$

                        Hint:




                        • If $x >0$, then we have $$frac{x}{1+|x|}=frac{1}{frac1x+1}$$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Hint:




                          • If $x >0$, then we have $$frac{x}{1+|x|}=frac{1}{frac1x+1}$$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Hint:




                            • If $x >0$, then we have $$frac{x}{1+|x|}=frac{1}{frac1x+1}$$






                            share|cite|improve this answer









                            $endgroup$



                            Hint:




                            • If $x >0$, then we have $$frac{x}{1+|x|}=frac{1}{frac1x+1}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 3 '18 at 13:16









                            Siong Thye GohSiong Thye Goh

                            100k1466117




                            100k1466117























                                0












                                $begingroup$

                                Since $f(1)>f(0)$ we can guess that $f(x)$ is strictly increasing, then we need to show that $forall a>0$



                                $$frac{x+a}{1+|x+a|}>frac{x}{1+|x|} iff (1+|x|)(x+a)>x(1+|x+a|) $$$$iff a+a|x|+x|x|-x|x+a|>0$$



                                then consider the cases





                                • $xge 0 implies |x|=x $ $$implies a+a|x|+x|x|-x|x+a|=a+ax+x^2-x^2-ax=a>0$$


                                • $x <0implies |x|=-x$ $$implies a+a|x|+x|x|-x|x+a|=a-ax-x^2+x^2+ax=a>0$$






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Why would $f(x+1)>f(x)$ imply that $f$ is strictly increasing?
                                  $endgroup$
                                  – Shubham Johri
                                  Dec 3 '18 at 13:31










                                • $begingroup$
                                  @ShubhamJohri Yes sorry, I don't know why I've used one, I fix that . Thanks
                                  $endgroup$
                                  – gimusi
                                  Dec 3 '18 at 13:32
















                                0












                                $begingroup$

                                Since $f(1)>f(0)$ we can guess that $f(x)$ is strictly increasing, then we need to show that $forall a>0$



                                $$frac{x+a}{1+|x+a|}>frac{x}{1+|x|} iff (1+|x|)(x+a)>x(1+|x+a|) $$$$iff a+a|x|+x|x|-x|x+a|>0$$



                                then consider the cases





                                • $xge 0 implies |x|=x $ $$implies a+a|x|+x|x|-x|x+a|=a+ax+x^2-x^2-ax=a>0$$


                                • $x <0implies |x|=-x$ $$implies a+a|x|+x|x|-x|x+a|=a-ax-x^2+x^2+ax=a>0$$






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  Why would $f(x+1)>f(x)$ imply that $f$ is strictly increasing?
                                  $endgroup$
                                  – Shubham Johri
                                  Dec 3 '18 at 13:31










                                • $begingroup$
                                  @ShubhamJohri Yes sorry, I don't know why I've used one, I fix that . Thanks
                                  $endgroup$
                                  – gimusi
                                  Dec 3 '18 at 13:32














                                0












                                0








                                0





                                $begingroup$

                                Since $f(1)>f(0)$ we can guess that $f(x)$ is strictly increasing, then we need to show that $forall a>0$



                                $$frac{x+a}{1+|x+a|}>frac{x}{1+|x|} iff (1+|x|)(x+a)>x(1+|x+a|) $$$$iff a+a|x|+x|x|-x|x+a|>0$$



                                then consider the cases





                                • $xge 0 implies |x|=x $ $$implies a+a|x|+x|x|-x|x+a|=a+ax+x^2-x^2-ax=a>0$$


                                • $x <0implies |x|=-x$ $$implies a+a|x|+x|x|-x|x+a|=a-ax-x^2+x^2+ax=a>0$$






                                share|cite|improve this answer











                                $endgroup$



                                Since $f(1)>f(0)$ we can guess that $f(x)$ is strictly increasing, then we need to show that $forall a>0$



                                $$frac{x+a}{1+|x+a|}>frac{x}{1+|x|} iff (1+|x|)(x+a)>x(1+|x+a|) $$$$iff a+a|x|+x|x|-x|x+a|>0$$



                                then consider the cases





                                • $xge 0 implies |x|=x $ $$implies a+a|x|+x|x|-x|x+a|=a+ax+x^2-x^2-ax=a>0$$


                                • $x <0implies |x|=-x$ $$implies a+a|x|+x|x|-x|x+a|=a-ax-x^2+x^2+ax=a>0$$







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited Dec 3 '18 at 13:38

























                                answered Dec 3 '18 at 13:18









                                gimusigimusi

                                92.9k94494




                                92.9k94494












                                • $begingroup$
                                  Why would $f(x+1)>f(x)$ imply that $f$ is strictly increasing?
                                  $endgroup$
                                  – Shubham Johri
                                  Dec 3 '18 at 13:31










                                • $begingroup$
                                  @ShubhamJohri Yes sorry, I don't know why I've used one, I fix that . Thanks
                                  $endgroup$
                                  – gimusi
                                  Dec 3 '18 at 13:32


















                                • $begingroup$
                                  Why would $f(x+1)>f(x)$ imply that $f$ is strictly increasing?
                                  $endgroup$
                                  – Shubham Johri
                                  Dec 3 '18 at 13:31










                                • $begingroup$
                                  @ShubhamJohri Yes sorry, I don't know why I've used one, I fix that . Thanks
                                  $endgroup$
                                  – gimusi
                                  Dec 3 '18 at 13:32
















                                $begingroup$
                                Why would $f(x+1)>f(x)$ imply that $f$ is strictly increasing?
                                $endgroup$
                                – Shubham Johri
                                Dec 3 '18 at 13:31




                                $begingroup$
                                Why would $f(x+1)>f(x)$ imply that $f$ is strictly increasing?
                                $endgroup$
                                – Shubham Johri
                                Dec 3 '18 at 13:31












                                $begingroup$
                                @ShubhamJohri Yes sorry, I don't know why I've used one, I fix that . Thanks
                                $endgroup$
                                – gimusi
                                Dec 3 '18 at 13:32




                                $begingroup$
                                @ShubhamJohri Yes sorry, I don't know why I've used one, I fix that . Thanks
                                $endgroup$
                                – gimusi
                                Dec 3 '18 at 13:32


















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