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












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












$begingroup$


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











        Your Answer





        StackExchange.ifUsing("editor", function () {
        return StackExchange.using("mathjaxEditing", function () {
        StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
        StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
        });
        });
        }, "mathjax-editing");

        StackExchange.ready(function() {
        var channelOptions = {
        tags: "".split(" "),
        id: "69"
        };
        initTagRenderer("".split(" "), "".split(" "), channelOptions);

        StackExchange.using("externalEditor", function() {
        // Have to fire editor after snippets, if snippets enabled
        if (StackExchange.settings.snippets.snippetsEnabled) {
        StackExchange.using("snippets", function() {
        createEditor();
        });
        }
        else {
        createEditor();
        }
        });

        function createEditor() {
        StackExchange.prepareEditor({
        heartbeatType: 'answer',
        autoActivateHeartbeat: false,
        convertImagesToLinks: true,
        noModals: true,
        showLowRepImageUploadWarning: true,
        reputationToPostImages: 10,
        bindNavPrevention: true,
        postfix: "",
        imageUploader: {
        brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
        contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
        allowUrls: true
        },
        noCode: true, onDemand: true,
        discardSelector: ".discard-answer"
        ,immediatelyShowMarkdownHelp:true
        });


        }
        });














        draft saved

        draft discarded


















        StackExchange.ready(
        function () {
        StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024044%2fhow-can-i-show-the-monotincity-of-fracx1x-without-differential-calculu%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        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


















                                draft saved

                                draft discarded




















































                                Thanks for contributing an answer to Mathematics Stack Exchange!


                                • Please be sure to answer the question. Provide details and share your research!

                                But avoid …



                                • Asking for help, clarification, or responding to other answers.

                                • Making statements based on opinion; back them up with references or personal experience.


                                Use MathJax to format equations. MathJax reference.


                                To learn more, see our tips on writing great answers.




                                draft saved


                                draft discarded














                                StackExchange.ready(
                                function () {
                                StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3024044%2fhow-can-i-show-the-monotincity-of-fracx1x-without-differential-calculu%23new-answer', 'question_page');
                                }
                                );

                                Post as a guest















                                Required, but never shown





















































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown

































                                Required, but never shown














                                Required, but never shown












                                Required, but never shown







                                Required, but never shown







                                -

                                Popular posts from this blog

                                Bundesstraße 106

                                Image
                                Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
                                Read more

                                Verónica Boquete

                                Image
                                Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
                                Read more

                                Ida-Boy-Ed-Garten

                                Image
                                Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
                                Read more