Expression of inverse function of $f(x)=frac{x}{sqrt{x+1}}$












1












$begingroup$


Here's what I did:
$$f(y)=x$$
$$frac{y}{sqrt{y+1}}=x$$
$$sqrt{y+1}=frac{y}{x}$$ $$(xneq0 ,and ,yneq0)$$
$$y+1=frac{y²}{x²}$$
$$y²-x²y²-x²=0$$
$$Delta=x^4+4x²$$
$$y=frac{x²-sqrt{x^4+4x²}}{2};or;y=frac{x²+sqrt{x^4+4x²}}{2}$$
$y=frac{x²-sqrt{x^4+4x²}}{2}$ impossible because $y>-1$
$\$Then we have:$$f^{-1}(x)=frac{x²+sqrt{x^4+4x²}}{2} forall{x}inBbb{R}$$
But the correct answer seems to be
$$f^{-1}(x)=frac{x²+xsqrt{x^2+4}}{2}$$






calculus







share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    Here's what I did:
    $$f(y)=x$$
    $$frac{y}{sqrt{y+1}}=x$$
    $$sqrt{y+1}=frac{y}{x}$$ $$(xneq0 ,and ,yneq0)$$
    $$y+1=frac{y²}{x²}$$
    $$y²-x²y²-x²=0$$
    $$Delta=x^4+4x²$$
    $$y=frac{x²-sqrt{x^4+4x²}}{2};or;y=frac{x²+sqrt{x^4+4x²}}{2}$$
    $y=frac{x²-sqrt{x^4+4x²}}{2}$ impossible because $y>-1$
    $\$Then we have:$$f^{-1}(x)=frac{x²+sqrt{x^4+4x²}}{2} forall{x}inBbb{R}$$
    But the correct answer seems to be
    $$f^{-1}(x)=frac{x²+xsqrt{x^2+4}}{2}$$






    calculus







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Here's what I did:
      $$f(y)=x$$
      $$frac{y}{sqrt{y+1}}=x$$
      $$sqrt{y+1}=frac{y}{x}$$ $$(xneq0 ,and ,yneq0)$$
      $$y+1=frac{y²}{x²}$$
      $$y²-x²y²-x²=0$$
      $$Delta=x^4+4x²$$
      $$y=frac{x²-sqrt{x^4+4x²}}{2};or;y=frac{x²+sqrt{x^4+4x²}}{2}$$
      $y=frac{x²-sqrt{x^4+4x²}}{2}$ impossible because $y>-1$
      $\$Then we have:$$f^{-1}(x)=frac{x²+sqrt{x^4+4x²}}{2} forall{x}inBbb{R}$$
      But the correct answer seems to be
      $$f^{-1}(x)=frac{x²+xsqrt{x^2+4}}{2}$$






      calculus







      share|cite|improve this question









      $endgroup$




      Here's what I did:
      $$f(y)=x$$
      $$frac{y}{sqrt{y+1}}=x$$
      $$sqrt{y+1}=frac{y}{x}$$ $$(xneq0 ,and ,yneq0)$$
      $$y+1=frac{y²}{x²}$$
      $$y²-x²y²-x²=0$$
      $$Delta=x^4+4x²$$
      $$y=frac{x²-sqrt{x^4+4x²}}{2};or;y=frac{x²+sqrt{x^4+4x²}}{2}$$
      $y=frac{x²-sqrt{x^4+4x²}}{2}$ impossible because $y>-1$
      $\$Then we have:$$f^{-1}(x)=frac{x²+sqrt{x^4+4x²}}{2} forall{x}inBbb{R}$$
      But the correct answer seems to be
      $$f^{-1}(x)=frac{x²+xsqrt{x^2+4}}{2}$$






      calculus




      calculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 9 '18 at 10:37









      AlexAlex

      153




      153






















          1 Answer
          1






          active

          oldest

          votes


















          3












          $begingroup$

          Your solution proposes that $f^{-1}(x)$ is always nonnegative, which is not true.



          From $f(x)=frac{x}{sqrt{x+1}}$, we can see that $x$ and its image share the same sign.



          Hence from $$y = frac{x^2pm sqrt{x^4+4x^2}}{2}$$



          We have begin{align}y&=frac{x^2+sign(x)sqrt{x^4+4x^2}}{2}\&=frac{x^2+sign(x)|x|sqrt{x^2+4}}{2}\&= frac{x^2+xsqrt{x^2+4}}{2}end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I understand but is there any other solution without sign function? because i've never seen it before
            $endgroup$
            – Alex
            Dec 9 '18 at 10:51










          • $begingroup$
            @Alex $sign(x)$ is the signum function. It is quite simple to understand: $sign(x)=begin{cases}1,&x>0\0,&x=0\-1,&x<0end{cases}$ or $sign(x)=begin{cases}frac{|x|}x,&xne0\0,&x=0end{cases}$
            $endgroup$
            – Shubham Johri
            Dec 9 '18 at 10:54













          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%2f3032249%2fexpression-of-inverse-function-of-fx-fracx-sqrtx1%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          Your solution proposes that $f^{-1}(x)$ is always nonnegative, which is not true.



          From $f(x)=frac{x}{sqrt{x+1}}$, we can see that $x$ and its image share the same sign.



          Hence from $$y = frac{x^2pm sqrt{x^4+4x^2}}{2}$$



          We have begin{align}y&=frac{x^2+sign(x)sqrt{x^4+4x^2}}{2}\&=frac{x^2+sign(x)|x|sqrt{x^2+4}}{2}\&= frac{x^2+xsqrt{x^2+4}}{2}end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I understand but is there any other solution without sign function? because i've never seen it before
            $endgroup$
            – Alex
            Dec 9 '18 at 10:51










          • $begingroup$
            @Alex $sign(x)$ is the signum function. It is quite simple to understand: $sign(x)=begin{cases}1,&x>0\0,&x=0\-1,&x<0end{cases}$ or $sign(x)=begin{cases}frac{|x|}x,&xne0\0,&x=0end{cases}$
            $endgroup$
            – Shubham Johri
            Dec 9 '18 at 10:54


















          3












          $begingroup$

          Your solution proposes that $f^{-1}(x)$ is always nonnegative, which is not true.



          From $f(x)=frac{x}{sqrt{x+1}}$, we can see that $x$ and its image share the same sign.



          Hence from $$y = frac{x^2pm sqrt{x^4+4x^2}}{2}$$



          We have begin{align}y&=frac{x^2+sign(x)sqrt{x^4+4x^2}}{2}\&=frac{x^2+sign(x)|x|sqrt{x^2+4}}{2}\&= frac{x^2+xsqrt{x^2+4}}{2}end{align}






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            I understand but is there any other solution without sign function? because i've never seen it before
            $endgroup$
            – Alex
            Dec 9 '18 at 10:51










          • $begingroup$
            @Alex $sign(x)$ is the signum function. It is quite simple to understand: $sign(x)=begin{cases}1,&x>0\0,&x=0\-1,&x<0end{cases}$ or $sign(x)=begin{cases}frac{|x|}x,&xne0\0,&x=0end{cases}$
            $endgroup$
            – Shubham Johri
            Dec 9 '18 at 10:54
















          3












          3








          3





          $begingroup$

          Your solution proposes that $f^{-1}(x)$ is always nonnegative, which is not true.



          From $f(x)=frac{x}{sqrt{x+1}}$, we can see that $x$ and its image share the same sign.



          Hence from $$y = frac{x^2pm sqrt{x^4+4x^2}}{2}$$



          We have begin{align}y&=frac{x^2+sign(x)sqrt{x^4+4x^2}}{2}\&=frac{x^2+sign(x)|x|sqrt{x^2+4}}{2}\&= frac{x^2+xsqrt{x^2+4}}{2}end{align}






          share|cite|improve this answer









          $endgroup$



          Your solution proposes that $f^{-1}(x)$ is always nonnegative, which is not true.



          From $f(x)=frac{x}{sqrt{x+1}}$, we can see that $x$ and its image share the same sign.



          Hence from $$y = frac{x^2pm sqrt{x^4+4x^2}}{2}$$



          We have begin{align}y&=frac{x^2+sign(x)sqrt{x^4+4x^2}}{2}\&=frac{x^2+sign(x)|x|sqrt{x^2+4}}{2}\&= frac{x^2+xsqrt{x^2+4}}{2}end{align}







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 9 '18 at 10:44









          Siong Thye GohSiong Thye Goh

          100k1466117




          100k1466117












          • $begingroup$
            I understand but is there any other solution without sign function? because i've never seen it before
            $endgroup$
            – Alex
            Dec 9 '18 at 10:51










          • $begingroup$
            @Alex $sign(x)$ is the signum function. It is quite simple to understand: $sign(x)=begin{cases}1,&x>0\0,&x=0\-1,&x<0end{cases}$ or $sign(x)=begin{cases}frac{|x|}x,&xne0\0,&x=0end{cases}$
            $endgroup$
            – Shubham Johri
            Dec 9 '18 at 10:54




















          • $begingroup$
            I understand but is there any other solution without sign function? because i've never seen it before
            $endgroup$
            – Alex
            Dec 9 '18 at 10:51










          • $begingroup$
            @Alex $sign(x)$ is the signum function. It is quite simple to understand: $sign(x)=begin{cases}1,&x>0\0,&x=0\-1,&x<0end{cases}$ or $sign(x)=begin{cases}frac{|x|}x,&xne0\0,&x=0end{cases}$
            $endgroup$
            – Shubham Johri
            Dec 9 '18 at 10:54


















          $begingroup$
          I understand but is there any other solution without sign function? because i've never seen it before
          $endgroup$
          – Alex
          Dec 9 '18 at 10:51




          $begingroup$
          I understand but is there any other solution without sign function? because i've never seen it before
          $endgroup$
          – Alex
          Dec 9 '18 at 10:51












          $begingroup$
          @Alex $sign(x)$ is the signum function. It is quite simple to understand: $sign(x)=begin{cases}1,&x>0\0,&x=0\-1,&x<0end{cases}$ or $sign(x)=begin{cases}frac{|x|}x,&xne0\0,&x=0end{cases}$
          $endgroup$
          – Shubham Johri
          Dec 9 '18 at 10:54






          $begingroup$
          @Alex $sign(x)$ is the signum function. It is quite simple to understand: $sign(x)=begin{cases}1,&x>0\0,&x=0\-1,&x<0end{cases}$ or $sign(x)=begin{cases}frac{|x|}x,&xne0\0,&x=0end{cases}$
          $endgroup$
          – Shubham Johri
          Dec 9 '18 at 10:54




















          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%2f3032249%2fexpression-of-inverse-function-of-fx-fracx-sqrtx1%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

          Berounka

          Image
          Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
          Read more

          Sphinx de Gizeh

          Image
          Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
          Read more

          Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

          Image
          3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
          Read more