Prove $2^{(h(x)-5)}+3x$ is not equal to $sin(x)$












1












$begingroup$


I am working on my CalcII class and have been doing reasonably well so far. I'm now stuck with this exercises:
Given $$ h(x): mathbb R+0 rightarrow mathbb R , h(x+Delta x) > h(x) , forall x in D$$
Prove that:



$$2 ^ {h(x)-5} + 3x neq sin(x) ,forall xgeq 0 $$



I am really at a loss here: I hate to say this, but I cannot see where to start here. I can see $h(x)$ is strictly increasing; however, I fail to see how this could help me to prove the inequality.



I'll be very grateful if anyone could point me in the right direction so I can work it out.



Lots of thanks in advance.






calculus







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    I am working on my CalcII class and have been doing reasonably well so far. I'm now stuck with this exercises:
    Given $$ h(x): mathbb R+0 rightarrow mathbb R , h(x+Delta x) > h(x) , forall x in D$$
    Prove that:



    $$2 ^ {h(x)-5} + 3x neq sin(x) ,forall xgeq 0 $$



    I am really at a loss here: I hate to say this, but I cannot see where to start here. I can see $h(x)$ is strictly increasing; however, I fail to see how this could help me to prove the inequality.



    I'll be very grateful if anyone could point me in the right direction so I can work it out.



    Lots of thanks in advance.






    calculus







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      I am working on my CalcII class and have been doing reasonably well so far. I'm now stuck with this exercises:
      Given $$ h(x): mathbb R+0 rightarrow mathbb R , h(x+Delta x) > h(x) , forall x in D$$
      Prove that:



      $$2 ^ {h(x)-5} + 3x neq sin(x) ,forall xgeq 0 $$



      I am really at a loss here: I hate to say this, but I cannot see where to start here. I can see $h(x)$ is strictly increasing; however, I fail to see how this could help me to prove the inequality.



      I'll be very grateful if anyone could point me in the right direction so I can work it out.



      Lots of thanks in advance.






      calculus







      share|cite|improve this question











      $endgroup$




      I am working on my CalcII class and have been doing reasonably well so far. I'm now stuck with this exercises:
      Given $$ h(x): mathbb R+0 rightarrow mathbb R , h(x+Delta x) > h(x) , forall x in D$$
      Prove that:



      $$2 ^ {h(x)-5} + 3x neq sin(x) ,forall xgeq 0 $$



      I am really at a loss here: I hate to say this, but I cannot see where to start here. I can see $h(x)$ is strictly increasing; however, I fail to see how this could help me to prove the inequality.



      I'll be very grateful if anyone could point me in the right direction so I can work it out.



      Lots of thanks in advance.






      calculus




      calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 25 '18 at 2:59









      Namaste

      1




      1










      asked Dec 25 '18 at 2:36









      ArtemArtem

      425




      425






















          3 Answers
          3






          active

          oldest

          votes


















          1












          $begingroup$

          The $$h(x+Delta x)>h(x) forall x $$
          tells us $h'(x) >0$ and therefore that $h(x)$ is strictly increasing. You saw this.
          Now, we have that $$xin Bbb R to sin x not> 1$$
          and that $$0<x<1 to sin x approx x$$
          Which implies $$3x>sin x$$



          All you need for the $2^{h(x)-5}$ is that:



          $$alpha in Bbb R implies 2^alpha > 0$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why is $h$ differentiable?
            $endgroup$
            – Michael Burr
            Dec 25 '18 at 3:07










          • $begingroup$
            I'm just using the definition of the derivative from first principles.
            $endgroup$
            – Rhys Hughes
            Dec 25 '18 at 3:31










          • $begingroup$
            In the statement, $Delta x$ is not quantified, so we don't know that the inequality holds for all $Delta x$'s. Even if it were universally quantified, the inequality would hold for a jump discontinuity, which isn't differentiable.
            $endgroup$
            – Michael Burr
            Dec 25 '18 at 11:57





















          6












          $begingroup$

          For $x > 0$, $3x > sin x$. Then we add something positive to the side that's already larger...



          At $x=0$, we're instead comparing (zero + positive) to (zero), and $3x+2^{text{whatever}} > sin x$. Combine the two parts, and $2^{h(x)-5} + 3x > sin x$ for all $xge 0$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Of course, @Ross Millikan. I considered using $ge$ signs instead, but wrote the strict inequality version instead. The full solution is probably a little simpler with the $ge$ signs, because then we don't have to handle zero separately.
            $endgroup$
            – jmerry
            Dec 25 '18 at 3:02










          • $begingroup$
            Point made. I've added to the answer to cover that case.
            $endgroup$
            – jmerry
            Dec 25 '18 at 3:19










          • $begingroup$
            Thanks. I had already upvoted, so can't do it again.
            $endgroup$
            – Ross Millikan
            Dec 25 '18 at 3:21



















          0












          $begingroup$

          Recall from geometric constructions that $sin x leq x$. Also, $x leq 3x < 3x + 2^y$ because $2^y > 0$ for all $yin Bbb R$. Combine these and you're done.






          share|cite|improve this answer









          $endgroup$














            Your Answer








            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%2f3051785%2fprove-2hx-53x-is-not-equal-to-sinx%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            The $$h(x+Delta x)>h(x) forall x $$
            tells us $h'(x) >0$ and therefore that $h(x)$ is strictly increasing. You saw this.
            Now, we have that $$xin Bbb R to sin x not> 1$$
            and that $$0<x<1 to sin x approx x$$
            Which implies $$3x>sin x$$



            All you need for the $2^{h(x)-5}$ is that:



            $$alpha in Bbb R implies 2^alpha > 0$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why is $h$ differentiable?
              $endgroup$
              – Michael Burr
              Dec 25 '18 at 3:07










            • $begingroup$
              I'm just using the definition of the derivative from first principles.
              $endgroup$
              – Rhys Hughes
              Dec 25 '18 at 3:31










            • $begingroup$
              In the statement, $Delta x$ is not quantified, so we don't know that the inequality holds for all $Delta x$'s. Even if it were universally quantified, the inequality would hold for a jump discontinuity, which isn't differentiable.
              $endgroup$
              – Michael Burr
              Dec 25 '18 at 11:57


















            1












            $begingroup$

            The $$h(x+Delta x)>h(x) forall x $$
            tells us $h'(x) >0$ and therefore that $h(x)$ is strictly increasing. You saw this.
            Now, we have that $$xin Bbb R to sin x not> 1$$
            and that $$0<x<1 to sin x approx x$$
            Which implies $$3x>sin x$$



            All you need for the $2^{h(x)-5}$ is that:



            $$alpha in Bbb R implies 2^alpha > 0$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why is $h$ differentiable?
              $endgroup$
              – Michael Burr
              Dec 25 '18 at 3:07










            • $begingroup$
              I'm just using the definition of the derivative from first principles.
              $endgroup$
              – Rhys Hughes
              Dec 25 '18 at 3:31










            • $begingroup$
              In the statement, $Delta x$ is not quantified, so we don't know that the inequality holds for all $Delta x$'s. Even if it were universally quantified, the inequality would hold for a jump discontinuity, which isn't differentiable.
              $endgroup$
              – Michael Burr
              Dec 25 '18 at 11:57
















            1












            1








            1





            $begingroup$

            The $$h(x+Delta x)>h(x) forall x $$
            tells us $h'(x) >0$ and therefore that $h(x)$ is strictly increasing. You saw this.
            Now, we have that $$xin Bbb R to sin x not> 1$$
            and that $$0<x<1 to sin x approx x$$
            Which implies $$3x>sin x$$



            All you need for the $2^{h(x)-5}$ is that:



            $$alpha in Bbb R implies 2^alpha > 0$$






            share|cite|improve this answer









            $endgroup$



            The $$h(x+Delta x)>h(x) forall x $$
            tells us $h'(x) >0$ and therefore that $h(x)$ is strictly increasing. You saw this.
            Now, we have that $$xin Bbb R to sin x not> 1$$
            and that $$0<x<1 to sin x approx x$$
            Which implies $$3x>sin x$$



            All you need for the $2^{h(x)-5}$ is that:



            $$alpha in Bbb R implies 2^alpha > 0$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 25 '18 at 2:58









            Rhys HughesRhys Hughes

            7,1341630




            7,1341630












            • $begingroup$
              Why is $h$ differentiable?
              $endgroup$
              – Michael Burr
              Dec 25 '18 at 3:07










            • $begingroup$
              I'm just using the definition of the derivative from first principles.
              $endgroup$
              – Rhys Hughes
              Dec 25 '18 at 3:31










            • $begingroup$
              In the statement, $Delta x$ is not quantified, so we don't know that the inequality holds for all $Delta x$'s. Even if it were universally quantified, the inequality would hold for a jump discontinuity, which isn't differentiable.
              $endgroup$
              – Michael Burr
              Dec 25 '18 at 11:57




















            • $begingroup$
              Why is $h$ differentiable?
              $endgroup$
              – Michael Burr
              Dec 25 '18 at 3:07










            • $begingroup$
              I'm just using the definition of the derivative from first principles.
              $endgroup$
              – Rhys Hughes
              Dec 25 '18 at 3:31










            • $begingroup$
              In the statement, $Delta x$ is not quantified, so we don't know that the inequality holds for all $Delta x$'s. Even if it were universally quantified, the inequality would hold for a jump discontinuity, which isn't differentiable.
              $endgroup$
              – Michael Burr
              Dec 25 '18 at 11:57


















            $begingroup$
            Why is $h$ differentiable?
            $endgroup$
            – Michael Burr
            Dec 25 '18 at 3:07




            $begingroup$
            Why is $h$ differentiable?
            $endgroup$
            – Michael Burr
            Dec 25 '18 at 3:07












            $begingroup$
            I'm just using the definition of the derivative from first principles.
            $endgroup$
            – Rhys Hughes
            Dec 25 '18 at 3:31




            $begingroup$
            I'm just using the definition of the derivative from first principles.
            $endgroup$
            – Rhys Hughes
            Dec 25 '18 at 3:31












            $begingroup$
            In the statement, $Delta x$ is not quantified, so we don't know that the inequality holds for all $Delta x$'s. Even if it were universally quantified, the inequality would hold for a jump discontinuity, which isn't differentiable.
            $endgroup$
            – Michael Burr
            Dec 25 '18 at 11:57






            $begingroup$
            In the statement, $Delta x$ is not quantified, so we don't know that the inequality holds for all $Delta x$'s. Even if it were universally quantified, the inequality would hold for a jump discontinuity, which isn't differentiable.
            $endgroup$
            – Michael Burr
            Dec 25 '18 at 11:57













            6












            $begingroup$

            For $x > 0$, $3x > sin x$. Then we add something positive to the side that's already larger...



            At $x=0$, we're instead comparing (zero + positive) to (zero), and $3x+2^{text{whatever}} > sin x$. Combine the two parts, and $2^{h(x)-5} + 3x > sin x$ for all $xge 0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Of course, @Ross Millikan. I considered using $ge$ signs instead, but wrote the strict inequality version instead. The full solution is probably a little simpler with the $ge$ signs, because then we don't have to handle zero separately.
              $endgroup$
              – jmerry
              Dec 25 '18 at 3:02










            • $begingroup$
              Point made. I've added to the answer to cover that case.
              $endgroup$
              – jmerry
              Dec 25 '18 at 3:19










            • $begingroup$
              Thanks. I had already upvoted, so can't do it again.
              $endgroup$
              – Ross Millikan
              Dec 25 '18 at 3:21
















            6












            $begingroup$

            For $x > 0$, $3x > sin x$. Then we add something positive to the side that's already larger...



            At $x=0$, we're instead comparing (zero + positive) to (zero), and $3x+2^{text{whatever}} > sin x$. Combine the two parts, and $2^{h(x)-5} + 3x > sin x$ for all $xge 0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Of course, @Ross Millikan. I considered using $ge$ signs instead, but wrote the strict inequality version instead. The full solution is probably a little simpler with the $ge$ signs, because then we don't have to handle zero separately.
              $endgroup$
              – jmerry
              Dec 25 '18 at 3:02










            • $begingroup$
              Point made. I've added to the answer to cover that case.
              $endgroup$
              – jmerry
              Dec 25 '18 at 3:19










            • $begingroup$
              Thanks. I had already upvoted, so can't do it again.
              $endgroup$
              – Ross Millikan
              Dec 25 '18 at 3:21














            6












            6








            6





            $begingroup$

            For $x > 0$, $3x > sin x$. Then we add something positive to the side that's already larger...



            At $x=0$, we're instead comparing (zero + positive) to (zero), and $3x+2^{text{whatever}} > sin x$. Combine the two parts, and $2^{h(x)-5} + 3x > sin x$ for all $xge 0$.






            share|cite|improve this answer











            $endgroup$



            For $x > 0$, $3x > sin x$. Then we add something positive to the side that's already larger...



            At $x=0$, we're instead comparing (zero + positive) to (zero), and $3x+2^{text{whatever}} > sin x$. Combine the two parts, and $2^{h(x)-5} + 3x > sin x$ for all $xge 0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 25 '18 at 3:18

























            answered Dec 25 '18 at 2:41









            jmerryjmerry

            17.1k11633




            17.1k11633












            • $begingroup$
              Of course, @Ross Millikan. I considered using $ge$ signs instead, but wrote the strict inequality version instead. The full solution is probably a little simpler with the $ge$ signs, because then we don't have to handle zero separately.
              $endgroup$
              – jmerry
              Dec 25 '18 at 3:02










            • $begingroup$
              Point made. I've added to the answer to cover that case.
              $endgroup$
              – jmerry
              Dec 25 '18 at 3:19










            • $begingroup$
              Thanks. I had already upvoted, so can't do it again.
              $endgroup$
              – Ross Millikan
              Dec 25 '18 at 3:21


















            • $begingroup$
              Of course, @Ross Millikan. I considered using $ge$ signs instead, but wrote the strict inequality version instead. The full solution is probably a little simpler with the $ge$ signs, because then we don't have to handle zero separately.
              $endgroup$
              – jmerry
              Dec 25 '18 at 3:02










            • $begingroup$
              Point made. I've added to the answer to cover that case.
              $endgroup$
              – jmerry
              Dec 25 '18 at 3:19










            • $begingroup$
              Thanks. I had already upvoted, so can't do it again.
              $endgroup$
              – Ross Millikan
              Dec 25 '18 at 3:21
















            $begingroup$
            Of course, @Ross Millikan. I considered using $ge$ signs instead, but wrote the strict inequality version instead. The full solution is probably a little simpler with the $ge$ signs, because then we don't have to handle zero separately.
            $endgroup$
            – jmerry
            Dec 25 '18 at 3:02




            $begingroup$
            Of course, @Ross Millikan. I considered using $ge$ signs instead, but wrote the strict inequality version instead. The full solution is probably a little simpler with the $ge$ signs, because then we don't have to handle zero separately.
            $endgroup$
            – jmerry
            Dec 25 '18 at 3:02












            $begingroup$
            Point made. I've added to the answer to cover that case.
            $endgroup$
            – jmerry
            Dec 25 '18 at 3:19




            $begingroup$
            Point made. I've added to the answer to cover that case.
            $endgroup$
            – jmerry
            Dec 25 '18 at 3:19












            $begingroup$
            Thanks. I had already upvoted, so can't do it again.
            $endgroup$
            – Ross Millikan
            Dec 25 '18 at 3:21




            $begingroup$
            Thanks. I had already upvoted, so can't do it again.
            $endgroup$
            – Ross Millikan
            Dec 25 '18 at 3:21











            0












            $begingroup$

            Recall from geometric constructions that $sin x leq x$. Also, $x leq 3x < 3x + 2^y$ because $2^y > 0$ for all $yin Bbb R$. Combine these and you're done.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              Recall from geometric constructions that $sin x leq x$. Also, $x leq 3x < 3x + 2^y$ because $2^y > 0$ for all $yin Bbb R$. Combine these and you're done.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                Recall from geometric constructions that $sin x leq x$. Also, $x leq 3x < 3x + 2^y$ because $2^y > 0$ for all $yin Bbb R$. Combine these and you're done.






                share|cite|improve this answer









                $endgroup$



                Recall from geometric constructions that $sin x leq x$. Also, $x leq 3x < 3x + 2^y$ because $2^y > 0$ for all $yin Bbb R$. Combine these and you're done.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 25 '18 at 7:21









                Lucas HenriqueLucas Henrique

                1,031414




                1,031414






























                    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%2f3051785%2fprove-2hx-53x-is-not-equal-to-sinx%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

                    Plaza Victoria

                    Image
                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
                    Read more

                    In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

                    Image
                    .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
                    Read more

                    How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

                    Image
                    1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
                    Read more