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

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          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









          $endgroup$














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            3 Answers
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            active

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            3 Answers
            3






            active

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            active

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            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






























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