Proof that $tan(x)leq{}x-frac{pi}{4}+tan(frac{pi}{4})$ using the Mean Value Theorem












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$begingroup$


We're asked to proof the above inequality $forall{}xin(frac{-pi}{2},frac{pi}{4}]$. Although am a bit confused with the fact that $tan(frac{-pi}{2})$ is undefined. And thus am stuck when applying the mean value theorem to proof the statement. What interval should I be taking in order to apply the theorem? Thanks in advance.



My attempt:



Let $f(t)=tan(t)$ such that $tin(frac{-pi}{2},x]$ whereas $xleq{}frac{pi}{4}$ then, because $f$ is continuous on the closed interval and differentiable on the open interval, $exists{}cin(-frac{pi}{2},x)$ such that the difference of the quotient is equal to $f'(c)$ but then again, I can't take $f(-frac{pi}{2})$






real-analysis calculus analysis proof-writing







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    0












    $begingroup$


    We're asked to proof the above inequality $forall{}xin(frac{-pi}{2},frac{pi}{4}]$. Although am a bit confused with the fact that $tan(frac{-pi}{2})$ is undefined. And thus am stuck when applying the mean value theorem to proof the statement. What interval should I be taking in order to apply the theorem? Thanks in advance.



    My attempt:



    Let $f(t)=tan(t)$ such that $tin(frac{-pi}{2},x]$ whereas $xleq{}frac{pi}{4}$ then, because $f$ is continuous on the closed interval and differentiable on the open interval, $exists{}cin(-frac{pi}{2},x)$ such that the difference of the quotient is equal to $f'(c)$ but then again, I can't take $f(-frac{pi}{2})$






    real-analysis calculus analysis proof-writing







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      We're asked to proof the above inequality $forall{}xin(frac{-pi}{2},frac{pi}{4}]$. Although am a bit confused with the fact that $tan(frac{-pi}{2})$ is undefined. And thus am stuck when applying the mean value theorem to proof the statement. What interval should I be taking in order to apply the theorem? Thanks in advance.



      My attempt:



      Let $f(t)=tan(t)$ such that $tin(frac{-pi}{2},x]$ whereas $xleq{}frac{pi}{4}$ then, because $f$ is continuous on the closed interval and differentiable on the open interval, $exists{}cin(-frac{pi}{2},x)$ such that the difference of the quotient is equal to $f'(c)$ but then again, I can't take $f(-frac{pi}{2})$






      real-analysis calculus analysis proof-writing







      share|cite|improve this question











      $endgroup$




      We're asked to proof the above inequality $forall{}xin(frac{-pi}{2},frac{pi}{4}]$. Although am a bit confused with the fact that $tan(frac{-pi}{2})$ is undefined. And thus am stuck when applying the mean value theorem to proof the statement. What interval should I be taking in order to apply the theorem? Thanks in advance.



      My attempt:



      Let $f(t)=tan(t)$ such that $tin(frac{-pi}{2},x]$ whereas $xleq{}frac{pi}{4}$ then, because $f$ is continuous on the closed interval and differentiable on the open interval, $exists{}cin(-frac{pi}{2},x)$ such that the difference of the quotient is equal to $f'(c)$ but then again, I can't take $f(-frac{pi}{2})$






      real-analysis calculus analysis proof-writing




      real-analysis calculus analysis proof-writing






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      share|cite|improve this question













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      edited Dec 9 '18 at 20:16







      kareem bokai

















      asked Dec 9 '18 at 20:01









      kareem bokaikareem bokai

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          2 Answers
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          $begingroup$

          If $x=fracpi4$, then $tan(x)=x-fracpi4+tanleft(fracpi4right)$. On the other and, if $xinleft(-fracpi2,fracpi4right)$, then$$frac{tan(x)-x-left(tanleft(fracpi4right)-fracpi4right)}{x-fracpi4}=tan'(c)-1,$$for some $c$ between $x$ and $fracpi4$. But $tan'(c)-1+=tan^2(c)>0$. Since $x-fracpi4<0$, you deduce that $tan(x)-x-left(tanleft(fracpi4right)-fracpi4right)<0$. In other words, $ tan(x)<x-fracpi4+tanleft(fracpi4right)$.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Why are we allowed to take $(x,frac{pi}{4})$ as an interval in this case? I see now that it leads to the right conclusion but am still confused on why we are capable of doing that. But thank you for your reply :)
            $endgroup$
            – kareem bokai
            Dec 9 '18 at 20:27



















          1












          $begingroup$

          This is just a variant on José Carlos Santos's answer, using the Mean Value Theorem on the function $tan x$ instead of $tan x-x$.



          Rewrite the inequality to be proved first as



          $${piover4}-xletanleft(piover4right)-tan x$$



          and then, since we're only concerned with $xlt{piover4}$ (after noting that the inequality is a trivial equality for $x={piover4}$), as



          $$1le{tanleft(piover4right)-tan xover{piover4}-x}$$



          Now the Mean Value Theorem tells us that $(f(b)-f(a))/(b-a)=f'(c)$ for some $cin(a,b)$, provided the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, which the tangent function is as long as $-pi/2lt x$ (and, here, less than $pi/4$), so we have



          $${tanleft(piover4right)-tan xover{piover4}-x}=sec^2c$$



          for some $cin(x,pi/4)$. Since $sec^2cge1$ for all $c$ (in the appropriate domain), we're done.






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            2 Answers
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            active

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            1












            $begingroup$

            If $x=fracpi4$, then $tan(x)=x-fracpi4+tanleft(fracpi4right)$. On the other and, if $xinleft(-fracpi2,fracpi4right)$, then$$frac{tan(x)-x-left(tanleft(fracpi4right)-fracpi4right)}{x-fracpi4}=tan'(c)-1,$$for some $c$ between $x$ and $fracpi4$. But $tan'(c)-1+=tan^2(c)>0$. Since $x-fracpi4<0$, you deduce that $tan(x)-x-left(tanleft(fracpi4right)-fracpi4right)<0$. In other words, $ tan(x)<x-fracpi4+tanleft(fracpi4right)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why are we allowed to take $(x,frac{pi}{4})$ as an interval in this case? I see now that it leads to the right conclusion but am still confused on why we are capable of doing that. But thank you for your reply :)
              $endgroup$
              – kareem bokai
              Dec 9 '18 at 20:27
















            1












            $begingroup$

            If $x=fracpi4$, then $tan(x)=x-fracpi4+tanleft(fracpi4right)$. On the other and, if $xinleft(-fracpi2,fracpi4right)$, then$$frac{tan(x)-x-left(tanleft(fracpi4right)-fracpi4right)}{x-fracpi4}=tan'(c)-1,$$for some $c$ between $x$ and $fracpi4$. But $tan'(c)-1+=tan^2(c)>0$. Since $x-fracpi4<0$, you deduce that $tan(x)-x-left(tanleft(fracpi4right)-fracpi4right)<0$. In other words, $ tan(x)<x-fracpi4+tanleft(fracpi4right)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Why are we allowed to take $(x,frac{pi}{4})$ as an interval in this case? I see now that it leads to the right conclusion but am still confused on why we are capable of doing that. But thank you for your reply :)
              $endgroup$
              – kareem bokai
              Dec 9 '18 at 20:27














            1












            1








            1





            $begingroup$

            If $x=fracpi4$, then $tan(x)=x-fracpi4+tanleft(fracpi4right)$. On the other and, if $xinleft(-fracpi2,fracpi4right)$, then$$frac{tan(x)-x-left(tanleft(fracpi4right)-fracpi4right)}{x-fracpi4}=tan'(c)-1,$$for some $c$ between $x$ and $fracpi4$. But $tan'(c)-1+=tan^2(c)>0$. Since $x-fracpi4<0$, you deduce that $tan(x)-x-left(tanleft(fracpi4right)-fracpi4right)<0$. In other words, $ tan(x)<x-fracpi4+tanleft(fracpi4right)$.






            share|cite|improve this answer









            $endgroup$



            If $x=fracpi4$, then $tan(x)=x-fracpi4+tanleft(fracpi4right)$. On the other and, if $xinleft(-fracpi2,fracpi4right)$, then$$frac{tan(x)-x-left(tanleft(fracpi4right)-fracpi4right)}{x-fracpi4}=tan'(c)-1,$$for some $c$ between $x$ and $fracpi4$. But $tan'(c)-1+=tan^2(c)>0$. Since $x-fracpi4<0$, you deduce that $tan(x)-x-left(tanleft(fracpi4right)-fracpi4right)<0$. In other words, $ tan(x)<x-fracpi4+tanleft(fracpi4right)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 9 '18 at 20:20









            José Carlos SantosJosé Carlos Santos

            158k22126228




            158k22126228












            • $begingroup$
              Why are we allowed to take $(x,frac{pi}{4})$ as an interval in this case? I see now that it leads to the right conclusion but am still confused on why we are capable of doing that. But thank you for your reply :)
              $endgroup$
              – kareem bokai
              Dec 9 '18 at 20:27


















            • $begingroup$
              Why are we allowed to take $(x,frac{pi}{4})$ as an interval in this case? I see now that it leads to the right conclusion but am still confused on why we are capable of doing that. But thank you for your reply :)
              $endgroup$
              – kareem bokai
              Dec 9 '18 at 20:27
















            $begingroup$
            Why are we allowed to take $(x,frac{pi}{4})$ as an interval in this case? I see now that it leads to the right conclusion but am still confused on why we are capable of doing that. But thank you for your reply :)
            $endgroup$
            – kareem bokai
            Dec 9 '18 at 20:27




            $begingroup$
            Why are we allowed to take $(x,frac{pi}{4})$ as an interval in this case? I see now that it leads to the right conclusion but am still confused on why we are capable of doing that. But thank you for your reply :)
            $endgroup$
            – kareem bokai
            Dec 9 '18 at 20:27











            1












            $begingroup$

            This is just a variant on José Carlos Santos's answer, using the Mean Value Theorem on the function $tan x$ instead of $tan x-x$.



            Rewrite the inequality to be proved first as



            $${piover4}-xletanleft(piover4right)-tan x$$



            and then, since we're only concerned with $xlt{piover4}$ (after noting that the inequality is a trivial equality for $x={piover4}$), as



            $$1le{tanleft(piover4right)-tan xover{piover4}-x}$$



            Now the Mean Value Theorem tells us that $(f(b)-f(a))/(b-a)=f'(c)$ for some $cin(a,b)$, provided the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, which the tangent function is as long as $-pi/2lt x$ (and, here, less than $pi/4$), so we have



            $${tanleft(piover4right)-tan xover{piover4}-x}=sec^2c$$



            for some $cin(x,pi/4)$. Since $sec^2cge1$ for all $c$ (in the appropriate domain), we're done.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              This is just a variant on José Carlos Santos's answer, using the Mean Value Theorem on the function $tan x$ instead of $tan x-x$.



              Rewrite the inequality to be proved first as



              $${piover4}-xletanleft(piover4right)-tan x$$



              and then, since we're only concerned with $xlt{piover4}$ (after noting that the inequality is a trivial equality for $x={piover4}$), as



              $$1le{tanleft(piover4right)-tan xover{piover4}-x}$$



              Now the Mean Value Theorem tells us that $(f(b)-f(a))/(b-a)=f'(c)$ for some $cin(a,b)$, provided the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, which the tangent function is as long as $-pi/2lt x$ (and, here, less than $pi/4$), so we have



              $${tanleft(piover4right)-tan xover{piover4}-x}=sec^2c$$



              for some $cin(x,pi/4)$. Since $sec^2cge1$ for all $c$ (in the appropriate domain), we're done.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                This is just a variant on José Carlos Santos's answer, using the Mean Value Theorem on the function $tan x$ instead of $tan x-x$.



                Rewrite the inequality to be proved first as



                $${piover4}-xletanleft(piover4right)-tan x$$



                and then, since we're only concerned with $xlt{piover4}$ (after noting that the inequality is a trivial equality for $x={piover4}$), as



                $$1le{tanleft(piover4right)-tan xover{piover4}-x}$$



                Now the Mean Value Theorem tells us that $(f(b)-f(a))/(b-a)=f'(c)$ for some $cin(a,b)$, provided the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, which the tangent function is as long as $-pi/2lt x$ (and, here, less than $pi/4$), so we have



                $${tanleft(piover4right)-tan xover{piover4}-x}=sec^2c$$



                for some $cin(x,pi/4)$. Since $sec^2cge1$ for all $c$ (in the appropriate domain), we're done.






                share|cite|improve this answer









                $endgroup$



                This is just a variant on José Carlos Santos's answer, using the Mean Value Theorem on the function $tan x$ instead of $tan x-x$.



                Rewrite the inequality to be proved first as



                $${piover4}-xletanleft(piover4right)-tan x$$



                and then, since we're only concerned with $xlt{piover4}$ (after noting that the inequality is a trivial equality for $x={piover4}$), as



                $$1le{tanleft(piover4right)-tan xover{piover4}-x}$$



                Now the Mean Value Theorem tells us that $(f(b)-f(a))/(b-a)=f'(c)$ for some $cin(a,b)$, provided the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, which the tangent function is as long as $-pi/2lt x$ (and, here, less than $pi/4$), so we have



                $${tanleft(piover4right)-tan xover{piover4}-x}=sec^2c$$



                for some $cin(x,pi/4)$. Since $sec^2cge1$ for all $c$ (in the appropriate domain), we're done.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 9 '18 at 20:47









                Barry CipraBarry Cipra

                59.4k653126




                59.4k653126






























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