Find the Range and Inverse of the function $f(x)=sec xtan x+sec^2 x$












0












$begingroup$


Find the Range and Inverse of the function $f(x)=sec xtan x+sec^2 x$



My try:



We have $$y=sec xleft(sec x+tan xright)$$



$$2y=left(sec x+tan x+sec x-tan xright)left(sec x+tan xright)$$



$$2y=left(sec x+tan xright)^2+1$$



So we get:



$$sec x+tan x=sqrt{2y-1}$$ Also



$$sec x-tan x-frac{1}{sqrt{2y-1}}$$



Adding above two results we get:



$$x=sec^{-1}left(frac{sqrt{2y-1}+frac{1}{sqrt{2y-1}}}{2}right)$$



Hence $$f^{-1}(x)=sec^{-1}left(frac{sqrt{2x-1}+frac{1}{sqrt{2x-1}}}{2}right)$$



How to find Range?






algebra-precalculus trigonometry optimization maxima-minima inverse-function







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Find the Range and Inverse of the function $f(x)=sec xtan x+sec^2 x$



    My try:



    We have $$y=sec xleft(sec x+tan xright)$$



    $$2y=left(sec x+tan x+sec x-tan xright)left(sec x+tan xright)$$



    $$2y=left(sec x+tan xright)^2+1$$



    So we get:



    $$sec x+tan x=sqrt{2y-1}$$ Also



    $$sec x-tan x-frac{1}{sqrt{2y-1}}$$



    Adding above two results we get:



    $$x=sec^{-1}left(frac{sqrt{2y-1}+frac{1}{sqrt{2y-1}}}{2}right)$$



    Hence $$f^{-1}(x)=sec^{-1}left(frac{sqrt{2x-1}+frac{1}{sqrt{2x-1}}}{2}right)$$



    How to find Range?






    algebra-precalculus trigonometry optimization maxima-minima inverse-function







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Find the Range and Inverse of the function $f(x)=sec xtan x+sec^2 x$



      My try:



      We have $$y=sec xleft(sec x+tan xright)$$



      $$2y=left(sec x+tan x+sec x-tan xright)left(sec x+tan xright)$$



      $$2y=left(sec x+tan xright)^2+1$$



      So we get:



      $$sec x+tan x=sqrt{2y-1}$$ Also



      $$sec x-tan x-frac{1}{sqrt{2y-1}}$$



      Adding above two results we get:



      $$x=sec^{-1}left(frac{sqrt{2y-1}+frac{1}{sqrt{2y-1}}}{2}right)$$



      Hence $$f^{-1}(x)=sec^{-1}left(frac{sqrt{2x-1}+frac{1}{sqrt{2x-1}}}{2}right)$$



      How to find Range?






      algebra-precalculus trigonometry optimization maxima-minima inverse-function







      share|cite|improve this question









      $endgroup$




      Find the Range and Inverse of the function $f(x)=sec xtan x+sec^2 x$



      My try:



      We have $$y=sec xleft(sec x+tan xright)$$



      $$2y=left(sec x+tan x+sec x-tan xright)left(sec x+tan xright)$$



      $$2y=left(sec x+tan xright)^2+1$$



      So we get:



      $$sec x+tan x=sqrt{2y-1}$$ Also



      $$sec x-tan x-frac{1}{sqrt{2y-1}}$$



      Adding above two results we get:



      $$x=sec^{-1}left(frac{sqrt{2y-1}+frac{1}{sqrt{2y-1}}}{2}right)$$



      Hence $$f^{-1}(x)=sec^{-1}left(frac{sqrt{2x-1}+frac{1}{sqrt{2x-1}}}{2}right)$$



      How to find Range?






      algebra-precalculus trigonometry optimization maxima-minima inverse-function




      algebra-precalculus trigonometry optimization maxima-minima inverse-function






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      asked Jan 8 at 4:51









      Umesh shankarUmesh shankar

      2,61931219




      2,61931219






















          1 Answer
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          3












          $begingroup$

          Hint:



          For $cos xne0iffsin xnepm1$ $$f(x)=dfrac1{1-sin x}$$



          Here $0<1-sin x<2iff f(x)>dfrac12$



          If $y=f(x)=cdots,$
          $$f^{-1}(y)=x=arcsinleft(1-dfrac1yright)$$






          share|cite|improve this answer











          $endgroup$













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            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3












            $begingroup$

            Hint:



            For $cos xne0iffsin xnepm1$ $$f(x)=dfrac1{1-sin x}$$



            Here $0<1-sin x<2iff f(x)>dfrac12$



            If $y=f(x)=cdots,$
            $$f^{-1}(y)=x=arcsinleft(1-dfrac1yright)$$






            share|cite|improve this answer











            $endgroup$


















              3












              $begingroup$

              Hint:



              For $cos xne0iffsin xnepm1$ $$f(x)=dfrac1{1-sin x}$$



              Here $0<1-sin x<2iff f(x)>dfrac12$



              If $y=f(x)=cdots,$
              $$f^{-1}(y)=x=arcsinleft(1-dfrac1yright)$$






              share|cite|improve this answer











              $endgroup$
















                3












                3








                3





                $begingroup$

                Hint:



                For $cos xne0iffsin xnepm1$ $$f(x)=dfrac1{1-sin x}$$



                Here $0<1-sin x<2iff f(x)>dfrac12$



                If $y=f(x)=cdots,$
                $$f^{-1}(y)=x=arcsinleft(1-dfrac1yright)$$






                share|cite|improve this answer











                $endgroup$



                Hint:



                For $cos xne0iffsin xnepm1$ $$f(x)=dfrac1{1-sin x}$$



                Here $0<1-sin x<2iff f(x)>dfrac12$



                If $y=f(x)=cdots,$
                $$f^{-1}(y)=x=arcsinleft(1-dfrac1yright)$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 8 at 5:05

























                answered Jan 8 at 5:00









                lab bhattacharjeelab bhattacharjee

                224k15156274




                224k15156274






























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