inverse function of a squared












-1












$begingroup$


i'm stuck on trying to find the inverse of the following function:



$f($$x) = {sqrt{x^2-4x}}$



What i did so far is:



$y = {sqrt{x^2-4x}}$



Swap



$x = {sqrt{y^2-4y}}$



Remove the root by squaring both sides



$x^2 = {y^2-4y}$



$y^2 - 4y - x^2 = 0$



and here is where i'm stuck, how should i proceed?






calculus inverse inverse-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    That is not a function, since you didn't tell us what its domain is.
    $endgroup$
    – José Carlos Santos
    Nov 24 '18 at 11:28
















-1












$begingroup$


i'm stuck on trying to find the inverse of the following function:



$f($$x) = {sqrt{x^2-4x}}$



What i did so far is:



$y = {sqrt{x^2-4x}}$



Swap



$x = {sqrt{y^2-4y}}$



Remove the root by squaring both sides



$x^2 = {y^2-4y}$



$y^2 - 4y - x^2 = 0$



and here is where i'm stuck, how should i proceed?






calculus inverse inverse-function







share|cite|improve this question









$endgroup$












  • $begingroup$
    That is not a function, since you didn't tell us what its domain is.
    $endgroup$
    – José Carlos Santos
    Nov 24 '18 at 11:28














-1












-1








-1





$begingroup$


i'm stuck on trying to find the inverse of the following function:



$f($$x) = {sqrt{x^2-4x}}$



What i did so far is:



$y = {sqrt{x^2-4x}}$



Swap



$x = {sqrt{y^2-4y}}$



Remove the root by squaring both sides



$x^2 = {y^2-4y}$



$y^2 - 4y - x^2 = 0$



and here is where i'm stuck, how should i proceed?






calculus inverse inverse-function







share|cite|improve this question









$endgroup$




i'm stuck on trying to find the inverse of the following function:



$f($$x) = {sqrt{x^2-4x}}$



What i did so far is:



$y = {sqrt{x^2-4x}}$



Swap



$x = {sqrt{y^2-4y}}$



Remove the root by squaring both sides



$x^2 = {y^2-4y}$



$y^2 - 4y - x^2 = 0$



and here is where i'm stuck, how should i proceed?






calculus inverse inverse-function




calculus inverse inverse-function






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 '18 at 11:23









user3106888user3106888

61




61












  • $begingroup$
    That is not a function, since you didn't tell us what its domain is.
    $endgroup$
    – José Carlos Santos
    Nov 24 '18 at 11:28


















  • $begingroup$
    That is not a function, since you didn't tell us what its domain is.
    $endgroup$
    – José Carlos Santos
    Nov 24 '18 at 11:28
















$begingroup$
That is not a function, since you didn't tell us what its domain is.
$endgroup$
– José Carlos Santos
Nov 24 '18 at 11:28




$begingroup$
That is not a function, since you didn't tell us what its domain is.
$endgroup$
– José Carlos Santos
Nov 24 '18 at 11:28










2 Answers
2






active

oldest

votes


















2












$begingroup$

To find its inverse, let's find the domain of $f(x)$ for start. Since it involves a square root, the root's argument must be non-negative, thus :



$$D_f = {x in mathbb R : x^2-4x geq 0 } = x in(-infty,0]cup[4,+infty) $$



Now, let's solve for $x$ :



$$f(x) = sqrt{x^2-4x} Leftrightarrow y = sqrt{x^2-4x} Leftrightarrow y^2 =x^2-4x $$



$$Leftrightarrow$$



$$x^2-4x-y^2 = 0$$



You can solve this equation with respect to $x$ now, by letting the determinant be $Delta = 4^2 +4y^2$ which is always positive, thus it has two unique and real solutions :



$$x_{1,2} = frac{4pmsqrt{4^2+4y^2}}{2}$$



Now, it is important to know the domain of your given function when you want to inverse. The domain will lead you to rejecting one of the two signs of the proposed solution. The domain of the inverse function will be the range of the initial function. Since a root is always non-negative, then the domain of the inverse must be $[0,infty)$ and that can lead you to determining $f^{-1}(x)$ by keeping this in mind and by interchaning $x$ and $y$ in the final expression yielded.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Assuming domain is $mathbb{R} setminus(0,4)$ and co-domian is $[0,infty) $ $y^2+4=x^2-4x+4implies y^2+4=(x-2)^2implies x=2pmsqrt{y^2+4}$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      You're missing out on a negative sign.
      $endgroup$
      – Rebellos
      Nov 24 '18 at 11:30






    • 1




      $begingroup$
      It should be a positive sign only, as range of the given function is positive.
      $endgroup$
      – Martund
      Nov 24 '18 at 11:34










    • $begingroup$
      @Crazyformaths Needs an elaboration, which is the exact point of inverses and domains.
      $endgroup$
      – Rebellos
      Nov 24 '18 at 11:34






    • 1




      $begingroup$
      $mathbb{R} setminus{(-4,4)}$ or $mathbb{R} setminus(-4,4)$ ?
      $endgroup$
      – Nosrati
      Nov 24 '18 at 11:35






    • 1




      $begingroup$
      @Rebellos: Sorry the domain was wrong.
      $endgroup$
      – Yadati Kiran
      Nov 24 '18 at 11:37











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    2 Answers
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    active

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    2 Answers
    2






    active

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    active

    oldest

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    active

    oldest

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    2












    $begingroup$

    To find its inverse, let's find the domain of $f(x)$ for start. Since it involves a square root, the root's argument must be non-negative, thus :



    $$D_f = {x in mathbb R : x^2-4x geq 0 } = x in(-infty,0]cup[4,+infty) $$



    Now, let's solve for $x$ :



    $$f(x) = sqrt{x^2-4x} Leftrightarrow y = sqrt{x^2-4x} Leftrightarrow y^2 =x^2-4x $$



    $$Leftrightarrow$$



    $$x^2-4x-y^2 = 0$$



    You can solve this equation with respect to $x$ now, by letting the determinant be $Delta = 4^2 +4y^2$ which is always positive, thus it has two unique and real solutions :



    $$x_{1,2} = frac{4pmsqrt{4^2+4y^2}}{2}$$



    Now, it is important to know the domain of your given function when you want to inverse. The domain will lead you to rejecting one of the two signs of the proposed solution. The domain of the inverse function will be the range of the initial function. Since a root is always non-negative, then the domain of the inverse must be $[0,infty)$ and that can lead you to determining $f^{-1}(x)$ by keeping this in mind and by interchaning $x$ and $y$ in the final expression yielded.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      To find its inverse, let's find the domain of $f(x)$ for start. Since it involves a square root, the root's argument must be non-negative, thus :



      $$D_f = {x in mathbb R : x^2-4x geq 0 } = x in(-infty,0]cup[4,+infty) $$



      Now, let's solve for $x$ :



      $$f(x) = sqrt{x^2-4x} Leftrightarrow y = sqrt{x^2-4x} Leftrightarrow y^2 =x^2-4x $$



      $$Leftrightarrow$$



      $$x^2-4x-y^2 = 0$$



      You can solve this equation with respect to $x$ now, by letting the determinant be $Delta = 4^2 +4y^2$ which is always positive, thus it has two unique and real solutions :



      $$x_{1,2} = frac{4pmsqrt{4^2+4y^2}}{2}$$



      Now, it is important to know the domain of your given function when you want to inverse. The domain will lead you to rejecting one of the two signs of the proposed solution. The domain of the inverse function will be the range of the initial function. Since a root is always non-negative, then the domain of the inverse must be $[0,infty)$ and that can lead you to determining $f^{-1}(x)$ by keeping this in mind and by interchaning $x$ and $y$ in the final expression yielded.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        To find its inverse, let's find the domain of $f(x)$ for start. Since it involves a square root, the root's argument must be non-negative, thus :



        $$D_f = {x in mathbb R : x^2-4x geq 0 } = x in(-infty,0]cup[4,+infty) $$



        Now, let's solve for $x$ :



        $$f(x) = sqrt{x^2-4x} Leftrightarrow y = sqrt{x^2-4x} Leftrightarrow y^2 =x^2-4x $$



        $$Leftrightarrow$$



        $$x^2-4x-y^2 = 0$$



        You can solve this equation with respect to $x$ now, by letting the determinant be $Delta = 4^2 +4y^2$ which is always positive, thus it has two unique and real solutions :



        $$x_{1,2} = frac{4pmsqrt{4^2+4y^2}}{2}$$



        Now, it is important to know the domain of your given function when you want to inverse. The domain will lead you to rejecting one of the two signs of the proposed solution. The domain of the inverse function will be the range of the initial function. Since a root is always non-negative, then the domain of the inverse must be $[0,infty)$ and that can lead you to determining $f^{-1}(x)$ by keeping this in mind and by interchaning $x$ and $y$ in the final expression yielded.






        share|cite|improve this answer









        $endgroup$



        To find its inverse, let's find the domain of $f(x)$ for start. Since it involves a square root, the root's argument must be non-negative, thus :



        $$D_f = {x in mathbb R : x^2-4x geq 0 } = x in(-infty,0]cup[4,+infty) $$



        Now, let's solve for $x$ :



        $$f(x) = sqrt{x^2-4x} Leftrightarrow y = sqrt{x^2-4x} Leftrightarrow y^2 =x^2-4x $$



        $$Leftrightarrow$$



        $$x^2-4x-y^2 = 0$$



        You can solve this equation with respect to $x$ now, by letting the determinant be $Delta = 4^2 +4y^2$ which is always positive, thus it has two unique and real solutions :



        $$x_{1,2} = frac{4pmsqrt{4^2+4y^2}}{2}$$



        Now, it is important to know the domain of your given function when you want to inverse. The domain will lead you to rejecting one of the two signs of the proposed solution. The domain of the inverse function will be the range of the initial function. Since a root is always non-negative, then the domain of the inverse must be $[0,infty)$ and that can lead you to determining $f^{-1}(x)$ by keeping this in mind and by interchaning $x$ and $y$ in the final expression yielded.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 '18 at 11:35









        RebellosRebellos

        14.5k31246




        14.5k31246























            1












            $begingroup$

            Assuming domain is $mathbb{R} setminus(0,4)$ and co-domian is $[0,infty) $ $y^2+4=x^2-4x+4implies y^2+4=(x-2)^2implies x=2pmsqrt{y^2+4}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              You're missing out on a negative sign.
              $endgroup$
              – Rebellos
              Nov 24 '18 at 11:30






            • 1




              $begingroup$
              It should be a positive sign only, as range of the given function is positive.
              $endgroup$
              – Martund
              Nov 24 '18 at 11:34










            • $begingroup$
              @Crazyformaths Needs an elaboration, which is the exact point of inverses and domains.
              $endgroup$
              – Rebellos
              Nov 24 '18 at 11:34






            • 1




              $begingroup$
              $mathbb{R} setminus{(-4,4)}$ or $mathbb{R} setminus(-4,4)$ ?
              $endgroup$
              – Nosrati
              Nov 24 '18 at 11:35






            • 1




              $begingroup$
              @Rebellos: Sorry the domain was wrong.
              $endgroup$
              – Yadati Kiran
              Nov 24 '18 at 11:37
















            1












            $begingroup$

            Assuming domain is $mathbb{R} setminus(0,4)$ and co-domian is $[0,infty) $ $y^2+4=x^2-4x+4implies y^2+4=(x-2)^2implies x=2pmsqrt{y^2+4}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              You're missing out on a negative sign.
              $endgroup$
              – Rebellos
              Nov 24 '18 at 11:30






            • 1




              $begingroup$
              It should be a positive sign only, as range of the given function is positive.
              $endgroup$
              – Martund
              Nov 24 '18 at 11:34










            • $begingroup$
              @Crazyformaths Needs an elaboration, which is the exact point of inverses and domains.
              $endgroup$
              – Rebellos
              Nov 24 '18 at 11:34






            • 1




              $begingroup$
              $mathbb{R} setminus{(-4,4)}$ or $mathbb{R} setminus(-4,4)$ ?
              $endgroup$
              – Nosrati
              Nov 24 '18 at 11:35






            • 1




              $begingroup$
              @Rebellos: Sorry the domain was wrong.
              $endgroup$
              – Yadati Kiran
              Nov 24 '18 at 11:37














            1












            1








            1





            $begingroup$

            Assuming domain is $mathbb{R} setminus(0,4)$ and co-domian is $[0,infty) $ $y^2+4=x^2-4x+4implies y^2+4=(x-2)^2implies x=2pmsqrt{y^2+4}$






            share|cite|improve this answer











            $endgroup$



            Assuming domain is $mathbb{R} setminus(0,4)$ and co-domian is $[0,infty) $ $y^2+4=x^2-4x+4implies y^2+4=(x-2)^2implies x=2pmsqrt{y^2+4}$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 24 '18 at 11:37

























            answered Nov 24 '18 at 11:28









            Yadati KiranYadati Kiran

            1,751619




            1,751619












            • $begingroup$
              You're missing out on a negative sign.
              $endgroup$
              – Rebellos
              Nov 24 '18 at 11:30






            • 1




              $begingroup$
              It should be a positive sign only, as range of the given function is positive.
              $endgroup$
              – Martund
              Nov 24 '18 at 11:34










            • $begingroup$
              @Crazyformaths Needs an elaboration, which is the exact point of inverses and domains.
              $endgroup$
              – Rebellos
              Nov 24 '18 at 11:34






            • 1




              $begingroup$
              $mathbb{R} setminus{(-4,4)}$ or $mathbb{R} setminus(-4,4)$ ?
              $endgroup$
              – Nosrati
              Nov 24 '18 at 11:35






            • 1




              $begingroup$
              @Rebellos: Sorry the domain was wrong.
              $endgroup$
              – Yadati Kiran
              Nov 24 '18 at 11:37


















            • $begingroup$
              You're missing out on a negative sign.
              $endgroup$
              – Rebellos
              Nov 24 '18 at 11:30






            • 1




              $begingroup$
              It should be a positive sign only, as range of the given function is positive.
              $endgroup$
              – Martund
              Nov 24 '18 at 11:34










            • $begingroup$
              @Crazyformaths Needs an elaboration, which is the exact point of inverses and domains.
              $endgroup$
              – Rebellos
              Nov 24 '18 at 11:34






            • 1




              $begingroup$
              $mathbb{R} setminus{(-4,4)}$ or $mathbb{R} setminus(-4,4)$ ?
              $endgroup$
              – Nosrati
              Nov 24 '18 at 11:35






            • 1




              $begingroup$
              @Rebellos: Sorry the domain was wrong.
              $endgroup$
              – Yadati Kiran
              Nov 24 '18 at 11:37
















            $begingroup$
            You're missing out on a negative sign.
            $endgroup$
            – Rebellos
            Nov 24 '18 at 11:30




            $begingroup$
            You're missing out on a negative sign.
            $endgroup$
            – Rebellos
            Nov 24 '18 at 11:30




            1




            1




            $begingroup$
            It should be a positive sign only, as range of the given function is positive.
            $endgroup$
            – Martund
            Nov 24 '18 at 11:34




            $begingroup$
            It should be a positive sign only, as range of the given function is positive.
            $endgroup$
            – Martund
            Nov 24 '18 at 11:34












            $begingroup$
            @Crazyformaths Needs an elaboration, which is the exact point of inverses and domains.
            $endgroup$
            – Rebellos
            Nov 24 '18 at 11:34




            $begingroup$
            @Crazyformaths Needs an elaboration, which is the exact point of inverses and domains.
            $endgroup$
            – Rebellos
            Nov 24 '18 at 11:34




            1




            1




            $begingroup$
            $mathbb{R} setminus{(-4,4)}$ or $mathbb{R} setminus(-4,4)$ ?
            $endgroup$
            – Nosrati
            Nov 24 '18 at 11:35




            $begingroup$
            $mathbb{R} setminus{(-4,4)}$ or $mathbb{R} setminus(-4,4)$ ?
            $endgroup$
            – Nosrati
            Nov 24 '18 at 11:35




            1




            1




            $begingroup$
            @Rebellos: Sorry the domain was wrong.
            $endgroup$
            – Yadati Kiran
            Nov 24 '18 at 11:37




            $begingroup$
            @Rebellos: Sorry the domain was wrong.
            $endgroup$
            – Yadati Kiran
            Nov 24 '18 at 11:37


















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