Range of function $f(x)=frac{x}{1-x^2}$












1












$begingroup$


How can we find the range of



$$f(x)=frac{x}{1-x^2}?$$



From the quadratic formula, I found that the range is $mathbb{R}setminus{0}.$ However, the correct one is $mathbb{R}.$






functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $f(0)=0{{{}}}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 3:41






  • 1




    $begingroup$
    It would be most helpful to see your work on how you used the quadratic formula to deduce the range is $mathbb{R}-{0}$. Then you could get a pointed diagnosis as to where you went wrong.
    $endgroup$
    – Randall
    Dec 28 '18 at 3:43












  • $begingroup$
    If you have learned some calculus, using $lim_{xto1^-}f(x)=+infty$ and $lim_{xto(-1)^+}f(x)=-infty$ and the fact $fin C(-1,1)$ can instantly deduce the range of $f$ is $mathbb R$.
    $endgroup$
    – Kemono Chen
    Dec 28 '18 at 3:50












  • $begingroup$
    What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
    $endgroup$
    – Masik Kara
    Dec 28 '18 at 4:15










  • $begingroup$
    Try $$x=tan t$$
    $endgroup$
    – lab bhattacharjee
    Dec 28 '18 at 4:46
















1












$begingroup$


How can we find the range of



$$f(x)=frac{x}{1-x^2}?$$



From the quadratic formula, I found that the range is $mathbb{R}setminus{0}.$ However, the correct one is $mathbb{R}.$






functions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $f(0)=0{{{}}}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 3:41






  • 1




    $begingroup$
    It would be most helpful to see your work on how you used the quadratic formula to deduce the range is $mathbb{R}-{0}$. Then you could get a pointed diagnosis as to where you went wrong.
    $endgroup$
    – Randall
    Dec 28 '18 at 3:43












  • $begingroup$
    If you have learned some calculus, using $lim_{xto1^-}f(x)=+infty$ and $lim_{xto(-1)^+}f(x)=-infty$ and the fact $fin C(-1,1)$ can instantly deduce the range of $f$ is $mathbb R$.
    $endgroup$
    – Kemono Chen
    Dec 28 '18 at 3:50












  • $begingroup$
    What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
    $endgroup$
    – Masik Kara
    Dec 28 '18 at 4:15










  • $begingroup$
    Try $$x=tan t$$
    $endgroup$
    – lab bhattacharjee
    Dec 28 '18 at 4:46














1












1








1





$begingroup$


How can we find the range of



$$f(x)=frac{x}{1-x^2}?$$



From the quadratic formula, I found that the range is $mathbb{R}setminus{0}.$ However, the correct one is $mathbb{R}.$






functions







share|cite|improve this question











$endgroup$




How can we find the range of



$$f(x)=frac{x}{1-x^2}?$$



From the quadratic formula, I found that the range is $mathbb{R}setminus{0}.$ However, the correct one is $mathbb{R}.$






functions




functions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 28 '18 at 3:48









JavaMan

11.1k12755




11.1k12755










asked Dec 28 '18 at 3:39









Masik KaraMasik Kara

395




395








  • 1




    $begingroup$
    $f(0)=0{{{}}}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 3:41






  • 1




    $begingroup$
    It would be most helpful to see your work on how you used the quadratic formula to deduce the range is $mathbb{R}-{0}$. Then you could get a pointed diagnosis as to where you went wrong.
    $endgroup$
    – Randall
    Dec 28 '18 at 3:43












  • $begingroup$
    If you have learned some calculus, using $lim_{xto1^-}f(x)=+infty$ and $lim_{xto(-1)^+}f(x)=-infty$ and the fact $fin C(-1,1)$ can instantly deduce the range of $f$ is $mathbb R$.
    $endgroup$
    – Kemono Chen
    Dec 28 '18 at 3:50












  • $begingroup$
    What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
    $endgroup$
    – Masik Kara
    Dec 28 '18 at 4:15










  • $begingroup$
    Try $$x=tan t$$
    $endgroup$
    – lab bhattacharjee
    Dec 28 '18 at 4:46














  • 1




    $begingroup$
    $f(0)=0{{{}}}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 28 '18 at 3:41






  • 1




    $begingroup$
    It would be most helpful to see your work on how you used the quadratic formula to deduce the range is $mathbb{R}-{0}$. Then you could get a pointed diagnosis as to where you went wrong.
    $endgroup$
    – Randall
    Dec 28 '18 at 3:43












  • $begingroup$
    If you have learned some calculus, using $lim_{xto1^-}f(x)=+infty$ and $lim_{xto(-1)^+}f(x)=-infty$ and the fact $fin C(-1,1)$ can instantly deduce the range of $f$ is $mathbb R$.
    $endgroup$
    – Kemono Chen
    Dec 28 '18 at 3:50












  • $begingroup$
    What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
    $endgroup$
    – Masik Kara
    Dec 28 '18 at 4:15










  • $begingroup$
    Try $$x=tan t$$
    $endgroup$
    – lab bhattacharjee
    Dec 28 '18 at 4:46








1




1




$begingroup$
$f(0)=0{{{}}}$.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 3:41




$begingroup$
$f(0)=0{{{}}}$.
$endgroup$
– Lord Shark the Unknown
Dec 28 '18 at 3:41




1




1




$begingroup$
It would be most helpful to see your work on how you used the quadratic formula to deduce the range is $mathbb{R}-{0}$. Then you could get a pointed diagnosis as to where you went wrong.
$endgroup$
– Randall
Dec 28 '18 at 3:43






$begingroup$
It would be most helpful to see your work on how you used the quadratic formula to deduce the range is $mathbb{R}-{0}$. Then you could get a pointed diagnosis as to where you went wrong.
$endgroup$
– Randall
Dec 28 '18 at 3:43














$begingroup$
If you have learned some calculus, using $lim_{xto1^-}f(x)=+infty$ and $lim_{xto(-1)^+}f(x)=-infty$ and the fact $fin C(-1,1)$ can instantly deduce the range of $f$ is $mathbb R$.
$endgroup$
– Kemono Chen
Dec 28 '18 at 3:50






$begingroup$
If you have learned some calculus, using $lim_{xto1^-}f(x)=+infty$ and $lim_{xto(-1)^+}f(x)=-infty$ and the fact $fin C(-1,1)$ can instantly deduce the range of $f$ is $mathbb R$.
$endgroup$
– Kemono Chen
Dec 28 '18 at 3:50














$begingroup$
What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
$endgroup$
– Masik Kara
Dec 28 '18 at 4:15




$begingroup$
What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
$endgroup$
– Masik Kara
Dec 28 '18 at 4:15












$begingroup$
Try $$x=tan t$$
$endgroup$
– lab bhattacharjee
Dec 28 '18 at 4:46




$begingroup$
Try $$x=tan t$$
$endgroup$
– lab bhattacharjee
Dec 28 '18 at 4:46










3 Answers
3






active

oldest

votes


















3












$begingroup$

Suppose $y$ is in $mathbb{R}$, and we wonder if $y$ is in the range. So we are wondering if there are $x$-values that solve $$y=frac{x}{1-x^2}$$ Of course, $x$ will not be $1$ or $-1$ or the right side is undefined. The equation is equivalent to $$yx^2+x-y=0$$ Note it is still the case that $x$ could not be $1$ or $-1$. We are still wondering if there are solutions in $x$ for our fixed value of $y$. Assuming $yneq0$, this is a quadratic equation in $x$, and then there are solutions if and only if the discriminant is $geq0$. The discriminant is $$1+4y^2$$ which is always $geq1$. So no matter what nonzero value $y$ is, there is an $x$ (two actually) with $f(x)=y$. Now what if $y=0$ and the equation is not quadratic? Then trivially observe that $x=0$ solves the first equation.



In the end, nothing in $mathbb{R}$ is excluded from the range.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Two usually, but yes. Nice. +1
    $endgroup$
    – Randall
    Dec 28 '18 at 4:01












  • $begingroup$
    @Randall Not just "usually", but always. The discriminant here is always greater than $1$ (so it's nonzero).
    $endgroup$
    – alex.jordan
    Dec 28 '18 at 4:11






  • 1




    $begingroup$
    I think I see what you were alluding to. I edited to handle $y=0$ separately.
    $endgroup$
    – alex.jordan
    Dec 28 '18 at 4:15










  • $begingroup$
    What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
    $endgroup$
    – Masik Kara
    Dec 28 '18 at 4:15






  • 1




    $begingroup$
    @MasikKara But if $y=0$ you did not have a quadratic equation in the first place and should not have applied the quadratic formula.
    $endgroup$
    – alex.jordan
    Dec 28 '18 at 4:16



















2












$begingroup$

A subtle aspect of the quadratic formula, that the solutions of $ax^2 + bx +c=0$ are $frac {-b pm {b^2 - 4ac}}{2a}$ are that in order to BE a quadratic, it is assumed that $a ne 0$.



If $a$ does equal $0$ then what you have is $bx + c=0$ and that is not a quadratic; it's a linear equation and its root is $-frac cb$.... (That is if it is assumed that $b ne 0$.)



(To carry this idea to an extreme if $a=0$ and $b=0$ then you have $c=0$ and that's simply a statment that is true for all reals if $c =0$ and not true for any reals if $c ne 0$. So if $c ne 0$ there are no "solutions". If $c=0$ then all $x in mathbb R$ are "solutions" because $0=0$ for all $x$ [which have nothing to do the statement]. ... But I digress...)



So to solve for $frac x{1-x^2} = y$ or $yx^2 + x - y = 0$ then solutions are:



$x = frac {-1 pm {1 + 4y^2}}{2y}$ if $y ne 0$. And that has solutions for all $yne 0$.



AND



if $y = 0$ then the equation is $0x^2 + x -0 = 0$ or $x = 0$ and that has solution... $x = 0$ if $y = 0$. And that is a solution for $y=0$.



So solutions exist for all $y$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Write it as:
    $$f(x)=frac x{1-x^2}=frac12left[frac1{1-x}-frac1{1+x}right]=frac12(g(x)-h(x)).$$
    The function $g(x)$ is decreasing and ranges $(-infty,0)cup (0,infty)$, $h(x)$ is increasing and ranges $(-infty,0)cup (0,infty)$ and $f(0)=0$. Hence, $f(x)$ ranges $(-infty,infty)$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But yoy are not sure that the function are continuous on $mathbb{R}$ so how can you proceed?
      $endgroup$
      – Rakesh Bhatt
      Dec 28 '18 at 5:15











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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Suppose $y$ is in $mathbb{R}$, and we wonder if $y$ is in the range. So we are wondering if there are $x$-values that solve $$y=frac{x}{1-x^2}$$ Of course, $x$ will not be $1$ or $-1$ or the right side is undefined. The equation is equivalent to $$yx^2+x-y=0$$ Note it is still the case that $x$ could not be $1$ or $-1$. We are still wondering if there are solutions in $x$ for our fixed value of $y$. Assuming $yneq0$, this is a quadratic equation in $x$, and then there are solutions if and only if the discriminant is $geq0$. The discriminant is $$1+4y^2$$ which is always $geq1$. So no matter what nonzero value $y$ is, there is an $x$ (two actually) with $f(x)=y$. Now what if $y=0$ and the equation is not quadratic? Then trivially observe that $x=0$ solves the first equation.



    In the end, nothing in $mathbb{R}$ is excluded from the range.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Two usually, but yes. Nice. +1
      $endgroup$
      – Randall
      Dec 28 '18 at 4:01












    • $begingroup$
      @Randall Not just "usually", but always. The discriminant here is always greater than $1$ (so it's nonzero).
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:11






    • 1




      $begingroup$
      I think I see what you were alluding to. I edited to handle $y=0$ separately.
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:15










    • $begingroup$
      What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
      $endgroup$
      – Masik Kara
      Dec 28 '18 at 4:15






    • 1




      $begingroup$
      @MasikKara But if $y=0$ you did not have a quadratic equation in the first place and should not have applied the quadratic formula.
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:16
















    3












    $begingroup$

    Suppose $y$ is in $mathbb{R}$, and we wonder if $y$ is in the range. So we are wondering if there are $x$-values that solve $$y=frac{x}{1-x^2}$$ Of course, $x$ will not be $1$ or $-1$ or the right side is undefined. The equation is equivalent to $$yx^2+x-y=0$$ Note it is still the case that $x$ could not be $1$ or $-1$. We are still wondering if there are solutions in $x$ for our fixed value of $y$. Assuming $yneq0$, this is a quadratic equation in $x$, and then there are solutions if and only if the discriminant is $geq0$. The discriminant is $$1+4y^2$$ which is always $geq1$. So no matter what nonzero value $y$ is, there is an $x$ (two actually) with $f(x)=y$. Now what if $y=0$ and the equation is not quadratic? Then trivially observe that $x=0$ solves the first equation.



    In the end, nothing in $mathbb{R}$ is excluded from the range.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Two usually, but yes. Nice. +1
      $endgroup$
      – Randall
      Dec 28 '18 at 4:01












    • $begingroup$
      @Randall Not just "usually", but always. The discriminant here is always greater than $1$ (so it's nonzero).
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:11






    • 1




      $begingroup$
      I think I see what you were alluding to. I edited to handle $y=0$ separately.
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:15










    • $begingroup$
      What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
      $endgroup$
      – Masik Kara
      Dec 28 '18 at 4:15






    • 1




      $begingroup$
      @MasikKara But if $y=0$ you did not have a quadratic equation in the first place and should not have applied the quadratic formula.
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:16














    3












    3








    3





    $begingroup$

    Suppose $y$ is in $mathbb{R}$, and we wonder if $y$ is in the range. So we are wondering if there are $x$-values that solve $$y=frac{x}{1-x^2}$$ Of course, $x$ will not be $1$ or $-1$ or the right side is undefined. The equation is equivalent to $$yx^2+x-y=0$$ Note it is still the case that $x$ could not be $1$ or $-1$. We are still wondering if there are solutions in $x$ for our fixed value of $y$. Assuming $yneq0$, this is a quadratic equation in $x$, and then there are solutions if and only if the discriminant is $geq0$. The discriminant is $$1+4y^2$$ which is always $geq1$. So no matter what nonzero value $y$ is, there is an $x$ (two actually) with $f(x)=y$. Now what if $y=0$ and the equation is not quadratic? Then trivially observe that $x=0$ solves the first equation.



    In the end, nothing in $mathbb{R}$ is excluded from the range.






    share|cite|improve this answer











    $endgroup$



    Suppose $y$ is in $mathbb{R}$, and we wonder if $y$ is in the range. So we are wondering if there are $x$-values that solve $$y=frac{x}{1-x^2}$$ Of course, $x$ will not be $1$ or $-1$ or the right side is undefined. The equation is equivalent to $$yx^2+x-y=0$$ Note it is still the case that $x$ could not be $1$ or $-1$. We are still wondering if there are solutions in $x$ for our fixed value of $y$. Assuming $yneq0$, this is a quadratic equation in $x$, and then there are solutions if and only if the discriminant is $geq0$. The discriminant is $$1+4y^2$$ which is always $geq1$. So no matter what nonzero value $y$ is, there is an $x$ (two actually) with $f(x)=y$. Now what if $y=0$ and the equation is not quadratic? Then trivially observe that $x=0$ solves the first equation.



    In the end, nothing in $mathbb{R}$ is excluded from the range.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 28 '18 at 4:15

























    answered Dec 28 '18 at 3:57









    alex.jordanalex.jordan

    39.5k560122




    39.5k560122












    • $begingroup$
      Two usually, but yes. Nice. +1
      $endgroup$
      – Randall
      Dec 28 '18 at 4:01












    • $begingroup$
      @Randall Not just "usually", but always. The discriminant here is always greater than $1$ (so it's nonzero).
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:11






    • 1




      $begingroup$
      I think I see what you were alluding to. I edited to handle $y=0$ separately.
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:15










    • $begingroup$
      What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
      $endgroup$
      – Masik Kara
      Dec 28 '18 at 4:15






    • 1




      $begingroup$
      @MasikKara But if $y=0$ you did not have a quadratic equation in the first place and should not have applied the quadratic formula.
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:16


















    • $begingroup$
      Two usually, but yes. Nice. +1
      $endgroup$
      – Randall
      Dec 28 '18 at 4:01












    • $begingroup$
      @Randall Not just "usually", but always. The discriminant here is always greater than $1$ (so it's nonzero).
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:11






    • 1




      $begingroup$
      I think I see what you were alluding to. I edited to handle $y=0$ separately.
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:15










    • $begingroup$
      What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
      $endgroup$
      – Masik Kara
      Dec 28 '18 at 4:15






    • 1




      $begingroup$
      @MasikKara But if $y=0$ you did not have a quadratic equation in the first place and should not have applied the quadratic formula.
      $endgroup$
      – alex.jordan
      Dec 28 '18 at 4:16
















    $begingroup$
    Two usually, but yes. Nice. +1
    $endgroup$
    – Randall
    Dec 28 '18 at 4:01






    $begingroup$
    Two usually, but yes. Nice. +1
    $endgroup$
    – Randall
    Dec 28 '18 at 4:01














    $begingroup$
    @Randall Not just "usually", but always. The discriminant here is always greater than $1$ (so it's nonzero).
    $endgroup$
    – alex.jordan
    Dec 28 '18 at 4:11




    $begingroup$
    @Randall Not just "usually", but always. The discriminant here is always greater than $1$ (so it's nonzero).
    $endgroup$
    – alex.jordan
    Dec 28 '18 at 4:11




    1




    1




    $begingroup$
    I think I see what you were alluding to. I edited to handle $y=0$ separately.
    $endgroup$
    – alex.jordan
    Dec 28 '18 at 4:15




    $begingroup$
    I think I see what you were alluding to. I edited to handle $y=0$ separately.
    $endgroup$
    – alex.jordan
    Dec 28 '18 at 4:15












    $begingroup$
    What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
    $endgroup$
    – Masik Kara
    Dec 28 '18 at 4:15




    $begingroup$
    What I found is $x=-1pm(1+4y^2)/2y.$ This means that $ynot=0.$
    $endgroup$
    – Masik Kara
    Dec 28 '18 at 4:15




    1




    1




    $begingroup$
    @MasikKara But if $y=0$ you did not have a quadratic equation in the first place and should not have applied the quadratic formula.
    $endgroup$
    – alex.jordan
    Dec 28 '18 at 4:16




    $begingroup$
    @MasikKara But if $y=0$ you did not have a quadratic equation in the first place and should not have applied the quadratic formula.
    $endgroup$
    – alex.jordan
    Dec 28 '18 at 4:16











    2












    $begingroup$

    A subtle aspect of the quadratic formula, that the solutions of $ax^2 + bx +c=0$ are $frac {-b pm {b^2 - 4ac}}{2a}$ are that in order to BE a quadratic, it is assumed that $a ne 0$.



    If $a$ does equal $0$ then what you have is $bx + c=0$ and that is not a quadratic; it's a linear equation and its root is $-frac cb$.... (That is if it is assumed that $b ne 0$.)



    (To carry this idea to an extreme if $a=0$ and $b=0$ then you have $c=0$ and that's simply a statment that is true for all reals if $c =0$ and not true for any reals if $c ne 0$. So if $c ne 0$ there are no "solutions". If $c=0$ then all $x in mathbb R$ are "solutions" because $0=0$ for all $x$ [which have nothing to do the statement]. ... But I digress...)



    So to solve for $frac x{1-x^2} = y$ or $yx^2 + x - y = 0$ then solutions are:



    $x = frac {-1 pm {1 + 4y^2}}{2y}$ if $y ne 0$. And that has solutions for all $yne 0$.



    AND



    if $y = 0$ then the equation is $0x^2 + x -0 = 0$ or $x = 0$ and that has solution... $x = 0$ if $y = 0$. And that is a solution for $y=0$.



    So solutions exist for all $y$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      A subtle aspect of the quadratic formula, that the solutions of $ax^2 + bx +c=0$ are $frac {-b pm {b^2 - 4ac}}{2a}$ are that in order to BE a quadratic, it is assumed that $a ne 0$.



      If $a$ does equal $0$ then what you have is $bx + c=0$ and that is not a quadratic; it's a linear equation and its root is $-frac cb$.... (That is if it is assumed that $b ne 0$.)



      (To carry this idea to an extreme if $a=0$ and $b=0$ then you have $c=0$ and that's simply a statment that is true for all reals if $c =0$ and not true for any reals if $c ne 0$. So if $c ne 0$ there are no "solutions". If $c=0$ then all $x in mathbb R$ are "solutions" because $0=0$ for all $x$ [which have nothing to do the statement]. ... But I digress...)



      So to solve for $frac x{1-x^2} = y$ or $yx^2 + x - y = 0$ then solutions are:



      $x = frac {-1 pm {1 + 4y^2}}{2y}$ if $y ne 0$. And that has solutions for all $yne 0$.



      AND



      if $y = 0$ then the equation is $0x^2 + x -0 = 0$ or $x = 0$ and that has solution... $x = 0$ if $y = 0$. And that is a solution for $y=0$.



      So solutions exist for all $y$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        A subtle aspect of the quadratic formula, that the solutions of $ax^2 + bx +c=0$ are $frac {-b pm {b^2 - 4ac}}{2a}$ are that in order to BE a quadratic, it is assumed that $a ne 0$.



        If $a$ does equal $0$ then what you have is $bx + c=0$ and that is not a quadratic; it's a linear equation and its root is $-frac cb$.... (That is if it is assumed that $b ne 0$.)



        (To carry this idea to an extreme if $a=0$ and $b=0$ then you have $c=0$ and that's simply a statment that is true for all reals if $c =0$ and not true for any reals if $c ne 0$. So if $c ne 0$ there are no "solutions". If $c=0$ then all $x in mathbb R$ are "solutions" because $0=0$ for all $x$ [which have nothing to do the statement]. ... But I digress...)



        So to solve for $frac x{1-x^2} = y$ or $yx^2 + x - y = 0$ then solutions are:



        $x = frac {-1 pm {1 + 4y^2}}{2y}$ if $y ne 0$. And that has solutions for all $yne 0$.



        AND



        if $y = 0$ then the equation is $0x^2 + x -0 = 0$ or $x = 0$ and that has solution... $x = 0$ if $y = 0$. And that is a solution for $y=0$.



        So solutions exist for all $y$.






        share|cite|improve this answer









        $endgroup$



        A subtle aspect of the quadratic formula, that the solutions of $ax^2 + bx +c=0$ are $frac {-b pm {b^2 - 4ac}}{2a}$ are that in order to BE a quadratic, it is assumed that $a ne 0$.



        If $a$ does equal $0$ then what you have is $bx + c=0$ and that is not a quadratic; it's a linear equation and its root is $-frac cb$.... (That is if it is assumed that $b ne 0$.)



        (To carry this idea to an extreme if $a=0$ and $b=0$ then you have $c=0$ and that's simply a statment that is true for all reals if $c =0$ and not true for any reals if $c ne 0$. So if $c ne 0$ there are no "solutions". If $c=0$ then all $x in mathbb R$ are "solutions" because $0=0$ for all $x$ [which have nothing to do the statement]. ... But I digress...)



        So to solve for $frac x{1-x^2} = y$ or $yx^2 + x - y = 0$ then solutions are:



        $x = frac {-1 pm {1 + 4y^2}}{2y}$ if $y ne 0$. And that has solutions for all $yne 0$.



        AND



        if $y = 0$ then the equation is $0x^2 + x -0 = 0$ or $x = 0$ and that has solution... $x = 0$ if $y = 0$. And that is a solution for $y=0$.



        So solutions exist for all $y$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 28 '18 at 5:42









        fleabloodfleablood

        72.8k22788




        72.8k22788























            0












            $begingroup$

            Write it as:
            $$f(x)=frac x{1-x^2}=frac12left[frac1{1-x}-frac1{1+x}right]=frac12(g(x)-h(x)).$$
            The function $g(x)$ is decreasing and ranges $(-infty,0)cup (0,infty)$, $h(x)$ is increasing and ranges $(-infty,0)cup (0,infty)$ and $f(0)=0$. Hence, $f(x)$ ranges $(-infty,infty)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But yoy are not sure that the function are continuous on $mathbb{R}$ so how can you proceed?
              $endgroup$
              – Rakesh Bhatt
              Dec 28 '18 at 5:15
















            0












            $begingroup$

            Write it as:
            $$f(x)=frac x{1-x^2}=frac12left[frac1{1-x}-frac1{1+x}right]=frac12(g(x)-h(x)).$$
            The function $g(x)$ is decreasing and ranges $(-infty,0)cup (0,infty)$, $h(x)$ is increasing and ranges $(-infty,0)cup (0,infty)$ and $f(0)=0$. Hence, $f(x)$ ranges $(-infty,infty)$.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              But yoy are not sure that the function are continuous on $mathbb{R}$ so how can you proceed?
              $endgroup$
              – Rakesh Bhatt
              Dec 28 '18 at 5:15














            0












            0








            0





            $begingroup$

            Write it as:
            $$f(x)=frac x{1-x^2}=frac12left[frac1{1-x}-frac1{1+x}right]=frac12(g(x)-h(x)).$$
            The function $g(x)$ is decreasing and ranges $(-infty,0)cup (0,infty)$, $h(x)$ is increasing and ranges $(-infty,0)cup (0,infty)$ and $f(0)=0$. Hence, $f(x)$ ranges $(-infty,infty)$.






            share|cite|improve this answer









            $endgroup$



            Write it as:
            $$f(x)=frac x{1-x^2}=frac12left[frac1{1-x}-frac1{1+x}right]=frac12(g(x)-h(x)).$$
            The function $g(x)$ is decreasing and ranges $(-infty,0)cup (0,infty)$, $h(x)$ is increasing and ranges $(-infty,0)cup (0,infty)$ and $f(0)=0$. Hence, $f(x)$ ranges $(-infty,infty)$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 28 '18 at 4:19









            farruhotafarruhota

            21.3k2841




            21.3k2841












            • $begingroup$
              But yoy are not sure that the function are continuous on $mathbb{R}$ so how can you proceed?
              $endgroup$
              – Rakesh Bhatt
              Dec 28 '18 at 5:15


















            • $begingroup$
              But yoy are not sure that the function are continuous on $mathbb{R}$ so how can you proceed?
              $endgroup$
              – Rakesh Bhatt
              Dec 28 '18 at 5:15
















            $begingroup$
            But yoy are not sure that the function are continuous on $mathbb{R}$ so how can you proceed?
            $endgroup$
            – Rakesh Bhatt
            Dec 28 '18 at 5:15




            $begingroup$
            But yoy are not sure that the function are continuous on $mathbb{R}$ so how can you proceed?
            $endgroup$
            – Rakesh Bhatt
            Dec 28 '18 at 5:15


















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