Determining range of a function












2












$begingroup$


I was trying to determine the domain and range of a function. The function is: $$y = frac{x}{x^2 + 1}$$
I determined the domain which is $mathbb{R}$. In this equation, when the value of $x$ is 0, the value of $y$ is $0$. Then, I tried to determine the range by this equation:
$$x = frac{1 pm sqrt{1 - 4y^2}}{2y}$$
I derived the equation from previous equation using quadratic equation formula. In this method we can see, the range is ${-0.5 leq y < 0}$ and ${0 < y leq 0.5}$. Zero cannot be a range because the denominator cannot be zero.
But if $0$ is not in the range, then we do not find any value for $x$ when $x$ is $0$ which is a contradiction of the definition of a function. Where is the fault?






functions







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  • $begingroup$
    Compare math.stackexchange.com/q/2199960/42969.
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    – Martin R
    Jan 13 at 12:31








  • 1




    $begingroup$
    For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – Martin R
    Jan 13 at 12:47










  • $begingroup$
    Note that: 1) this is not a quadratic function, (2) $0$ is clearly in the range: take $x = 0$ to obtain $y = 0/(0^2 + 1) = 0$, your reason why it can't be makes no sense and simply has nothing to do with the range of the function, and (3) taking $x = 1$ gives $y = frac{1}{2}$, so clearly the rest of your reasoning is not valid.
    $endgroup$
    – user3482749
    Jan 13 at 14:13
















2












$begingroup$


I was trying to determine the domain and range of a function. The function is: $$y = frac{x}{x^2 + 1}$$
I determined the domain which is $mathbb{R}$. In this equation, when the value of $x$ is 0, the value of $y$ is $0$. Then, I tried to determine the range by this equation:
$$x = frac{1 pm sqrt{1 - 4y^2}}{2y}$$
I derived the equation from previous equation using quadratic equation formula. In this method we can see, the range is ${-0.5 leq y < 0}$ and ${0 < y leq 0.5}$. Zero cannot be a range because the denominator cannot be zero.
But if $0$ is not in the range, then we do not find any value for $x$ when $x$ is $0$ which is a contradiction of the definition of a function. Where is the fault?






functions







share|cite|improve this question











$endgroup$












  • $begingroup$
    Compare math.stackexchange.com/q/2199960/42969.
    $endgroup$
    – Martin R
    Jan 13 at 12:31








  • 1




    $begingroup$
    For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – Martin R
    Jan 13 at 12:47










  • $begingroup$
    Note that: 1) this is not a quadratic function, (2) $0$ is clearly in the range: take $x = 0$ to obtain $y = 0/(0^2 + 1) = 0$, your reason why it can't be makes no sense and simply has nothing to do with the range of the function, and (3) taking $x = 1$ gives $y = frac{1}{2}$, so clearly the rest of your reasoning is not valid.
    $endgroup$
    – user3482749
    Jan 13 at 14:13














2












2








2





$begingroup$


I was trying to determine the domain and range of a function. The function is: $$y = frac{x}{x^2 + 1}$$
I determined the domain which is $mathbb{R}$. In this equation, when the value of $x$ is 0, the value of $y$ is $0$. Then, I tried to determine the range by this equation:
$$x = frac{1 pm sqrt{1 - 4y^2}}{2y}$$
I derived the equation from previous equation using quadratic equation formula. In this method we can see, the range is ${-0.5 leq y < 0}$ and ${0 < y leq 0.5}$. Zero cannot be a range because the denominator cannot be zero.
But if $0$ is not in the range, then we do not find any value for $x$ when $x$ is $0$ which is a contradiction of the definition of a function. Where is the fault?






functions







share|cite|improve this question











$endgroup$




I was trying to determine the domain and range of a function. The function is: $$y = frac{x}{x^2 + 1}$$
I determined the domain which is $mathbb{R}$. In this equation, when the value of $x$ is 0, the value of $y$ is $0$. Then, I tried to determine the range by this equation:
$$x = frac{1 pm sqrt{1 - 4y^2}}{2y}$$
I derived the equation from previous equation using quadratic equation formula. In this method we can see, the range is ${-0.5 leq y < 0}$ and ${0 < y leq 0.5}$. Zero cannot be a range because the denominator cannot be zero.
But if $0$ is not in the range, then we do not find any value for $x$ when $x$ is $0$ which is a contradiction of the definition of a function. Where is the fault?






functions




functions






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share|cite|improve this question













share|cite|improve this question




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edited Jan 14 at 12:12









N. F. Taussig

45.4k103358




45.4k103358










asked Jan 13 at 12:18







user634326



















  • $begingroup$
    Compare math.stackexchange.com/q/2199960/42969.
    $endgroup$
    – Martin R
    Jan 13 at 12:31








  • 1




    $begingroup$
    For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – Martin R
    Jan 13 at 12:47










  • $begingroup$
    Note that: 1) this is not a quadratic function, (2) $0$ is clearly in the range: take $x = 0$ to obtain $y = 0/(0^2 + 1) = 0$, your reason why it can't be makes no sense and simply has nothing to do with the range of the function, and (3) taking $x = 1$ gives $y = frac{1}{2}$, so clearly the rest of your reasoning is not valid.
    $endgroup$
    – user3482749
    Jan 13 at 14:13


















  • $begingroup$
    Compare math.stackexchange.com/q/2199960/42969.
    $endgroup$
    – Martin R
    Jan 13 at 12:31








  • 1




    $begingroup$
    For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – Martin R
    Jan 13 at 12:47










  • $begingroup$
    Note that: 1) this is not a quadratic function, (2) $0$ is clearly in the range: take $x = 0$ to obtain $y = 0/(0^2 + 1) = 0$, your reason why it can't be makes no sense and simply has nothing to do with the range of the function, and (3) taking $x = 1$ gives $y = frac{1}{2}$, so clearly the rest of your reasoning is not valid.
    $endgroup$
    – user3482749
    Jan 13 at 14:13
















$begingroup$
Compare math.stackexchange.com/q/2199960/42969.
$endgroup$
– Martin R
Jan 13 at 12:31






$begingroup$
Compare math.stackexchange.com/q/2199960/42969.
$endgroup$
– Martin R
Jan 13 at 12:31






1




1




$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Martin R
Jan 13 at 12:47




$begingroup$
For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– Martin R
Jan 13 at 12:47












$begingroup$
Note that: 1) this is not a quadratic function, (2) $0$ is clearly in the range: take $x = 0$ to obtain $y = 0/(0^2 + 1) = 0$, your reason why it can't be makes no sense and simply has nothing to do with the range of the function, and (3) taking $x = 1$ gives $y = frac{1}{2}$, so clearly the rest of your reasoning is not valid.
$endgroup$
– user3482749
Jan 13 at 14:13




$begingroup$
Note that: 1) this is not a quadratic function, (2) $0$ is clearly in the range: take $x = 0$ to obtain $y = 0/(0^2 + 1) = 0$, your reason why it can't be makes no sense and simply has nothing to do with the range of the function, and (3) taking $x = 1$ gives $y = frac{1}{2}$, so clearly the rest of your reasoning is not valid.
$endgroup$
– user3482749
Jan 13 at 14:13










2 Answers
2






active

oldest

votes


















2












$begingroup$

A reason you get a contradiction is that you got too far ahead before looking to see where you were going. Indeed if we want to solve for $x$ in
$$y = frac{x}{x^2 + 1},$$
the first steps could be
$$y (x^2 + 1) = x,$$
$$y x^2 + - x + y = 0.$$
Now if $y neq 0$ this is a quadratic equation that can be solved by the quadratic formula, but if $y = 0$ then it is a linear equation that can be solved much more simply:
$$ -x = 0, $$
$$ x = 0.$$



Indeed, you should simply take note of the obvious fact that $0$ is in the range of the function (since $y= 0$ when $x = 0$) and look for additional possible non-zero values of $y,$
that is, to look for any possible ways that $y neq 0.$
Your quadratic formula then finds the rest of the range for you.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    An idea what to do with this kind of questions: assume $;ainBbb R;$ belongs to the function's range, then there exists $;xinBbb R;$ (the function's domain) s.t.



    $$a=f(x)=frac x{x^2+1}implies ax^2-x+a=0implies Delta=1-4a^2ge0implies |a|lefrac12$$



    and thus you get that it must be that $;-frac12le alefrac12;$ , and the range of the function is waaaay different from what you thought.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      I do not understand why you say the range is way different from the one proposed by the OP. The OP found all of the same values but was confused by his method.
      $endgroup$
      – N. F. Taussig
      Jan 14 at 18:55










    • $begingroup$
      @N.F.Taussig Perhaps, The OP's original post wasn't written with LaTeX and I thought I read something involving intervals with $;5;$ ...That's why I wrote what I wrote.
      $endgroup$
      – DonAntonio
      Jan 14 at 19:37














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    2 Answers
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    active

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    2 Answers
    2






    active

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    active

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    2












    $begingroup$

    A reason you get a contradiction is that you got too far ahead before looking to see where you were going. Indeed if we want to solve for $x$ in
    $$y = frac{x}{x^2 + 1},$$
    the first steps could be
    $$y (x^2 + 1) = x,$$
    $$y x^2 + - x + y = 0.$$
    Now if $y neq 0$ this is a quadratic equation that can be solved by the quadratic formula, but if $y = 0$ then it is a linear equation that can be solved much more simply:
    $$ -x = 0, $$
    $$ x = 0.$$



    Indeed, you should simply take note of the obvious fact that $0$ is in the range of the function (since $y= 0$ when $x = 0$) and look for additional possible non-zero values of $y,$
    that is, to look for any possible ways that $y neq 0.$
    Your quadratic formula then finds the rest of the range for you.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      A reason you get a contradiction is that you got too far ahead before looking to see where you were going. Indeed if we want to solve for $x$ in
      $$y = frac{x}{x^2 + 1},$$
      the first steps could be
      $$y (x^2 + 1) = x,$$
      $$y x^2 + - x + y = 0.$$
      Now if $y neq 0$ this is a quadratic equation that can be solved by the quadratic formula, but if $y = 0$ then it is a linear equation that can be solved much more simply:
      $$ -x = 0, $$
      $$ x = 0.$$



      Indeed, you should simply take note of the obvious fact that $0$ is in the range of the function (since $y= 0$ when $x = 0$) and look for additional possible non-zero values of $y,$
      that is, to look for any possible ways that $y neq 0.$
      Your quadratic formula then finds the rest of the range for you.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        A reason you get a contradiction is that you got too far ahead before looking to see where you were going. Indeed if we want to solve for $x$ in
        $$y = frac{x}{x^2 + 1},$$
        the first steps could be
        $$y (x^2 + 1) = x,$$
        $$y x^2 + - x + y = 0.$$
        Now if $y neq 0$ this is a quadratic equation that can be solved by the quadratic formula, but if $y = 0$ then it is a linear equation that can be solved much more simply:
        $$ -x = 0, $$
        $$ x = 0.$$



        Indeed, you should simply take note of the obvious fact that $0$ is in the range of the function (since $y= 0$ when $x = 0$) and look for additional possible non-zero values of $y,$
        that is, to look for any possible ways that $y neq 0.$
        Your quadratic formula then finds the rest of the range for you.






        share|cite|improve this answer









        $endgroup$



        A reason you get a contradiction is that you got too far ahead before looking to see where you were going. Indeed if we want to solve for $x$ in
        $$y = frac{x}{x^2 + 1},$$
        the first steps could be
        $$y (x^2 + 1) = x,$$
        $$y x^2 + - x + y = 0.$$
        Now if $y neq 0$ this is a quadratic equation that can be solved by the quadratic formula, but if $y = 0$ then it is a linear equation that can be solved much more simply:
        $$ -x = 0, $$
        $$ x = 0.$$



        Indeed, you should simply take note of the obvious fact that $0$ is in the range of the function (since $y= 0$ when $x = 0$) and look for additional possible non-zero values of $y,$
        that is, to look for any possible ways that $y neq 0.$
        Your quadratic formula then finds the rest of the range for you.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 14:48









        David KDavid K

        55.9k345121




        55.9k345121























            0












            $begingroup$

            An idea what to do with this kind of questions: assume $;ainBbb R;$ belongs to the function's range, then there exists $;xinBbb R;$ (the function's domain) s.t.



            $$a=f(x)=frac x{x^2+1}implies ax^2-x+a=0implies Delta=1-4a^2ge0implies |a|lefrac12$$



            and thus you get that it must be that $;-frac12le alefrac12;$ , and the range of the function is waaaay different from what you thought.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I do not understand why you say the range is way different from the one proposed by the OP. The OP found all of the same values but was confused by his method.
              $endgroup$
              – N. F. Taussig
              Jan 14 at 18:55










            • $begingroup$
              @N.F.Taussig Perhaps, The OP's original post wasn't written with LaTeX and I thought I read something involving intervals with $;5;$ ...That's why I wrote what I wrote.
              $endgroup$
              – DonAntonio
              Jan 14 at 19:37


















            0












            $begingroup$

            An idea what to do with this kind of questions: assume $;ainBbb R;$ belongs to the function's range, then there exists $;xinBbb R;$ (the function's domain) s.t.



            $$a=f(x)=frac x{x^2+1}implies ax^2-x+a=0implies Delta=1-4a^2ge0implies |a|lefrac12$$



            and thus you get that it must be that $;-frac12le alefrac12;$ , and the range of the function is waaaay different from what you thought.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              I do not understand why you say the range is way different from the one proposed by the OP. The OP found all of the same values but was confused by his method.
              $endgroup$
              – N. F. Taussig
              Jan 14 at 18:55










            • $begingroup$
              @N.F.Taussig Perhaps, The OP's original post wasn't written with LaTeX and I thought I read something involving intervals with $;5;$ ...That's why I wrote what I wrote.
              $endgroup$
              – DonAntonio
              Jan 14 at 19:37
















            0












            0








            0





            $begingroup$

            An idea what to do with this kind of questions: assume $;ainBbb R;$ belongs to the function's range, then there exists $;xinBbb R;$ (the function's domain) s.t.



            $$a=f(x)=frac x{x^2+1}implies ax^2-x+a=0implies Delta=1-4a^2ge0implies |a|lefrac12$$



            and thus you get that it must be that $;-frac12le alefrac12;$ , and the range of the function is waaaay different from what you thought.






            share|cite|improve this answer









            $endgroup$



            An idea what to do with this kind of questions: assume $;ainBbb R;$ belongs to the function's range, then there exists $;xinBbb R;$ (the function's domain) s.t.



            $$a=f(x)=frac x{x^2+1}implies ax^2-x+a=0implies Delta=1-4a^2ge0implies |a|lefrac12$$



            and thus you get that it must be that $;-frac12le alefrac12;$ , and the range of the function is waaaay different from what you thought.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 13 at 12:22









            DonAntonioDonAntonio

            180k1495233




            180k1495233












            • $begingroup$
              I do not understand why you say the range is way different from the one proposed by the OP. The OP found all of the same values but was confused by his method.
              $endgroup$
              – N. F. Taussig
              Jan 14 at 18:55










            • $begingroup$
              @N.F.Taussig Perhaps, The OP's original post wasn't written with LaTeX and I thought I read something involving intervals with $;5;$ ...That's why I wrote what I wrote.
              $endgroup$
              – DonAntonio
              Jan 14 at 19:37




















            • $begingroup$
              I do not understand why you say the range is way different from the one proposed by the OP. The OP found all of the same values but was confused by his method.
              $endgroup$
              – N. F. Taussig
              Jan 14 at 18:55










            • $begingroup$
              @N.F.Taussig Perhaps, The OP's original post wasn't written with LaTeX and I thought I read something involving intervals with $;5;$ ...That's why I wrote what I wrote.
              $endgroup$
              – DonAntonio
              Jan 14 at 19:37


















            $begingroup$
            I do not understand why you say the range is way different from the one proposed by the OP. The OP found all of the same values but was confused by his method.
            $endgroup$
            – N. F. Taussig
            Jan 14 at 18:55




            $begingroup$
            I do not understand why you say the range is way different from the one proposed by the OP. The OP found all of the same values but was confused by his method.
            $endgroup$
            – N. F. Taussig
            Jan 14 at 18:55












            $begingroup$
            @N.F.Taussig Perhaps, The OP's original post wasn't written with LaTeX and I thought I read something involving intervals with $;5;$ ...That's why I wrote what I wrote.
            $endgroup$
            – DonAntonio
            Jan 14 at 19:37






            $begingroup$
            @N.F.Taussig Perhaps, The OP's original post wasn't written with LaTeX and I thought I read something involving intervals with $;5;$ ...That's why I wrote what I wrote.
            $endgroup$
            – DonAntonio
            Jan 14 at 19:37




















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