Finding the stationary point of a type of hyperbola?












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$begingroup$


I know that to find stationary points on a function, we need to differentiate the function and set that = 0.



But how can we find the stationary points of the below function?



$$y=frac{1}{x} + frac{1}{x^2} - frac{1}{x^3}$$



When I differentiate it, I get fractions with $x$ in the denominator... I try to factorize and I get a quadratic equation with non-real roots and then $frac{1}{x^4}$ which $= 0$. But the only way for $frac{1}{x} = 0$ is if $x=infty$.



I thought maybe there are no stationary points but this function does have stationary points that can be seen here:



https://www.wolframalpha.com/input/?i=1%2Fx+%2B+1%2Fx%5E2+-+1%2Fx%5E3



The maximum is at $x=1$ and the minimum at $x=-3$.



How can we solve this problem? Or more specifically, perhaps the question is how to solve the equation $frac{1}{x^n} = 0$ ?






derivatives stationary-point







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    I know that to find stationary points on a function, we need to differentiate the function and set that = 0.



    But how can we find the stationary points of the below function?



    $$y=frac{1}{x} + frac{1}{x^2} - frac{1}{x^3}$$



    When I differentiate it, I get fractions with $x$ in the denominator... I try to factorize and I get a quadratic equation with non-real roots and then $frac{1}{x^4}$ which $= 0$. But the only way for $frac{1}{x} = 0$ is if $x=infty$.



    I thought maybe there are no stationary points but this function does have stationary points that can be seen here:



    https://www.wolframalpha.com/input/?i=1%2Fx+%2B+1%2Fx%5E2+-+1%2Fx%5E3



    The maximum is at $x=1$ and the minimum at $x=-3$.



    How can we solve this problem? Or more specifically, perhaps the question is how to solve the equation $frac{1}{x^n} = 0$ ?






    derivatives stationary-point







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      I know that to find stationary points on a function, we need to differentiate the function and set that = 0.



      But how can we find the stationary points of the below function?



      $$y=frac{1}{x} + frac{1}{x^2} - frac{1}{x^3}$$



      When I differentiate it, I get fractions with $x$ in the denominator... I try to factorize and I get a quadratic equation with non-real roots and then $frac{1}{x^4}$ which $= 0$. But the only way for $frac{1}{x} = 0$ is if $x=infty$.



      I thought maybe there are no stationary points but this function does have stationary points that can be seen here:



      https://www.wolframalpha.com/input/?i=1%2Fx+%2B+1%2Fx%5E2+-+1%2Fx%5E3



      The maximum is at $x=1$ and the minimum at $x=-3$.



      How can we solve this problem? Or more specifically, perhaps the question is how to solve the equation $frac{1}{x^n} = 0$ ?






      derivatives stationary-point







      share|cite|improve this question











      $endgroup$




      I know that to find stationary points on a function, we need to differentiate the function and set that = 0.



      But how can we find the stationary points of the below function?



      $$y=frac{1}{x} + frac{1}{x^2} - frac{1}{x^3}$$



      When I differentiate it, I get fractions with $x$ in the denominator... I try to factorize and I get a quadratic equation with non-real roots and then $frac{1}{x^4}$ which $= 0$. But the only way for $frac{1}{x} = 0$ is if $x=infty$.



      I thought maybe there are no stationary points but this function does have stationary points that can be seen here:



      https://www.wolframalpha.com/input/?i=1%2Fx+%2B+1%2Fx%5E2+-+1%2Fx%5E3



      The maximum is at $x=1$ and the minimum at $x=-3$.



      How can we solve this problem? Or more specifically, perhaps the question is how to solve the equation $frac{1}{x^n} = 0$ ?






      derivatives stationary-point




      derivatives stationary-point






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 28 at 15:58









      xihiro

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      asked Jan 28 at 15:44









      user961627user961627

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          $begingroup$

          Hmm.
          begin{align*}
          y&=frac{1}{x}+frac{1}{x^2}-frac{1}{x^3} \
          &=frac{x^2+x-1}{x^3} \
          y'&=frac{x^3(2x+1)-3x^2(x^2+x-1)}{x^6} \
          &=frac{x(2x+1)-3(x^2+x-1)}{x^4} \
          &=frac{2x^2+x-3x^2-3x+3}{x^4} \
          &=frac{-x^2-2x+3}{x^4}.
          end{align*}
          Setting this equal to zero is tantamount to solving $x^2+2x-3=0,$ with solutions
          $$x=frac{-2pmsqrt{4+4(3)}}{2}=frac{-2pm 4}{2}={1, -3}. $$
          You can probably use the Second Derivative Test to show which of these is a local min, and which a local max.






          share|cite|improve this answer









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            $begingroup$

            Hmm.
            begin{align*}
            y&=frac{1}{x}+frac{1}{x^2}-frac{1}{x^3} \
            &=frac{x^2+x-1}{x^3} \
            y'&=frac{x^3(2x+1)-3x^2(x^2+x-1)}{x^6} \
            &=frac{x(2x+1)-3(x^2+x-1)}{x^4} \
            &=frac{2x^2+x-3x^2-3x+3}{x^4} \
            &=frac{-x^2-2x+3}{x^4}.
            end{align*}
            Setting this equal to zero is tantamount to solving $x^2+2x-3=0,$ with solutions
            $$x=frac{-2pmsqrt{4+4(3)}}{2}=frac{-2pm 4}{2}={1, -3}. $$
            You can probably use the Second Derivative Test to show which of these is a local min, and which a local max.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              Hmm.
              begin{align*}
              y&=frac{1}{x}+frac{1}{x^2}-frac{1}{x^3} \
              &=frac{x^2+x-1}{x^3} \
              y'&=frac{x^3(2x+1)-3x^2(x^2+x-1)}{x^6} \
              &=frac{x(2x+1)-3(x^2+x-1)}{x^4} \
              &=frac{2x^2+x-3x^2-3x+3}{x^4} \
              &=frac{-x^2-2x+3}{x^4}.
              end{align*}
              Setting this equal to zero is tantamount to solving $x^2+2x-3=0,$ with solutions
              $$x=frac{-2pmsqrt{4+4(3)}}{2}=frac{-2pm 4}{2}={1, -3}. $$
              You can probably use the Second Derivative Test to show which of these is a local min, and which a local max.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Hmm.
                begin{align*}
                y&=frac{1}{x}+frac{1}{x^2}-frac{1}{x^3} \
                &=frac{x^2+x-1}{x^3} \
                y'&=frac{x^3(2x+1)-3x^2(x^2+x-1)}{x^6} \
                &=frac{x(2x+1)-3(x^2+x-1)}{x^4} \
                &=frac{2x^2+x-3x^2-3x+3}{x^4} \
                &=frac{-x^2-2x+3}{x^4}.
                end{align*}
                Setting this equal to zero is tantamount to solving $x^2+2x-3=0,$ with solutions
                $$x=frac{-2pmsqrt{4+4(3)}}{2}=frac{-2pm 4}{2}={1, -3}. $$
                You can probably use the Second Derivative Test to show which of these is a local min, and which a local max.






                share|cite|improve this answer









                $endgroup$



                Hmm.
                begin{align*}
                y&=frac{1}{x}+frac{1}{x^2}-frac{1}{x^3} \
                &=frac{x^2+x-1}{x^3} \
                y'&=frac{x^3(2x+1)-3x^2(x^2+x-1)}{x^6} \
                &=frac{x(2x+1)-3(x^2+x-1)}{x^4} \
                &=frac{2x^2+x-3x^2-3x+3}{x^4} \
                &=frac{-x^2-2x+3}{x^4}.
                end{align*}
                Setting this equal to zero is tantamount to solving $x^2+2x-3=0,$ with solutions
                $$x=frac{-2pmsqrt{4+4(3)}}{2}=frac{-2pm 4}{2}={1, -3}. $$
                You can probably use the Second Derivative Test to show which of these is a local min, and which a local max.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 28 at 15:56









                Adrian KeisterAdrian Keister

                5,28171933




                5,28171933






























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