Finding the stationary point of a type of hyperbola?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 = 1 x + 1 x 2 − 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 1 x 4 which = 0. But the only way for 1 x = 0 is if x = ∞.

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Finding the stationary point of a type of hyperbola?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  =      1    x    +      1          x      2        −      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       1          x      4       which   =  0. But the only way for       1    x    =  0 is if   x  =  ∞.

Laura Fletcher · Accepted Answer

Step 1                    y                            =                  1          x                +                  1                      x            2                          −                  1                      x            3                                                            =                                            x              2                        +            x            −            1                                x            3                                                        y          ′                                    =                                            x              3                        (            2            x            +            1            )            −            3                          x              2                        (                          x              2                        +            x            −            1            )                                x            6                                                            =                              x            (            2            x            +            1            )            −            3            (                          x              2                        +            x            −            1            )                                x            4                                                            =                              2                          x              2                        +            x            −            3                          x              2                        −            3            x            +            3                                x            4                                                            =                              −                          x              2                        −            2            x            +            3                                x            4                          .            Step 2Setting this equal to zero is tantamount to solving       x    2    +  2  x  −  3  =  0  , with solutions  x  =            −      2      ±              4        +        4        (        3        )              2    =            −      2      ±      4        2    =  {  1  ,  −  3  }  .

Finding the stationary point of a type of hyperbola? I know that to find stationary points on a function, we need to differentiate the function and set that = 0.

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