Find m to the equation: { 2 x 3 − ( y + 2 ) x 2 + x y = m ( 1 ) x 2 + x − y = 1 − 2 m ( 2 ) My try: From ( 1 ) and ( 2 ) ⇒: 4 x 3 − 2 ( y + 2 ) x 2 + 2 x y + x 2 + x − y = 1 ⇔ 4 x 3 − 3 x 2 + x − 1 = y ( 2 x 2 − 2 x + 1 ) ⇔ y = 4 x 3 − 3 x 2 + x − 1 2 x 2 − 2 x + 1 ⇔ y = 2 x + 1 2 − 3 2 ( 2 x 2 − 2 x + 1 ) From ( 2 ) ⇒ 2 m = 1 + y − x 2 − x = 1 + 2 x + 1 2 − 3 2 ( 2 x 2 − 2 x + 1 ) − x 2 − x = − x 2 + x + 3 2 − 3 2 ( 2 x 2 − 2 x + 1 ) And I don't know how to contine,The result is: m ≤ 1 − 3 2

Question

Find m to the equation:      {                            2                      x            3                    −                      (            y            +            2            )                                x            2                    +          x          y          =          m                              (          1          )                                                  x            2                    +          x          −          y          =          1          −          2          m                              (          2          )                        My try: From   (  1  ) and   (  2  )    ⇒:  4      x    3    −  2      (    y    +    2    )        x    2    +  2  x  y  +      x    2    +  x  −  y  =  1    ⇔  4      x    3    −  3      x    2    +  x  −  1  =  y      (    2          x      2        −    2    x    +    1    )      ⇔  y  =                    4                  x          3                −        3                  x          2                +        x        −        1                    2                  x          2                −        2        x        +        1                ⇔  y  =  2  x  +            1      2        −            3              2                  (          2                      x            2                    −          2          x          +          1          )                    From   (  2  )    ⇒  2  m  =  1  +  y  −      x    2    −  x    =  1  +  2  x  +            1      2        −            3              2                  (          2                      x            2                    −          2          x          +          1          )                      −      x    2    −  x    =  −      x    2    +  x  +            3      2        −            3              2                  (          2                      x            2                    −          2          x          +          1          )                    And I don&#039;t know how to contine,The result is:   m  ≤  1  −                    3            2

Diego Mathews · Accepted Answer

From your last expression, after division by   2, the value of   m is  m  =            −              x        4            +      2              x        3            −      x              2              x        2            −      2      x      +      1        .To maximize this we find its derivative and get      m    ′    (  x  )  =            −      (      2      x      −      1      )      (      2              x        4            −      4              x        3            +      4              x        2            −      2      x      −      1      )              (      2              x        2            −      2      x      +      1              )        2              .So there is a critical point at   x  =  1  /  2 and we need to set the fourth degree polynomial in the numerator to zero. [The denominator is positive for all   x].It is maybe just luck, but if we substitute   x  =  1  /  2  +  t into the fourth degree polynomial it becomes   2      t    4    +      t    2    −  11  /  8 which is quadratic in  and leads to       t    2    =  (  −  1  +      12    )  /  4. [We cannot use the negative sign choice here since we need   1  /  2. The negative choice corresponds to the pair of nonreal conjugate complex solutions to the fourth degree equation.] So we now have three critical points for   m, namely 1/2 and   1  /  2  ±      −    1    +          12        /  2.. The value of m is largest among these when   x  =  1  /  2  +      −    1    +          12        /  2  , and for this   x we have   m  =  1  −      3    /  2. Note also that since clearly   m  →  −  ∞ as   x  →  ±  ∞ we can be sure the maximum of   m must occur at one of the above critical points.After a closer look it turns out the maximum   m  =  1  −      3    /  2 occurs at both of the critical points   x  =  1  /  2  ±      −    1    +          12        /  2.. This seems unusual since the expression for   m is not symmetrical around   x  =  1  /  2  ,, but somehow the two critical points, and their   m values, are symmetrical around    x  =  1  /  2.

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