Is there a rational function f ( x ) ∈ Q ( x ) such that 2 ≤ f ( x ) ≤ 2 x 3 for all x ≥ 2 ?My thoughts : it is easy to find such an f if we relax the conditions to f ( x ) ∈ R ( x ) (take f constant equal to 2 ), however no easy perturbation of this solution seems to solve the original problem. Clearly f must have zero degree in x and can be written in the form x + ( x 2 − 2 ) g ( x ) where g is another rational function. Then I am stuck.

Question

Is there a rational function   f  (  x  )  ∈            Q        (  x  ) such that       2    ≤  f  (  x  )  ≤            2      x        3   for all   x  ≥      2  ?My thoughts : it is easy to find such an f if we relax the conditions to   f  (  x  )  ∈            R        (  x  ) (take f constant equal to       2  ), however no easy perturbation of this solution seems to solve the original problem. Clearly f must have zero degree in x and can be written in the form   x  +  (      x    2    −  2  )  g  (  x  ) where g is another rational function. Then I am stuck.

massetereqe · Accepted Answer

The simplest function I can find seems   f  (  x  )  =                    2        x        +        2                    x        +        2            . It is easily verified this satisfies the double inequality for   x  ≥      2  .This was obtained by looking at linear approximants, and then setting the conditions   f  (      2    )  =      2    ,  f  (  x  )  ≥      2   and finally the RHS inequality.Even among linear functions, there are of course many choices, many of which can be got from the form  f  (  x  )  =                    a        x        +        2        b                    b        x        +        a            , with  (  a  ,  b  )  ∈  {  (  3  ,  2  )  ,  (  5  ,  2  )  ,  (  5  ,  3  )  ,  (  7  ,  3  )  ,  (  7  ,  4  )  ,  .  .  .  }... which leads to the thought                     (        2        n        ±        1        )        x        +        2        n                    n        x        +        2        n        ±        1             could work for   1  &amp;lt;  n  ∈      N   (not checked).