I have a function f ( a , b ) = a b ( a + b 2 ) 2 , and (to me) it has some cool properties (e.g f ( a , b ) = f ( b , a ), f ( x , 0 ) = 0, f ( x , x ) = 1, 0 ≤ f ≤ 1, etc.). Now I wanted to know the inverse function for this. It's a rational function, but I couldn't figure out how to do this. y = x b ( x + b 2 ) 2 = 4 x b ( x + b ) 2 x = 4 y b ( y + b ) 2 ⟹ x ( y + b ) 2 = 4 y b ⟹ ( y + b ) 2 y = 4 b x y 2 + b 2 y = 4 b x − 2 b ⟹ y + b 2 y = 4 b x − 2 bAt this point, I have no idea where to go. I think that I'm missing out on some fundamental rule of solving this, but I don't have a clue about where to go. How do I simplify the y + b 2 y ?NOTE: The final function f − 1 ( y , b ) is such that f − 1 ( f ( x , b ) , b ) = x for any x and b such that x and b are not both 0. One can be zero, but not both (since f ( 0 , 0 ) is undefined).

Question

I have a function   f  (  a  ,  b  )  =            a      b              (                        a          +          b                2                    )        2            , and (to me) it has some cool properties (e.g   f  (  a  ,  b  )  =  f  (  b  ,  a  ),   f  (  x  ,  0  )  =  0,   f  (  x  ,  x  )  =  1,   0  ≤  f  ≤  1, etc.). Now I wanted to know the inverse function for this. It&#039;s a rational function, but I couldn&#039;t figure out how to do this.  y  =            x      b              (                        x          +          b                2                    )        2              =            4      x      b              (      x      +      b              )        2                x  =            4      y      b              (      y      +      b              )        2                ⟹    x  (  y  +  b      )    2    =  4  y  b    ⟹              (      y      +      b              )        2              y    =            4      b        x                        y        2            +              b        2              y    =            4      b        x    −  2  b    ⟹    y  +            b      2        y    =            4      b        x    −  2  bAt this point, I have no idea where to go. I think that I&#039;m missing out on some fundamental rule of solving this, but I don&#039;t have a clue about where to go. How do I simplify the   y  +            b      2        y  ?NOTE: The final function       f          −      1        (  y  ,  b  ) is such that       f          −      1        (  f  (  x  ,  b  )  ,  b  )  =  x for any   x and   b such that   x and   b are not both 0. One can be zero, but not both (since   f  (  0  ,  0  ) is undefined).

soosenhc · Accepted Answer

You can write  y  =            4      x      b              (      x      +      b              )        2                    x    2    y  +  2  b  x  y  +  −  4  b  x  +      b    2    y  =  0    x  =            4      b      −      2      b      y      ±              (        4        b        −        2        b        y                  )          2                −        4                  b          2                          y          2                            2      y          x  =            2      b      −      b      y      ±              4                  b          2                −        4        b        y              y

I have a function f ( a , b ) = a b </mrow>

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