How do you algebraically demostrate that f ( x ) = x 3 − x is not a one to one function?So, I'm trying to prove that the function its not one to one, but I'm stuck: f ( x 1 ) = f ( x 2 ) → x 1 = x 2 then is a one to oneSo x 1 3 − x 1 = x 2 3 − x 2 x 1 3 − x 2 3 − x 1 + x 2 = 0 ( x 1 − x 2 ) ( x 1 2 + x 1 x 2 + x 2 2 ) − ( x 1 − x 2 ) = 0 ( x 1 − x 2 ) ( x 1 2 + x 1 x 2 + x 2 2 − 1 ) = 0Which is true if x 1 = x 2 but how do I go about proving that is not a one to one function?

Question

How do you algebraically demostrate that   f  (  x  )  =      x    3    −  x is not a one to one function?So, I&#039;m trying to prove that the function its not one to one, but I&#039;m stuck:   f  (      x    1    )  =  f  (      x    2    )  →      x    1    =      x    2   then is a one to oneSo       x    1    3    −      x    1    =      x    2    3    −      x    2        x    1    3    −      x    2    3    −      x    1    +      x    2    =  0  (      x    1    −      x    2    )  (      x    1    2    +      x    1        x    2    +      x    2    2    )  −  (      x    1    −      x    2    )  =  0  (      x    1    −      x    2    )  (      x    1    2    +      x    1        x    2    +      x    2    2    −  1  )  =  0Which is true if       x    1    =      x    2   but how do I go about proving that is not a one to one function?

Nia Molina · Accepted Answer

Explanation:You can assume f to be one to one and show contradictions.  ∃     0  ,  1  ∈      R   such that       0    3    −  0  =  0 and       1    3    −  1  =  0 which contradicts f being a one to one function. Hence, f is not one to one.

fabios3 · Accepted Answer

Explanation:This is also true if       x    1    2    +      x    1    ⋅      x    2    +      x    2    2    −  1  =  0With the p-q formula, treating       x    1   as the variable, we get      x    1    =  −            x      2        2    +                    x        2        2            4        −          x      2      2        +    1    =  −            x      2        2    +      1    −                  3                  x          2          2                    4      and       x    1    =  −            x      2        2    +      1    −                  3                  x          2          2                    4      Thus, there are solutions when       x    1    ≠      x    2   and you are not one to one.