How to find all zeros of f ( x ) = 3 x 3 - 12 x 2 + 3 x ?

Question

How to find all zeros of             f              (        x        )              =    3          x      3        -    12          x      2        +    3    x  ?

Karbamidjts · Accepted Answer

We must first determine the derivative of a function before we can determine its zeros. This one is simple; we'll just use the power rule to evaluate each phrase individually.

f (x) = 3 x^{3} - 12 x^{2} + 3 x

f' (x) = 9 x^{2} - 24 x + 3

Then we set the derivative equal to zero and solve for x

9 x^{2} - 24 x + 3 = 0

Given that this function cannot be factored, we can enter this into our calculator to determine the two x values. Obtaining two values

x = 2.535

x = .131

udahnulizwk · Accepted Answer

First, factor out an x:            f              (        x        )              =    x          (      3              x        2            -      12      x      +      3      x      )      Already, you can figure out that when x zero, f(x) is zero, so       x    =    0   is one solution.Then, attempt to factor the quadratic. Unfortunately, in this case, you can&#039;t use factoring, so plug the numbers into the quadratic formula, which is       x    =                  -        b        ±                                            b              2                        -            4            a            c                              2        a   in case you don&#039;t have it memorized:      x    =                  -                  (          -          12          )                ±                                                            (                -                12                )                            2                        -            4                          (              3              )                                      (              3              )                                                  2                  (          3          )                          x    =                  12        ±                              144            -            36                                      2                  (          3          )                          x    =                  12        ±                              108                                      6            (The square root can be simplified to       6                  3            , because                     108              =                  6        ⋅        6        ⋅        3              =                            6                    2        ⋅                  3              =    6    ⋅                  3            )      x    =                  12        ±        6                              3                                      6            A six can also be cancelled out:      x    =          (      2      ±                        3                    )      That means the zeroes are       x    =    0    ,    x    =    2    +                  3              ,                and              x    =    2    -                  3

How to find all zeros of f(x)=3x^3-12x^2+3x?

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