We have f '(x) = 6x2 + 6x − 432, so f ''(x) = 12x+6 correct: your answer is correct. , which equals 0 when

The derivative of the first derivative given to be, f'(x) = 6x² + 6x - 432 is given to be,
    f''(x) = 12x + 6

As asked in the given, the value of the second derivative becomes zero when the value of x is, 
   f"(x) = 0 = 12x + 6

    12x + 6 = 0

Subtract 6 from both sides of the equation such that,
   12x = -6

Divide both sides of the equation by 12 giving us an answer of 
  x = -1/2.

Hence, x = -1/2. 

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