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

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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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