<PRE><B>Question 557658</B>
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ 3x^4\ -\ 6x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f&#39;(x)\ =\ 12x^3\ -\ 12x]


The three zeros of *[tex \Large f&#39;(x)] are 0, 1, and -1


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f&#39;&#39;(x)\ =\ 36x^2\ -\ 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f&#39;&#39;(0)\ =\ -12\ &lt;\ 0]


Therefore maximum


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f&#39;&#39;(1)\ =\ 24\ &gt;\ 0]


Therefore minimum


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f&#39;&#39;(-1)\ =\ 24\ &gt;\ 0]


Therefore minimum


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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