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关于方程,f(x) = 2x^4 − 6x^2 怎么推出这个的 f has a relative max.at x = 0 and a relative min.at x = \x06正负根号下3/2
人气:458 ℃ 时间:2020-10-02 02:23:06
解答
f'=8x^3-12x=4x(2x^2-3)
令f'=0
得x=0,\x06正负根号下3/2
f''=24x^2-12
f''(0)0
所以 f has a relative max.at x = 0 and a relative min.at x = \x06正负根号下3/2
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