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
问题描述：
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
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