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13问答网 > 若函数y=ax+x恰有三个单调区间,求实数a的取值范围,并求出单调区间

若函数y=ax+x恰有三个单调区间,求实数a的取值范围,并求出单调区间

Y=AX3+X恰有三个单调区间，求实数a的取值范围，并求出单调区间

2025-03-16 13:50:29

推荐回答（2个）

回答1：

因为y=ax³+x 恰有三个单调区间，
所以y'=3ax²+1=0有两个不相等的实数根x=±1/√(-3a)，因此a<0。
在区间(-∞,√(-3a)/3a)和(-√(-3a)/3a,+∞)上，函数y单调递减，
在区间(√(-3a)/3a,-√(-3a)/3a)上，函数y单调递增。

回答2：

是“y=ax+x^3“吧
y'=3x^2+a=0
x^2=-a/3>0
a<0
x=±√（-3a)/3
单增区间(-∞，-√（-3a)/3),(√（-3a)/3,+∞）
单减区间[-√（-3a)/3,√（-3a)/3]

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