来自黄若宏的问题
已知函数f(x)=1−x1+x2ex.(Ⅰ)求f(x)的单调区间;(Ⅱ)证明:当f(x1)=f(x2)(x1≠x2)时,x1+x2<0.
已知函数f(x)=1−x1+x2ex.
(Ⅰ)求f(x)的单调区间;
(Ⅱ)证明:当f(x1)=f(x2)(x1≠x2)时,x1+x2<0.
1回答
2020-12-2720:53
已知函数f(x)=1−x1+x2ex.(Ⅰ)求f(x)的单调区间;(Ⅱ)证明:当f(x1)=f(x2)(x1≠x2)时,x1+x2<0.
已知函数f(x)=1−x1+x2ex.
(Ⅰ)求f(x)的单调区间;
(Ⅱ)证明:当f(x1)=f(x2)(x1≠x2)时,x1+x2<0.
(I)易知函数的定义域为R.
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