设函数f(x)=x•ekx(k≠0)((ekx)′=kekx) (1)求曲线y=f(x)在点(0,f(0))处的切线方程; (2)求函数f(x)的单调区间.

问题描述:

设函数f(x)=x•ekx(k≠0)((ekx)′=kekx
(1)求曲线y=f(x)在点(0,f(0))处的切线方程;
(2)求函数f(x)的单调区间.

(1)f′(x)=ekx+kxekx=(1+kx)ekx(x∈R),且f′(0)=1,∴切线斜率为1,又f(0)=0,∴曲线y=f(x)在点(0,f(0))处的切线方程为x-y=0.(2)f′(x)=(kx+1)ekx(x∈k),令f′(x)=0,得x=-1k,①若...