Question

求f(X)=(ax+1)×e^x单调区间

Answer 1

f'(x)=a*e^x+(ax+1)*e^x=(ax+a+1)*e^x 又因为e^x>0

令f'(x)=0,则x=-(a+1)/a

a>0,当x属于（负无穷,-(a+1)/a）,f'(x)0,f(x)单调递增

当x属于（-(a+1)/a,正无穷）,f'(x)