设a∈R,函数f(x)=((e^-x)/2)(ax^2+a+1),其中其中e是自然对数的底数.①判断f(x)在R上的单调性②判断f(x)在R上的单调性当-1＜a＜0时,求f(x)在[1,2]上的最小值

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设a∈R,函数f(x)=((e^-x)/2)(ax^2+a+1),其中其中e是自然对数的底数.①判断f(x)在R上的单调性②判断f(x)在R上的单调性当-1＜a＜0时,求f(x)在[1,2]上的最小值
设a∈R,函数f(x)=((e^-x)/2)(ax^2+a+1),其中其中e是自然对数的底数.
①判断f(x)在R上的单调性
②判断f(x)在R上的单调性当-1＜a＜0时,求f(x)在[1,2]上的最小值

设a∈R,函数f(x)=((e^-x)/2)(ax^2+a+1),其中其中e是自然对数的底数.①判断f(x)在R上的单调性②判断f(x)在R上的单调性当-1＜a＜0时,求f(x)在[1,2]上的最小值
(1）当a不等于0时.对f(x)求导得
f(x)’=（-0.5e^-x)(ax^2-2ax+a+1),令f(x)’=0所以f(x)’=（-0.5e^-x)(ax^2-2ax+a+1)=0 设K(x)=-0.5e^-x可知K(x)＜0
且在R上单调递增.又设H(x)=ax^2-2ax+a+1=0
由△=4a^2-4a（a+1)讨论得,《1》当△＞0时,a＜0即f(x)’=0的两根为(a+（-a)^0.5)/a和
（a-(-a)^0.5)/a.所以当x∈【-∞,(a+（-a)^0.5)/a】时f(x)’＞0即此时f(x)单调递增.当x∈【(a+（-a)^0.5)/a,（a-(-a)^0.5)/a】时f(x)’＜0即此时f(x)单调递减.当x∈【（a-(-a)^0.5)/a,+∞】时f(x)’＞0即此时f(x)单调递增.《2》当△＜0时f(x)’＜0即此时f(x)在x∈R上单调递减.
当a=0时,f(x)=0.5e^-x可知f(x)在x∈R上单调递增.
（2）你根据a的范围去求x是范围.再看看（1）中a＜0时的单调区间看看当x在[1,2]时是否单调,如果单调就将x=1和x=2带入f(x)对求出的两个值比较大小,小的就是最小值.如果不单调就把在[1,2]内的极值点和x=1和x=2带入f(x)比较三个值的大小,最小的就是答案.
解的不好见谅!第二问我没时间给你往这写了.你按那个过程试着做做就有答案了.

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