Question

已知a＞1,函数f(x)=求函数f(x)在x∈［1,2］时的最小值.

Answer 1

解:(1)当1＜a≤2时,

∵x∈［1,a］时,f′(x)≤0,                                                                                    ?

x∈(a,2］时,f′(x)＞0,                                                                                        ?

∴f(x)_Min=f(a)=0.                                                                                                     ?

(2)当a＞2时,∵x∈［1,2］,?

∴x＜a.∴f(x)=x²·E^-ax,f′(x)=(2x-ax²)·E^-ax＜0.                                                        ?

∴f(x)在x∈［1,2］时是减函数.                                                                       ?

∴f(x)_Min=f(2)=4E^-2a.