Question

已知函数fx=x2-x的绝对值-ax，当a≤-1,求fx在(-2，2)上的最小值

Answer 1

f(x)=x2-|x|-ax, a<=-1
f(x)=x2-|x|+|a|x
(1) x>=0
f(x)=x2+(|a|-1)x>=0
(2)x<0
f(x)=x2-(|a|-1)|x|
=[|x|-(|a|-1)/2]^2-(|a|-1)^2/4
当|a|-1<=4
f(x)min=-(|a|-1)^2/4