一道数学题目..急
若2次函数F(X)X=X2-(A-2)X+5在区间(0.5,1)上是增函数,则F(2)的取值范围是
参考答案:F(X)=X2-(A-2)X+5
=(x-(A-2)/2)^2+5-B
开口向上,所以(A-2)/2<=0.5
A<=3
F(2)=13-2A>=7
若2次函数F(X)X=X2-(A-2)X+5在区间(0.5,1)上是增函数,则F(2)的取值范围是
参考答案:F(X)=X2-(A-2)X+5
=(x-(A-2)/2)^2+5-B
开口向上,所以(A-2)/2<=0.5
A<=3
F(2)=13-2A>=7