Question

若函数f(x)=3x的平方-(2a+6)x+a+3的值域为闭区间0到正无穷,求实数a的范围； 若y=3x的平方-(2a+

Answer 1

f(x)=3x^2-(2a+6)x+a+3

=3(x^2-(2(a+3)/3)x+((a+3)/3)^2)-((a+3)^2/3)+(a+3)

=3(x-(a+3)/3)^2-((a+3)^2/3)+(a+3)

f(x)的值域为[0,∞),所以

-((a+3)^2/3)+(a+3)>=0

(a+3)^2-3(a+3)