Question

若函数f(x)=x²+x-1/4的定义域为[-1,2]，求f(x)的值域

要有详细过程

Answer 1

楼上算错啦。f(x)=x²+x+1/4-1/2
=(x+1/2)²-1/2
-1<=x<=2
-1/2<=x+1/2<=5/2
1/4<=(x+1/2)2(这个2是平方）<=25/4
-1/4<=(x+1/2)2(这个2是平方）-1/2<=23/4
so,值域是[-1/4,23/4]

Answer 2

f(x)=x²+x+1/4-1/2
=(x+1/2)²-1/2
对称轴x=-1/2
-1<=x<=2
所以x=-1/2,最小值=-1/2
x=2，最大=23/4
值域[-1/2,23/4]