函数奇偶性与值域问题若函数f(x)=a(x^2)+(a+1)x+2是定义域[-2,2]上的偶函数,求此函数的值域

Answer 1

偶函数
所以f(x)=f(-x)
所以ax^2+(a+1)x+2=a(-x)^2+(a+1)(-x)+2
ax^2+(a+1)x+2=ax^2-(a+1)x+2
(a+1)x=-(a+1)x
2(a+1)x=0
所以a+1=0
a=-1
f(x)=-x^2+2
-2<=x<=2
0<=x^2<=4
-4<=-x^2<=0
-2<=-x^2+2<=2
所以值域[-2,2]