函数y=f(x)=ax^2+2ax+4(0
问题描述：
函数y=f(x)=ax^2+2ax+4(0
答
应该是0
f(x2)-f(x1)=a(x2^2-x1^2)+2a(x2-x1)
=a(x2+x1)(x2-x1)+2a(x2-x1)
=a(1-a)(x2-x1)+2a(x2-x1)
=-a(a-3)(x2-x1)
∵0x1
∴f(x2)-f(x1)>0
f(x2)>f(x1)