已知函数f（x）=axˇ+2ax+4（0∠a∠3）,若x1∠ x2 且x1 +x2=1-a,则判断f（x1）与f（x2）

x1+x2=1-a--->x1=1-a-x2 f(x1)-f(x2) = a(x1²-x2²)+2a(x1-x2)+(4-4) = a(x1+x2)(x1-x2)+2a(x1-x2) = [a(1-a)+2a](x1-x2) = a(3-a)(x1-x2) 0＜a＜3--->a(3-a)＞0 x1＜x2 --->f(x1)＜f(x2)