Question

急求，急求！！！！高一数学函数。

Answer 1

f(x)=x^2+ax+3-a抛物线f(x)开口向上，对称轴x=-a/21）当对称轴x=-a/2<=-2即a>=4时f(x)在[-2,2]上是增函数f(x)>=f(-2)=4-2a+3-a=7-3a>=2解得：a<=5/3矛盾，不符合2）当对称轴-2<=x=-a/2<=2即-4<a<4时f(x)在对称轴上取得最小值f(x)>=f(-a/2)=a^2/4-a^2/2+3-a=-a^2/4-a+3>=2a^2+4a-4<=0解得：-2-2√2<=a<=-2+2√2所以：-4<a<=-2+2√23）当对称轴x=-a/2>=2即a<=-4时f(x)在[-2,2]上是减函数f(x)>=f(2)=4+2a+3-a=7+a>=2a>=-5所以：-5<=a<=-4综上所述，-5<=a<=-2+2√2