Question

已知函数f(x)=x^2+ax+1,当fx属于[-3,1]时,f(x)大于等于-3恒成立,求实数a的取值范

Answer 1

解：（i）当-a/2<-3,即a>6时,
f(x)min=f(-3)=9-3a+1>=-3，得a<=13/3(与a>6不合，舍去);
（ii）当-3<=-a/2<1，即2<a<6时,
f(x)min=f(-a/2)=a^2/4-a^2/2+1=-a^2/4+1>=-3，得-4<=a<=4，即2<a<=4;
(iii)当-a/2>=1，即a<=2时,
f(x)min=f(1)=1+a+1=2+a>=-3，得a>=-5，即-5<=a<=2。
综上：实数a的取值范围为[-5,4]

Answer 2

f(0)=1>-3;
x∈（0,1]时x^2+ax+1>=-3,变为
a>=-(x+4/x),
-(x+4/x)|max=-5,
∴a>=-5;
x∈[-3,0)时a<=-(x+4/x),
-(x+4/x)|min=4,
∴x<=4.
综上，-5<=a<=4.