已知函数f=x^2+mx+3,当x∈[-2,2]时

函数f=x^2+mx+3,当x∈[-2,2]时
f大于等于m恒成立,需f(x)min≥m
f(x)=(x+m/2)²+3-m²/4
当-m/24时,
f(x)min=f(-2)=7-2m≥m ==>m≤7/3
无解
当-2≤-m/2≤2即-4≤m≤4时,
f(x)min=f(-m/2)=3-m²/4≥m
==>m²+4m-12≤0
==>-6≤m≤2
∴-4≤m≤2
当-m/2>2即m

①-m/2≤-2
f(-2)≥m
m≥4
4-2m+3≥m
7≥3m
m≤7/3
m无解
②-m/2≥2
f(2)≥m
m≤-4
4+2m+3≥m
m≥-7
-7≤m≤-4
③-2<-m/2<2
-4f（x）-m=x^2+mx+3-m
△≤0
m^2-4...

无解
没办法的