(1)f(x)=x^2+ax+bM>=|f(0)|=|b|M>=|f(1)|=|1+a+b|M>=|f(-1)|=|1-a+b|4M>=2|b|+|1+a+b|+|1-a+b|>=|(-2b)+(1+a+b)+(1-a+b)|=2M>=1/2[-b,1+a+b,1-a+b同号时取等](2)I.若-b,1+a+b,1-a+b均>=0,M=1/2,则:1+a+b<=1/2..............①1-a+b<=1/2..............②-b<=1/2.................③①+②:2+2b<=1,b<=-1/2③:b>=-1/2∴b=-1/2代回①:a<=0,②:a>=0∴a=0f(x)=x^2-1/2II.若-b,1+a+b,1-a+b均<0,M=1/2,则:0>1+a+b>=-1/2..............①0>1-a+b>=-1/2..............②0>-b>=-1/2.................③①+③:0>1+a>=-1,-2<=a<-1②+③:0>1-a>=-1,1<a<=2无解综上:f(x)=x^2-1/2