【1】-x²+2x+3≥0.===>x²-2x-3≤0.===>(x+1)(x-3)≤0.===>-1≤x≤3.∴函数定义域为[-1,3].【2】当x∈[-1,3]时,-x²+2x+3=4-(x-1)².此时易知,0≤-x²+2x+3≤4.===>0≤y≤2.∴函数值域为[0,2]
【1】-x²+2x+3≥0.===>x²-2x-3≤0.===>(x+1)(x-3)≤0.===>-1≤x≤3.∴函数定义域为[-1,3].【2】当x∈[-1,3]时,-x²+2x+3=4-(x-1)².此时易知,0≤-x²+2x+3≤4.===>0≤y≤2.∴函数值域为[0,2]