问题 填空题 函数f(x)=-2x,(x≤0)x2+1,(x>0),则f[f(-2)]=______. 答案 ∵f(x)=-2x,(x≤0)x2+1,(x>0),∴f(-2)=-2×(-2)=4,∴f[f(-2)]=f(4)=42+1=17.故答案为:17.