已知二次函数f(x)满足f(3x+1)=9x^2-6x+5,当X∈[a-1,a],求f(x)最小值解析：∵函数f(x)满足f(3x+1)=9x^2-6x+5,设t=3x+1==>x=(t-1)/3f(t)=9[(t-1)/3]^2-6*(t-1)/3+5=t^2-4t+8∴f(x)=x^2-4x+8其图像为开口向上的抛物线,...