14. 已知二次函数f(x)f(x)f(x)满足f(x)=f(4−x)f(x)=f(4-x)f(x)=f(4−x),fff(2)=f=f=f(1)−1-1−1,若不等式f(x)⩽−2x+2f(x)\leqslant -2x+2f(x)⩽−2x+2有唯一实数解.(1)求函数f(x)f(x)f(x)的解析式;(2)若函数f(x)f(x)f(x)在[t[t[t,t+1]t+1]t+1]上的最小值为g(t)g(t)g(t).①求g(t)g(t)g(t);②解不等式g(m)<g(2m−12)g(m)<g(2m-\frac{1}{2})g(m)<g(2m−21)