高一数学题，帮帮忙啊

设f(x)=x^2-6tx+10t^2在闭区间[-1,1]上的最大值是M(t),最小值是m(t)，求 m(t)和M(t)的表达式
看清楚哦，不是[t-1,t+1]!

f(x)=x^2-6tx+10t^2=(x-3t)^2+t^2

假设t>=0,若3t>=1 即t>=1/3 则最大值取x=-1,M(t)=f(x)max=10t^2+6t+1

                            最小值取x=1,m(t)=f(x)min=10t^2-6t+1

         若3t<1 即0<t<1/3 则最大值取x=-1,M(t)=f(x)max=10t^2+6t+1

                            最小值取x=3t,m(t)=f(x)min=t^2

假设t<0,若3t<=-1 即t<=-1/3 则最大值取x=1,M(t)=f(x)max=10t^2-6t+1

                             最小值取x=-1,m(t)=f(x)min=10t^2+6t+1

        若3t>-1 即-1/3<t<0 则最大值取x=1,M(t)=f(x)max=10t^2-6t+1

                             最小值取x=3t,m(t)=f(x)min=t^2