f(x)=
2x-m (x≥m)2m-x(x<m),则f(x)的值域应是g(x)的值域的子集.
①当4≤m≤8时,f(x)在(-∞,4]上单调减,
故f(x)≥f(4)=2m-4,g(x)在[4,m]上单调减,[m,+∞)上单调增,
故g(x)≥g(m)=2m-8,
所以2m-4≥2m-8,解得4≤m≤5或m≥6.
②当m>8时,f(x)在(-∞,4]上单调减,
故f(x)≥f(4)=2m-4,g(x)在[4,
m2]单调增,[
m2,m]上单调减,[m,+∞)上单调增,g(4)=4m-16>g(m)=2m-8
故g(x)≥g(m)=2m-8,所以2m-4≥2m-8,解得4≤m≤5或m≥6.
③0<m<4时,f(x)在(-∞,m]上单调减,[m,4]上单调增,故f(x)≥f(m)=1.g(x)在[4,+∞)上单调增,故g(x)≥g(4)=8-2m,
所以8-2m≤1,即72≤m<4.
④m≤0时,f(x)在(-∞,m]上单调减,[m,4]上单调增,故f(x)≥f(m)=1.g(x)在[4,+∞)上单调增,
故g(x)≥g(4)=8-2m,所以8-2m≤1,即m≥
72.(舍去)
综上,m的取值范围是[
72,5]∪[6,+∞).