已知函数f(x)=2^∣x-m∣和函数g(x)=x∣x-m∣+2m-8

2个回答

  • 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,+∞).