已知函数f(x)=2的x次方/2x+m,且f(1)=2/3,(1)求m的值(2)求函数f(x)的值域

1个回答

  • f(x)=2^x/(2x+m)

    f(1)=2/3

    2/3 = 2/(2+m)

    m=1

    f(x)=2^x/(2x+1)

    定义域x≠-1/2

    f'(x) = {(2x+1)*2^xln2-2*2^x}/(2x+1)^2=2ln2*2^x{x+1/2-1/ln2}/(2x+1)^2

    x<1/ln2-1/2时单调减;x>1/ln2-1/2时单调增

    x=1/ln2-1/2时有极小值:2^(1/ln2-1/2)/(2/ln2) = √2*ln2*2^(1/ln2)/4

    值域【√2*ln2*2^(1/ln2)/4,+∞)

    f(x)=2^x/(2x+m)

    f(2)=2/3

    2/3 = 2^2/(2*2+m)

    m=2

    f(x)=2^x/(2x+2)

    定义域x≠-1

    f'(x) = {2(x+1)*2^xln2-2*2^x}/(2x+2)^2=ln2*2^x{x+1-1/ln2}/{2(x+1)^2}

    x<1/ln2-1时单调减;x>1/ln2-1时单调增

    x=1/ln2-1时有极小值:2^(1/ln2-1)/(2/ln2) = 2^(1/ln2)ln2/4

    值域【2^(1/ln2)ln2/4,+∞)