已知函数f(x)=a•e x + a+1 x -2(a+1)(a>0) .

1个回答

  • (Ⅰ)当a=1时,f(x)=e x+

    2

    x -4,∴f′(x)=e x-

    2

    x 2 ,∴f′(1)=e-2,

    ∵f(1)=e-2,

    ∴f(x)在点(1,f(1))处的切线方程为:(e-2)x-y=0.

    (Ⅱ)∵f(x)=a•e x+

    a+1

    x -2(a+1)(a>0) .

    ∴f′(x)=

    a x 2 e x -(a+1)

    x 2 ,

    令g(x)=ax 2e x-(a+1),则g′(x)=ax(2+x)e x>0,

    ∴g(x)在(0,+∞)上单调递增,

    ∵g(0)=-(a+1)<0,当x→+∞时,g(x)>0,

    ∴存在x 0∈(0,+∞),使g(x 0)=0,且f(x)在(0,x 0)上单调递减,f(x)在(x 0,+∞)上单调递增,

    ∵g(x 0)= a x 0 2 e x 0 -(a+1)=0,∴ a x 0 2 e x 0 =a+1,即 a e x 0 =

    a+1

    x 0 2 ,

    ∵对于任意的x∈(0,+∞),恒有f(x)≥0成立,

    ∴f(x) min=f(x 0)= a e x 0 +

    a+1

    x 0 -2(a+1)≥0,∴

    a+1

    x 0 2 +

    a+1

    x 0 -2(a+1)≥0,

    1

    x 0 2 +

    1

    x 0 -2≥0 ,∴ 2 x 0 2 - x 0 -1≤ 0,解得-

    1

    2 ≤x 0≤1,

    ∵ a x 0 2 e x 0 =a+1,∴ x 0 2 e x 0 =

    a+1

    a >1,

    令h(x 0)= x 0 2 e x 0 ,而h(0)=0,当x 0→+∞时,h(x 0)→+∞,

    ∴存在m∈(0,+∞),使h(m)=1,

    ∵h(x 0)= x 0 2 e x 0 在(0,+∞)上,∴x 0>m,

    ∴m<x 0≤1,

    ∵h(x 0)= x 0 2 e x 0 在(m,1]上∴h(m)<h(x 0)≤h(1),

    ∴1<

    a+1

    a ≤e,∴a≥

    1

    e-1 .