(1) When m=e, find the minimum value of f(x);

(2) Discuss the number of zeros of the function g(x)=f'(x)-x/3;

(3) If there is a b>a>0, [f (b)-f (a)]/(b-a) <; 1 holds, and find the range of m.

(1) Analysis: When m=e and f(x)=lnx+e/x,

Let f ′ (x) = (x-e)/x2 = 0 = = > x = e;

When x∈(0, e), f ′ (x) < 0, and f(x) is a decreasing function on (0, e) ∴;

When x∈(e, +∞), f ′ (x) > 0, and f(x) is the increasing function on (e, +∞);

When x = e, f(x) takes the minimum value f (e) = lne+e/e = 2;

(2) analysis: ∫ function g (x) = f ′ (x)-x/3 =1/x-m/x2-x/3 (x > 0),

Let g(x)=0 and get m =-1/3x3+x (x > 0);

Let φ (x) =- 1/3x 3+x (x ≥ 0),

∴φ′(x)=-x^2+ 1=-(x- 1)(x+ 1)；

When x∈(0, 1), φ′ (x) > 0, and φ(x) is increasing function at (0, 1).

When x∈( 1, +∞), φ′ (x) < 0, and φ(x) is a decreasing function on (1, +∞);

∴φ(x) takes the maximum at x= 1, where x= 1 is the maximum point of φ(x), and φ (1) = 2/3;

φ(0)=0, combined with the image of y=φ(x), as shown in the figure;

Understand:

When m > 2/3, the function g(x) has no zero;

When m=2/3, the function g(x) has one and only one zero;

When 0 < m < 2/3, the function g(x) has two zeros;

When m≤0, the function g(x) has one and only one zero;

To sum up:

When m > 2/3, the function g(x) has no zero;

When m=2/3 or m≤0, the function g(x) has one and only one zero;

When 0 < m < 2/3, the function g(x) has two zeros;

(3 analysis: ∫ for any b > a > 0, [f (b)-f (a)]/(b-a)

Equivalent to f (b)-b < f (a)-a constant;

Let h (x) = f (x)-x = lnx+m/x-x (x > 0),

∴h(x) monotonically decreases at (0, +∞);

∫ h ′ (x) =1/x-m/x2-1≤ 0 always holds when (0, +∞),

∴m≥-x^2+x=-(x- 1/2)^2+ 1/4(x>0)，

∴m≥ 1/4；

For m= 1/4, h ′ (x) = 0 only holds when x= 1/2;

The range of ∴m is [1/4, +∞).