Given the function f (x) = 2 / x + alnx, a ∈ R (1) If a = 4, find the monotone interval of function f (x) (2) If the function f (x) increases monotonically on [1, + 00], find the value range of real number a

Given the function f (x) = 2 / x + alnx, a ∈ R (1) If a = 4, find the monotone interval of function f (x) (2) If the function f (x) increases monotonically on [1, + 00], find the value range of real number a

1）a=4,f(x)=2/x+4lnx
f'(x)=-2/x^2+4/x=2(2x-1)/x^2
The minimum point is x = 1 / 2
When 0 = 0
That is AX-2 > = 0, a > = 2 / X
The maximum value of 2 / X is 2 / x = 2 when x = 1
So there is a > = 2

The known function f (x) = x ^ 2 + 2x + alnx
(1) When a = - 4, find the extremum of function f (x)
(2) The monotone increasing interval of function f (x) is discussed
(3) When t ≥ 1, the inequality f (2t-1) ≥ 2F (T) - 3 tries to find the value range of A

1) F '(x) = 2x + 2 + A / x = (2x ^ 2 + 2x + a) / x = (2x ^ 2 + 2X-4) / x = 0 (x > 0) x ^ 2 + X-2 = 0 (x-1) (x + 2) = 0, so there is a unique stationary point x = 1, the left derivative is less than 0, and the right derivative is greater than 0. That is to say, take the minimum value f (1) = 3.2) f' (x) = 2x + 2 + A / x = (2x ^ 2 + 2x + a) / X. because x > 0, the sign of F '(x) is changed from quadratic

Given the function f (x) = X-1 / x, G (x) = alnx (a ∈ R)
H (x) = f (x) + G (x), and H (x) has two extreme points X1 X2, where x1 ∈ (0,1 / 2], find the minimum value of H (x1) - H (x2)

H (x) = f (x) + G (x) = X-1 / x + alnx (x > 0) H '(x) = 1 + 1 / x ^ 2 + A / x = (x ^ 2 + ax + 1) / x ^ 2H (x) has two extreme points, let H' (x) = 0, that is, x ^ 2 + ax + 1 = 0, then the equation has two unequal positive roots, x1, X2 and x1 ∈ (0,1 / 2], then {Δ = a ^ 2-4 > 0 {X1 + x2 = - a > 0 {x1x2 = 1} a