Let f (x) = xlnx, X ∈ [e ^ - 2, e], then the maximum value of F (x) is

Let f (x) = xlnx, X ∈ [e ^ - 2, e], then the maximum value of F (x) is

f(x)=xlnx,x∈[e^-2,e]
f'(x)=lnx+x*1/x=1+lnx
Let f (x) = 0, that is, 1 + LNX = 0
The solution is x = e ^ (- 1)
So when x ∈ [e ^ (- 2), e ^ (- 1)], f '(x)

Given that f (x) = xlnx, if f (x) > = - x ^ 2 + ax-6 is constant on (0, positive infinity), the value range of real number a is

F (x) > = - x ^ 2 + ax-6 is constant on (0, positive infinity),
That is, xlnx ≥ - x ^ 2 + ax-6 is constant on (0, positive infinity),
The results show that ax ≤ xlnx + X & # 178; + 6 is constant on (0, positive infinity),
The results show that a ≤ LNX + X + 6 / X is constant on (0, positive infinity),
Constructor g (x) = LNX + X + 6 / X
The minimum value of a ≤ g (x)
g'(x)=1/x+1-6/x²=(x²+x-6)/x²=(x-2)(x+3)/x²
When x > 2, G '(x) > 0, G (x) is an increasing function
0

Given the function f (x) = e ^ x + ax, G (x) = e ^ xlnx. (2), if f (x) > 0 holds for any real genus x ≥ 0, the value range of a is obtained
(3) When a = - 1, is there a real number XO ∈ [1, e] such that the tangent of the curve C: y = g (x) - f (x) at the point x = x0 is perpendicular to the y-axis? If it exists, calculate the value of x0; if not, explain the reason

(2) If any real number x ≥ 0, f (x) > 0 is constant
f'(x)=e^x+a
a> When f (x) = e ^ x + a > 0, f (x) increases
So f (x) > = f (0) = 1 > 0 holds
a=1
So there won't be y '(x) = 0
That is to say, there is no x0 in line with the meaning of the question

If xlnx ≥ ax-a holds for all x ∈ [e, + ∞), then the value range of real number a is?

Given: X ∈ [e, + ∞), then: X-1 > 0
If xlnx ≥ ax-a, then A0 is required (considering the range of x), u = x - (LNX) - 1 > 0
So y '= [x - (LNX) - 1] / (x-1) ^ 2 > 0, the function y is monotone increasing function in its domain
Substituting the minimum value of x = e into y, the minimum value of X is equal to E / (E-1)
So: when a