Given the function f (x) = alnx - (1 + a) x + 0.5x ^ 2, (a belongs to R), if f (x) > = 0, it holds for any X in the domain of definition, and the value range of a is obtained

Given the function f (x) = alnx - (1 + a) x + 0.5x ^ 2, (a belongs to R), if f (x) > = 0, it holds for any X in the domain of definition, and the value range of a is obtained


F (x) = alnx - (1 + a) x + 0.5x ^ 2 > = 0 holds on x > 0
A (lnx-x) > = x-0.5x * x is obtained
And lnx-x0 (x = 2)
When x



If the definition field of function = 1 / LG (X & # 178; + BX + 2b) is r, then the value range of B is?





Given that the function f (x) = x2 + 2 (A-1) x + 2 is an increasing function on [4, + ∞), then the value range of real number a is______ .


∵ f (x) = x2 + 2 (A-1) x + 2 is an increasing function on [4, + ∞], and the symmetry axis 1-A ≤ 4 is a ≥ - 3, so the answer is: [- 3, + ∞)



Given that the function y = x ^ 2 - 2x + 3 has a maximum value of 3 and a minimum value of 2 in the closed interval [0, M], what is the range of M?
The answer is [1.2], why not (0.2), and why can't I choose between 0 and 1


First, y = x & # 178; - 2x + 3 is sorted into y = (x-1) &# 178; + 2,
We can see that y = (x-1) ² + 2, when x is a real number, its minimum value is 2, and its maximum value is infinite,
Because y = (x-1) &# 178; + 2, its symmetry axis is x = 1, and the minimum value is obtained if and only if x = 1
When x = 0 or x = 2, the value of the function is 3,
So when x is in the interval [0,2], the minimum value of function y is 2 and the maximum value is 3
So the value of M is [1.2]
(if M