The known function f (x) = x & # 178; + e ^ X-1 / 2 (x)

The known function f (x) = x & # 178; + e ^ X-1 / 2 (x)

∵ f (x) and G (x) are not even functions. The symmetric point of ∵ f (x) on Y-axis can not be on its own image, but can only be on another function image. The definition domain of ∵ f (x) is X & lt; 0. The symmetric point of ∵ G (x) image can only exist when X & gt; 0

The known function f (x) = 1 / 2x & # 178; - (a + 1) x + a ln x + 4 (a > 0)
(1) Finding monotone decreasing interval of function f (x)
(2) When a = 2, the function y = f (x) has zeros in [e ^ n, + ∞) (n ∈ z), and the maximum value of n is obtained
1 / 2 is the coefficient of X & #

The domain of F (x) is x ∈ (0, + ∞)
F & # 180; (x) = x - (a + 1) + A / x, let F & # 180; (x) = 0, that is, X & # 178; - (a + 1) x + a = 0, that is, (x-a) (x-1) = 0
Question 1
(1) When a = 1, the two stationary points coincide,
F & # 180; (x) = X-2 + 1 / x = (√ X-1 / √ x) &# 178; ≥ 0 (if and only if x = 1, "=" holds),
So the function increases monotonically in the whole domain;
(2) When a < 1, F & # 180; (x) = x - (a + 1) + A / x = (x-a) (x-1) / x,
F (x) increases monotonically in (0, a) ∪ (1, + ∞) and decreases monotonically in (a, 1);
(3) When a ＞ 1, F & # 180; (x) = x - (a + 1) + A / x = (x-a) (x-1) / x,
F (x) increases monotonically in (0,1) ∪ (a, + ∞) and decreases monotonically in (1, a);
Question 2
When a = 2, y = f (x) = 1 / 2x & # 178; - 3x + 2lnx + 4, F & # 180; (x) = (X-2) (x-1) / X
It is known from (3) of 1 that f (x) increases monotonically in (0,1) ∪ (2, + ∞) and decreases monotonically in (1,2);
The maximum f (1) = 3 / 2, the minimum f (2) = 2ln2, so f (x) has a unique zero point in (0,1),
So e ^ n < 1, n < ln1 = 0, so the maximum value of n is - 1