The function f (x) = ln (1 + x), when x > 0, the inequality f (x) > KX / K + X (K ≥ 0) holds, and the value range of real number k is obtained

The function f (x) = ln (1 + x), when x > 0, the inequality f (x) > KX / K + X (K ≥ 0) holds, and the value range of real number k is obtained

Let g (x) = f (x) - KX / (K + x)
→g(x)＝ln(1+x)-kx/(k+x).
Obviously, G (0) = 0
F (x) > KX / (x + k) is constant when x > 0 and K ≥ 0,
Then G (x) > G (0), that is, G (x) increases when x > 0
∴g'(x)＝1/(1+x)-k^2/(k+x)>0
In this paper, the author analyzes the relationship between X (1 + 2k-k ^ 2) > 0 and X (1 + 2k-k ^ 2) > 0,
1 + 2k-k ^ 2 > 0 and K ≥ 0,
The solution is 0 ≤ K

Given a ∈ R, function f (x) = (- x2 + ax) E-X (x ∈ R, e is the base of natural logarithm). (I) when a = - 2, find the monotone decreasing interval of function f (x); (II) if function f (x) monotone decreasing in (- 1,1), find the value range of a; (III) if function f (x) is monotone function on R, if not, find the value range of a: if not, please explain the reason

(I) when a = - 2, f (x) = (- x2-2x) E-X; F ′ (x) = (x2-2) E-X, Let f ′ (x) ＜ 0, then the monotone decreasing interval of x2-2 ＜ 0, х - 2 ＜ x ＜ 2 х f (x) is (- 2,2); (II) f ′ (x) = [X2 - (a + 2) x + a] E-X, if f (x) decreases monotonically in (- 1,1), that is, when - 1 ＜ x ＜ 1, f ′ (x) ≤ 0, that is, X2 - (a + 2) x + a ≤ 0 is constant to X ∈ (- 1,1) Let g (x) = X2 - (a + 2) x + A, then G (- 1) ≤ 0g (1) ≤ 0  1 + (a + 2) + a ≤ 01 - (a + 2) + a ≤ 0, the solution is a ≤ - 32; (III) f '(x) = [X2 - (a + 2) x + a] E-X, its positive and negative depends on the quadratic formula X2 - (a + 2) x + A, the value of the quadratic formula (the first term is positive) can not always be negative, that is to say, the original function can not be a monotone decreasing function on the whole real number field; if it wants to be monotone increasing function, it can be a monotone increasing function Function, then X2 - (a + 2) x + a ≥ 0 holds for X ∈ R ∵ △ = (a + 2) 2-4a = A2 + 4 ＞ 0 ∵ function can not monotonically increase on R. in conclusion, function f (x) can not be monotone function on R

Given the function f (x) = ex-x (E is the base of natural logarithm)
(1) Finding the minimum of F (x)
(2) Let the solution set of inequality f (x) > ax be p, and "x l0 ＜ equal to X ＜ equal to 2" be included in P, and find the value range of real number a

(1) F (x) = e ^ x-xf '(x) = e ^ X-1 = 0x = 0f' '(x) = e ^ x E ^ 0 = 1 > 0, so when x = 0, the minimum value is e ^ 0-0 = 1 (2) f (x) - ax > 0, then G (x) = f (x) - AX = e ^ x-x-axg' (x) = e ^ x-1-ai when 1 + a > 0, that is, when a > - 1, e ^ x-1-a = 0, x = ln (1 + a) when x > ln (1 + a), then G (x) = f (x) - AX = e ^ x-x-axg '(x) = e ^ x-1-a

Given the function f (x) = ex KX, X belongs to R (E is the base of natural logarithm)
If k = e, find the extremum of the function. (2) if K belongs to R, find the monotone interval of the function

f(x)=e^x-kx
one
k=e
f(x)=e^x-ex
f'(x)=e^x-e=0,x=1
When x > 1, f '(x) > 0, f (x) rises monotonically
When xlnk, f '(x) > 0, f (x) rises monotonically
When x