The known function f (x) = log (1-x) + loga (x + 3), where 0

The known function f (x) = log (1-x) + loga (x + 3), where 0

When - 3 = 0 or M = 0 is obtained from 1-x > 0 and X + 3 > 0,
In (- 3,1), the inequality - x2 + 2mx-m2 + 2M1 and G (1) > = 0, that is, M > 1 and m ^ 2-4m + 2 > = 0, M > = 2 + √ 2
From M = 0, i.e. M = 0, M = 2 + √ 2 or M = 2 can be obtained

Given the function f (x) = log2 (x + 1), G (x) = log2 (1-x) (1), find the definition field of function f (x) + G (x)

F (x) = log2 (x + 1), G (x) = log2 (1-x), the domain of function f (x) + G (x) is x + 1 > 0; 1-x > 0, the domain of function f (x) + G (x) is {x | - 1

Given the function y = f (x) = loga (1-ax) (a > 0 and a ≠ 1); (1) find the domain of definition and value of F (x); (2) prove that f (x) is a decreasing function in the domain of definition

(1) From 1-ax ＞ 0, we get ax ＜ 1. (1 score) when a ＞ 1, X ＜ 0; (2 score) when 0 ＜ a ＜ 1, X ＞ 0. (3 score) so the definition domain of F (x) is when 0 ＜ a ＜ 1, X ∈ (0, + ∞); when a ＞ 1, X ∈ (- ∞, 0). (4 score) when a ＞ 1, X ＜ 0, {1 ＞ 1-ax ＞ 0, {loga (1-ax) ＜ 0, that is, the value domain of function is (- ∞, 0), Therefore, the range of F (x) is y ∈ (0, + ∞) when 0 ＜ a ＜ 1; y ∈ (- ∞, 0) when a ＞ 1. (2) when 0 ＜ a ＜ 1, take any x1, X2 ∈ (0, + ∞), and x1 ＜ X2, (5 points), then ax1 ＞ AX2, so 1 − ax1 ＜ 1 − AX2. (6 points) because 0 ＜ a ＜ 1, so Loga (1 − ax1) > loga (1 − AX2), that is, f (x1) > F (x2). (8 points) so when 0 ＜ a ＜ 1, f (x) is a decreasing function on (0, + ∞). (9 points) similarly, when a ＞ 1, any x1, X2 ∈ (- ∞, 0), and x1 ＜ X2, (10 points) can be obtained. When a ＞ 1, f (x) is also a decreasing function on (- ∞, 0). (14 points)