Find the domain of the function y = 2 ^ X-1 / 2 ^ x + 1, judge the monotonicity of the function and prove it by definition

Find the domain of the function y = 2 ^ X-1 / 2 ^ x + 1, judge the monotonicity of the function and prove it by definition

The domain is r
y=(2^x-1)/(2^x+1)
=[(2^x+1)-2]/(2^x+1)
=1-2/(2^x+1)
It's incremental
Certificate: Ling x1

Let f (x) be a function over the domain R. for any x, y belongs to R. f (x + y) = f (x) times f (y). When x is greater than 0, f (x) is greater than 0 and less than 1
It is proved that f (0) = 1, and when x is less than 0, f (x) is greater than 1
It is proved that f (x) decreases monotonically on R

Let x = y = 0, f (0) = f (0) times f (0), f (0) = 0 or 1,
If x > 0 and y = 0, f (x) = f (x) times f (0),
If f (0) = 0, then f (x) = 0, which is contradictory to f (x) greater than 0 when x is greater than 0, so f (0) = 1,
If x > 0, y = - x1,
Pay attention to this

Let f (x) be an odd function whose domain is r, and G (x) be a function whose domain is R and is constant greater than 0. When x > 0, f '(x) * g (x) < f (x) g'（

Let y = f (x) g (x)

(2007 · Xian'an District simulation) Let f (x) be an odd function with the domain of R, and G (x) be a function with the domain of R, and if x > 0, f '(x) g (x) < f (x) g' (x). If f (1) = 0, then the solution set of inequality f (x) > 0 is ()
A. （-∞，-1）∪（1，+∞）B. （-1，0）∪（0，1）C. （-∞，-1）∪（0，1）D. （-1，0）∪（1，+∞）

First of all, because g (x) is a function whose domain of definition is r, the solution set of F (x) > 0 is equivalent to the solution set of F (x) g (x) > 0. Next, we focus on the functional properties of F (x) g (x). Because when x > 0, f '(x) g (x) < f (x) g' (x), when x > 0, (f (x) g (x)) / < 0, that is, f (x) g (x), when x > 0, is decreasing. From F (1) = 0, f (1) g (1) = 0 If f (x) is an odd function, f (- 1) = 0, X ＜ - 1, f (x) g (x) = − f (− x) g (x) ＞ 0, the solution set of F (x) ＞ 0 is (- ∞, - 1) ∪ (0,1), so C

Given the function f (x) = √ 3sin2x + 2cos ^ 2x + m, its definition field is [0, Pai / 2], and the maximum value is 6,
1、 Find the value of constant M; 2. Find the monotone increasing interval of function f (x); note: ^ 2 means square, pie is the symbol of PI

1 f(x)=√3sin2x+cos2x+1+m
=2sin(2x+π/6)+1+m
0=

If the function y = (f) x is [0,1], given 0

∵ f (x) is defined as [0,1]
In F (x) = f (x + a) + F (2x + a)
{ 0≤x+a≤1
{0≤2x+a≤1
==>
{-a≤x≤1-a
{-a/2≤x≤(1-a)/2
(∵0

When x is less than 0, f (x) = 1 + 2x, and when x is greater than 0, f (x)?

f(x)=1-2x

Let f (x) be a periodic function with period 2, which is defined as
When - 1

It converges to [f (- 1) + F (1)] / 2 = 1 / 2

Given that the function f (x) satisfies f (x + y) + F (X-Y) = 2F (x) · f (y) & nbsp; (x ∈ R, y ∈ R), and f (0) ≠ 0, we try to prove that f (x) is an even function

It is proved that: let x = y = 0 ∵ f (x + y) + F (X-Y) = 2F (x) · f (y) ∵ f (0) ∵ f (0) ≠ 0, ∵ f (0) = 1, let x = 0 ∵ f (x + y) + F (X-Y) = 2F (x) · f (y) ∵ f (y) + F (- y) = 2F (0) · f (y) ∵ f (- y) = f (y), that is, f (x) is even function

Let f [x] be an even function, G [x] be an odd function, the domain of definition is x not equal to ± 1, and f [x] + G [x] = 1 / [X-1] find f [x] and G [x]

Substituting - x for X f [- x] + G [- x] = 1 / [- X-1] is equal to the equation f [x] - G [x] = 1 / [- X-1]
The equation f [x] + G [x] = 1 / [X-1]
f[x]-g[x]=1/[-x－1]
F (x) can be obtained by adding two formulas
We can find g (x) by subtraction
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