It is known that f (x) is an even function defined on R, and an odd function g (x) defined on R passes through the point (- 1,1) If G (x) = f (x-1), then f (2007) + F (2008) =?

Odd function g (x)
g(0)=0,g(1)=-g(-1)=-1
Odd function g (x)
g(-x)=f(-x-1)=-f(x-1)
F (x) even, f (- x-1) = f (x + 1) = - f (x-1)
So f (x) = - f (X-2)
f(x+2)=-f(x+2-2)=-f(x)=f(x-2)
So 4 is the period of F (x)
f(2007)+f(2008)=f(-1)+f(0)=f(0-1)+f(1-1)=g(0)+g(1)=-1

Let f (x) and G (x) be odd and even functions defined on R, respectively

F (x) is an odd function, f (- x) = - f (x)
G (x) is even function, G (- x) = g (x)
Therefore, f (- x) g (- x) = - f (x) g (x)
That is, f (x) g (x) is an odd function
When x < 0, f (x) g (x) is an increasing function
Therefore, when x > 0, f (x) g (x) is also an increasing function
g(-3)=g(3)=0
Therefore, f (- 3) g (- 3) = f (3) g (3) = 0
Therefore, f (x) g (x) < 0 is: F (x) g (x) < f (- 3) g (- 3) or F (x) g (x) < f (3) g (3)
When x < 0, f (x) g (x) is an increasing function
Therefore, the solution set of F (x) g (x) < f (- 3) g (- 3) is x ＜ - 3
When x > 0, f (x) g (x) is also an increasing function
Therefore, the solution set of F (x) g (x) < f (3) g (3) is 0 < x < 3
To sum up, the solution set of the inequality f (x) g (x) < 0 is: x < 3 or 0 < x < 3

Given that f (x) is an odd function defined on R, G (x) is an even function defined on R, and f (x) - G (x) = 1-x ^ 2-x3, find g (x)

If f (x) - G (x) = 1-x ^ 2-x ^ 3f (- x) - G (- x) = 1-x ^ 2 + x ^ 3, the sum of the two formulas gives [f (x) + F (- x)] + [g (x) + G (- x)] = 2-2x ^ 2 because f (x) is an odd function defined on R, and G (x) is an even function defined on R, so f (x) + F (- x) = 0g (x) + G (- x) = 2g (x), so 2g (x) = 2-2x ^ 2, G (x) = 1-x ^

If even function f (x) and odd function g (x) defined on R satisfy f (x) + G (x) = ex, then G (x) = () A. ex-e-x B. 1 2（ex+e-x） C. 1 2（e-x-ex） D. 1 2（ex-e-x）

∵ f (x) is an even function defined on R
∴f（-x）=f（x）
And ∵ g (x) is an odd function defined on R
g（-x）=-g（x）
By F (x) + G (x) = ex,
∴f（-x）+g（-x）=f（x）-g（x）=e-x，
∴g（x）=1
2（ex-e-x）
Therefore, D is selected

If the even function f (x) defined on R is an increasing function on [0, positive infinity), and f (1 / 3) = 0, then the value range of X satisfying f (log (1 / 27) x) > 0 is I'm not very good at math,

That is, log (1 / 27) x > 1 / 3
Log (1 / 27) x is a decreasing function in the domain of definition
When log (1 / 27) x = 1 / 3, x = 1 / 3
That is, x < 1 / 3

If the even function f (x) defined on R is a decreasing function on [- ∞, 0), and f (1 / 2) = 0, then the value range of X for f [log4 (x)] > 0 is satisfied

It's about the same as that one
Even function f (x) defined on R is a decreasing function on (- ∞, 0]
So f (x) is an increasing function on [0, + ∞)
And f (1 / 2) = 0
Drawing pictures shows that
f(x)>0
Launch X1 / 2
So Log1 / 4 (x) 1 / 2
Zero

For an even function defined on R, f (x) is a decreasing function on [0, positive infinity), and f (log (1 / 8) x) > 0, what is the value range of X?

If it is an even function and f (x) is a minus function on [0, positive infinity), then
The maximum value is f (0). What if f (0) is less than 0. Then how can f (log (1 / 8) x) be greater than 0? There is a problem with this problem

F (x) is an even function on R and a decreasing function on [0, + ∞), f (1 / 2) = 0. Solving the inequality f ((Log1 / 4) x)

If a zero point of even function f (x) is 1 / 2 and the interval [0, + ∞) is an increasing function, then f (x) < 0 means - 1 / 2 < X

Let f (x) be an even function defined on R. when x < 0, f (x) XF ′ (x) < 0 and f (- 4) = 0, then the solution set of inequality XF (x) > 0 is

Let g (x) = XF (x), then G '(x) = [XF (x)]' = x'f (x) + XF '(x) = XF' (x) + F (x) < 0,
The function g (x) is a decreasing function on the interval (- ∞, 0),
∵ f (x) is an even function defined on R,
/ / g (x) = XF (x) is an odd function on R,
The function g (x) is a decreasing function on the interval (0, + ∞),
∵f（-4）=0,
∴f（4）=0；
That is, G (4) = 0, G (- 4) = 0
﹥ XF (x) ﹥ 0 becomes g (x) ﹥ 0,
Let x > 0, so the inequality is g (x) > G (4), that is, 0 < x < 4
Let x < 0, so the inequality is g (x) > G (- 4), that is, X ＜ - 4
So the solution set is (- ∞, - 4) ∪ (0,4)

Let f (x) be an even function defined on R. when x > 0, f (x) + XF '(x) > 0, and f (1) = 0, then the solution set of inequality XF (x) > 0 is? Seek detailed analysis

When x > 0, f (x) + XF '(x) > 0
That is [XF (x)] '> 0 [[XF (x)]' = x'f (x) + XF '(x) = f (x) + XF' (x)]
The function XF (x) is an increasing function on (0, + ∞)
∵ f (x) is an even function
/ / XF (x) is an odd function
ν x (f (x) is an increasing function on (- ∞, 0)
∵f(1)=f(-1)=0
The solution set of XF (x) > 0
Is (- 1,0) U (1, + ∞)

It is known that f (x) is an even function defined on R, and an odd function g (x) defined on R passes through the point (- 1,1) If G (x) = f (x-1), then f (2007) + F (2008) =?