It is known that the function f (x) is an even function defined on R, and when x ≤ 0, f (x) = - x2 + X. The expression of F (x) and the analytic expression of F (x) when x > 0 are obtained

It is known that the function f (x) is an even function defined on R, and when x ≤ 0, f (x) = - x2 + X. The expression of F (x) and the analytic expression of F (x) when x > 0 are obtained

If (x) - x = (- x) - x, then (x) - x = - x = - 0

Even function f (x) defined on R satisfies: for any x1, X2 belongs to [0, positive infinity) (x1 is not equal to x2), there is (f (x2) - f (x1)) / (x2-x1)

For any x1, X2 belongs to [0, positive infinity) (x1 is not equal to x2), there is (f (x2) - f (x1)) / (x2-x1)

The function f (x) defined on R is an increasing function on (- ∞, a), and the function y = f (x + a) is an even function. When X1 < A, X2 > A, and | x1-a | x2-a |, Then the relationship between F (2a-x1) and f (x2) is

If y = f (x + a) is an even function, then f (- x + a) = f (x + a)
So f (x) is symmetric about x = a
It is an increasing function on (- ∞, a), so it is a decreasing function on (a, + ∞)
X1 < A, X2 > A, and | x1-a | x2-a|
Remove the absolute value a-x1a
Let f (2a-x1) > F (x2) be a decreasing function on (a, + ∞)

If the domain of F (x), G (x) is r, f (x) is odd function, G (x) is even function, and f (x) + G (x) = 1 / (x ^ 2-x + 1), find the expression of F (x)

f(-x)+g(-x)=1/(x^2+x+1)
Because f (x) is odd and G (x) is even
So f (- x) = - f (x) g (- x) = g (x)
So - f (x) + G (x) = 1 / (x ^ 2 + X + 1) 2
The formula (x + x) is further reduced by (x) + 1 times (f)

If f (x) is an even function and G (x) is an odd function, they have the same domain of definition, and f (x) + G (x) = the last half of X-1, find the expressions of F (x), G (x)

f(x)+g(x)=1/(x-1)…… 1 formula F (- x) = f (x), G (- x) = - G (x) f (- x) + G (- x) = - 1 / (x + 1) = f (x) - G (x) 2 Formula 1 + 2, 2f (x) = 1 / (x-1) - 1 / (x + 1) = 2 / (x ^ 2-1) f (x) = 1 / (x ^ 2-1) 1 / (x + 1) = 2g (x) = 1 / (x-1) + 1 / (x + 1) = 2x / (x ^ 2-x) g (x) = x / (x ^ 2-1)

If f (x) is odd, G (x) is even, and f (x) + G (x) = 1 X − 1, then f (x)=______ .

∵f（x）+g（x）=1
x−1，①
∴f(−x)+g(−x)＝1
−x−1，
∵ f (x) is an odd function, G (x) is an even function,
∴−f(x)+g(x)＝1
−x−1，②
① 2 g (x) = 1
x−1+1
−x−1=2
x2−1，
∴g(x)＝1
x2−1．
∴f(x)＝1
x−1−1
x2−1=x
x2−1．
So the answer is: X
x2−1．

It is known that f (x) is an even function and G (x) is an odd function. There is f (x) + G (x) = 1 / (x + 1). The expressions of F (x) and G (x) are obtained

If f (- x) = f (x) g (- x) = f (x) g (- x) = - G (x) f (x) f (x) + G (x) = 1 / (x + 1) ① take x as the - XF (- x) + G (- x) = 1 / (1-x) f (x) f (x) - G (x) = 1 / (1-x) f (x) - G (x) = 1 / (1-x) ② ① + ② get 2F (x) = 1 / (x + 1) + 1 / (1-x) = 2 / (1-x ^ 2) f (x) = 1 / (1-x ^ 2) f (x) = 1 / (1-x ^ 2) f (x) = 1 / (1-x ^ 2) 1 (1-x ^ 2) 1 (1-x ^ 2) ① - ② get 2g (x) = 1 / (x + 1) - 1 / (x 1 / (1-x) = - 2x / (1

Let f (x) and G (x) be odd functions with the same definition domain. How about (1) is the function f (x) = f (x) + G (x) odd or even? Why? (2)

Let f (x), G (x) be odd functions with the same domain
Then f (- x) = - f (x)
g(-x)=-g(x)
Then f (- x) = f (- x) + G (- x)
=-f(x)-g(x)
=-(f(x)+g(x))
=-F (x) (because f (x) = f (x) + G (x))
So it's an odd function

If the function f (x) and G (x) are even functions with the same domain, is f (x) = f (x) + G (x) even? Odd? Why

F(-x)=f(-x)+g(-x)=f(x)+g(x)=F(x)
So the addition is even

Given that f (x) is an even function defined on R, G (x) is an odd function defined on R, and G (x) = f (x-1), then the value of F (2013) + F (2015) is () A. -1 B. 1 C. 0 D. Unable to calculate

∵ f (- x-1) = g (- x) = - G (x) = - f (x-1), and f (x) is even function
/ / F (x + 1) = f [- (x + 1)] = f (- x-1), so f (x + 1) = - f (x-1)
∴f（x+1）+f（x-1）=0．
∴f（2013）+f（2015）=f（2014-1）+f（2014+1）=0，
Therefore, C