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It is known that even function f (x) defined on R satisfies f (x + 1) = f (1-x). It is proved that f (x) is a periodic function
It is proved that if t = X-1, then there is
f(t+1)=f(1-t)
That is, f (x) = f (2-x)
And f (x) is even function
∴f(2-x)=f(x-2)
∴f(x)=f(x-2)
Let a = x + 2
f(a)=f(a-2)
That is, f (x) = f (x + 2)
ν f (x) is a periodic function with 2 as the minimum period
If the function f (x) whose domain is R is a decreasing function on (8, + 00) and f (x + 8) is even, then a.f (6) > F (7) B.F (6) > F (9) c.f It is known that the function f (x) whose domain is R is a decreasing function on (8, + 00) and the function f (x + 8) is an even function A.f(6)>f(7) B.f(6)>f(9) c.f(7)>f(9) d.f(7)>f(10)
Let g (x) = f (x + 8) ∵ f (x) be a decreasing function on (8, + ∞)
F (x) is an odd function defined on R, G (x) is an even function defined on R, G (x) = f (x-1), find f (2009) + F (2010) It's f (2009) + F (2011) (sorry)
G (x) = f (x-1), G (x) is an even function defined on R
G (- x) = f (- x-1) = f (x-1), f (x-1) = f [- 2 - (x-1)] that is, f (x) = f (- 2-x)
F (x) is an odd function defined on R,
f(x)=f(-2-x)=-f(x+2)=-f[-2-(x+2)]=-f(-4-x)=f(x+4) T=4
F (2009) + F (2011) = f (1) + F (3) = f (- 3) + F (3) = 0 when f (x) = f (- 2-x)
If the even function f (x) defined on R satisfies that f (x-1) is an odd function, then f (2009) = () A. 0 B. 2008 C. 2009 D. -2008
Because f (x-1) is an odd function, we get f (- x-1) = - f (x-1), and f (x) is even function, so f (- x-1) = - f (x-1) = f (x + 1),
That is, f (x + 2) = - f (x), so f (x + 4) = f (x), that is, the period of the function is 4
So f (2009) = f (2008 + 1) = f (1), when x = - 1. F (- 1 + 2) = - f (- 1), that is, f (1) = - f (1), so f (1) = 0
So f (2009) = 0,
Therefore, a
Given that a function is even on R, there is f (x + 2) = f (x) for x > = 0, and if x ∈ [0,2), f (x) = log2 (x + 1), then f (- 2008) + F (2009) =?
If f (x) is even function, then f (- 2008) = f (2008) f (x + 2) = f (x), indicating that the function is a periodic function with period 2. Then f (2008) = f (2006) =. = f (0) and f (2009) = f (1), then f (- 2008) + F (2009) = f (0) + F (1) and X belongs to [0,2], f (x) = log2 (x + 1) f (0) = log2 (0 + 1) = log2 (1)
It is known that f (x) is an even function of R, f (x-1) = f (- x-1). If f (0) = 2, then the values of F (2008) and f (2009) are obtained Can't count it out! That's it
According to f (x-1) = f (- x-1)
F (x-1) = f (- (x + 1)) = f (x + 1)
So the period of F (x) is t = 2
So f (2008) = f (0) = 2
F (2009) = f (1), it doesn't seem to work out
It is known that the function f (x) = log2 (4 ^ x + 1) + KX, (K ∈ R) is an even function to find the value of K
Are you from Qingzhong
(1) ∵ the function f (x) = log2 (4x + 1) + KX (K ∈ R) is an even function
﹤ f (- x) = log2 (4-x + 1) - KX = f (x) = log2 (4x + 1) + KX is always true
In other words, log2 (4x + 1) - 2x KX = log2 (4x + 1) + KX holds
K = - 1
（2）∵a＞0
The definition domain of the function g (x) = log2 (a * 2x-43a) is (log243, + ∞)
That is 2x > 43
The graph of function f (x) and G (x) has and only one intersection point,
The equation log2 (4x + 1) - x = log2 (a * 2x-43a) has and only one solution in (log243, + ∞)
That is, the equation 4x + 12x = a * 2x-43a has only one solution on (log243, + ∞)
Let 2x = t, then t > 43, which is equivalent to that the equation (A-1) t2-43at-1 = 0 (*) has only one solution on (43, + ∞)
When a = 1, t = - 34 ∉ (43, + ∞) is obtained;
When 0 < a < 1, let H (T) = (A-1) t2-43at-1, and its symmetry axis t = 2A3 (A-1) < 0
The function H (T) = (A-1) t2-43at-1 decreases on (0, + ∞), while H (0) = - 1
The equation (*) has no solution at (43, + ∞)
When a > 1, let H (T) = (A-1) t2-43at-1, and its symmetry axis t = 2A3 (A-1) > 0
Therefore, if h (43) < 0, i.e. 169 (A-1) - 169a-1 < 0, this constant holds
The range of a is a > 1
To sum up, the value range of a is a > 1
Even function f (x) defined on R satisfies f (x + 1) = f (1-x). If f (x) = 2 ^ x when 0 ≤ x < 1, then f (log2 (6)) =?
f(log2（6））=f(log2(2*3))=f(1+log2(3))=f(1-log2(3))
=f(log2(3)-1)=2^(log2(3)-1)=3/2
It is known that f (x) is an even function defined on R and satisfies f (x) = - f (x + (2 / 3)), f (- 1) = 1, f (0) = - 2. Find f (1) + F (2) + F (3). + F (2008)
The answer is 0
If the even function f (x) defined on R satisfies f (x) = - 1 / F (x + 3) and f [4) = 2, then f (2008) =? Why is f (2008) = f (4),
f(x)=-1/f(x+3)
f(x+3)=-1/f(x+6)
f(x)=f(x+6)
Therefore, f (x) = f (x + 334x6)
f(4)=f(2008)=-2
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