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It is known that the function f (x) is an even function defined on the interval [- 2,2]. When x ∈ [0,2], f (x) is a decreasing function. If the inequality f (1-m) < f (m) holds, then the value range of real number m is () A. [−1，1 2) B. [1，2] C. [0，1 2） D. （−1，1 2）
Even function f (x) is a decreasing function on [0,2],
It is an increasing function on (- 2, 0). It can be concluded that the smaller the absolute value of the independent variable is, the larger the function value is
The inequality f (1-m) < f (m) can be changed into
|1−m|＞|m|
−2≤m≤2
−2≤1−m≤2
M ∈ [- 1,1) is obtained
2）
Therefore, a
Given that f (x) is an even function defined on R, if G (x) passes through point (1,3) and FG (x) = f (x-1), then f (2007) + F (2008) Is f (x-1) = g (x) = f (- x-1) f (x) not even? Why can't it change like this on the left
If f (x) is an even function defined on R, then f (x) = f (- x), so f (x-1) = f (- x + 1);
If f (x-1) is an even function on R, then f (x-1) = f (- x-1), which is different from the above
Let f (x) and G (x) be defined by X ∈ R and X ≠± 1, f (x) be even function, G (x) be odd function, and f (x) + G (x) = 1 Find the analytic expressions of F (x) and G (x)
∵ f (x) is an even function, G (x) is an odd function, ᙽ f (- x) = f (x), and G (- x) = - G (x) is obtained by F (x) + G (x) = 1 x − 1, that is, f (− x) + G (− x) = 1 − x − 1
It is known that f (x) is an even function and G (x) is an odd function. Their definition domain is {x ∈ R and X ≠ ± 1}. If f (x) + G (x) = 1 / (x-1), then f (x) =? G (x) =? G (x) =?
f(-x)+g(-x)=1/(-x-1)
therefore
f(x)-g(x)=1/(-x-1)
f(x)+g(x）=1/（x-1)
From the above two formulas, we can find f (x), G (x)
Given that f (x) is an even function on R, G (x) is an odd function on R, and G (x) = f (x-1), and G (- 1) = 2, then f (2008) =? 2?
f（2008）=f(2004) =.=f(0)=f(1-1)=g(1)=-g(1-)=-2
Given that f (x-1) is odd function, f (x + 1) is even function, f (2008) = 1, then f (4)=______ .
∵ f (x-1) is an odd function and f (x + 1) is an even function
﹤ f (- x-1) = - f (x-1), f (- x + 1) = f (x + 1) is always true
In F (- x-1) = - f (x-1), if t = X-1, then x = t + 1, so there is f (- T-2) = - f (T) ①
In F (- x + 1) = f (x + 1), if t = x + 1, then x = T-1, so there is f (T) = f (- t + 2) ②
From (1) and (2) - f (- T-2) = f (- t + 2) 3,
If M = - t + 2, then t = - M + 2, then f (m) = - f (M-4) = f (M-8) can be obtained by substituting (3), so that the period of the function is 8
2008 = 251 × 8
Therefore, f (2008) = f (0) = 1
F (4) = - f (0) = - 1
So the answer is - 1
If the odd function f (x) defined on R satisfies f (x + 1) + F (x-3) = 0, then f (2008)=
Odd function f (x) defined on R
f(0)=0
t=x-3,x=t+3
f(x+1)+f(x-3)=0
f(t+4)+f(t)=0
f(t+4)=-f(t)
f(2008)=-f(2004)=(-1)^2*f(2000)=(-1)^(2008/4)*f(0)=0
Given that the function f (x) defined on R is odd, and the period of function f (3x + 1) is 3 and f (1) = 5, then the value of F (2007) + F (2008) is () A. 0 B. 5 C. 2 D. -5
∵ the period of the function f (3x + 1) is 3, f [3 (x + 3) + 1] = f (3x + 1), that is, f [(3x + 1) + 9] = f (3x + 1), f (x) is a function with period 9; and f (x) is an odd function on R,
Given that f (x) is an even function defined on R, if the odd function g (x) defined on R passes through the point (- 1,3) and G (x) = f (x-1), then f (2009) + F (2010)=______ .
∵ the function f (x) is an even function defined on R, so f (- x) = f (x), the odd function g (x) defined on R, and G (x) = f (x-1), so there is f (x-1) = - f (- x-1) = - f (x + 1) = f (x + 3), so t = 4, the odd function g (x) defined on R passes through the point (- 1,3), ∵ g (- 1) = 3, G (...)
It is known that the function f (x) whose domain is R is a decreasing function on (2010, + ∞) and the function y = f (x + 2010) is even function f(2008)>f(2009) f(2008)>f(2011) f(2009)>f(2011) f(2009)>f(2012)
If (- 2010, + ∞) is a minus function, or y = f (x-2010) is an even function
RELATED INFORMATIONS
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