Even function f (x) defined on R satisfies f (x + 1) = - f (x) when x ∈ [- 1,0], f (x) = (1 / 2) is x power, then f (log28) is equal to () A.3 B.1/8 C.-2 D.2

Even function f (x) defined on R satisfies f (x + 1) = - f (x) when x ∈ [- 1,0], f (x) = (1 / 2) is x power, then f (log28) is equal to () A.3 B.1/8 C.-2 D.2

By replacing x with · x + 1, f (x + 2) = f (x), f (x) is a periodic function, t = 2, log 28 = 3, f (3) = f (1) = f (- 1) = 2

Even function f (x) defined on R satisfies f (x + 1) = - f (x), and when x ∈ [- 1,0], f (x) = (1) 2) Then f (log28) is equal to () A. 3 B. 1 Eight C. -2 D. 2

Let f (x + 1) = - f (x), let x = x + 1
f（x+2）=-f（x+1）
f（x+2）=-（-f（x））=f（x），
Then the function f (x) is a periodic function with period 2,
∴f（log28）=f（3log22）=f（3）=f（3-4）=f（-1）．
When x ∈ [- 1,0], f (x) = (1)
2)x，
∴f（log28）=f（-1）=(1
2)−1＝2．
Therefore, D

If y = f (x) is an even function defined on R and satisfies f (x) = - f (3 / 2), f (- 1) = 1, f (0) = - 2, then the value of F (1) + F (2) +... + F (2008) is

If f (x) is an even function, then it is axisymmetric about the y-axis, that is, f (x) = f (- x) and f (x) = - f (x + 3 / 2) = f (x + 3)
Therefore, f (1) = f (- 1) = 1
f(2)=-f(1/2+3/2)=f(1/2)=-f(-1+3/2)=f(-1)=1
f(3)=f(0)=-2
That is, the original formula = [f (1) +... " +f(2007)]+f(2008)=1

F (x) is defined in (- ∞, + ∞). Why is y = f (x ^ 2) even Ask for instruction~

Even function must be f (- x) = f (x)
And y = f (x 2) = f [(- x) 2] is even function

Even function f defined on R satisfies f = F If x ∈ (0,1), f = 2 ^ (x) - 1, then f is equal to A 1/4 B 1/2 C 5/8 D 1 F = 2 to the power of X-1

Because f = f, f (x) is a periodic function because log2 (8 / 3) = log2 (8) - log2 (3) = 3-log2 (3) ≈ 1.415, so f (log2 (8 / 3)) = f (- 1-log2 (3)) because f (x) is an even function, so = f (1 + log2 (3)) = f (log2 (3) - 1) = f (log2 (3 / 2)) = 2 ^ (log2 (3 / 2)) - 1 = 3 / 2-1 = 1 / 2 select B

Let the even function f (x) defined on R satisfy f (x + 1) + F (x) = 1, and when x belongs to [1,2], f (x) = 2-x, then f (- 2 Let the even function f (x) defined on R satisfy f (x + 1) + F (x) = 1, and when x belongs to [1,2], f (x) = 2-x, then f (- 2004.5) =? Must have a process Don't answer if you don't understand!

From F (x + 1) + F (x) = 1, f (x + 1) = 1-f (x) = 1 - [1-f (x-1)] = f (x-1), if f (x + 2) = f (x), then the period of the function is 2, so f (x + 2004) = f (x), because it is even function, f (- x-2004) = f (- x), x = 0.5, f (- 0.5) = f (2-0.5) = f (1.5) = 2-1.5 = 0.5

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 even, G (x) is odd, ᙽ f (- x) = f (x), and G (- x) = - G (x)
By F (x) + G (x) = 1
x−1             ①
F (− x) + G (− x) = 1
−x−1，
That is, f (x) - G (x) = 1
−x−1＝−1
x+1   ②
It is found that f (x) = 1
x2−1，g(x)＝x
x2−1．

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 even, G (x) is odd, ᙽ f (- x) = f (x), and G (- x) = - G (x)
By F (x) + G (x) = 1
x−1             ①
F (− x) + G (− x) = 1
−x−1，
That is, f (x) - G (x) = 1
−x−1＝−1
x+1   ②
It is found that f (x) = 1
x2−1，g(x)＝x
x2−1．

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 even, G (x) is odd, ᙽ f (- x) = f (x), and G (- x) = - G (x)
By F (x) + G (x) = 1
x−1             ①
F (− x) + G (− x) = 1
−x−1，
That is, f (x) - G (x) = 1
−x−1＝−1
x+1   ②
It is found that f (x) = 1
x2−1，g(x)＝x
x2−1．

It is known that f (x) is even function and G (x) is odd function. The domain of definition is {XLX ∈ R and X ≠ ± 1}, and f (x) + G (x) = 1 / 1 of X-1. Find f (x), G (x) Same as above

According to the known conditions, there are f (x) = f (- x), G (- x) = - G (x);
So for f (x) + G (x) = 1 / (x-1) (equation 1), there are
F (- x) + G (- x) = 1 / (- x-1)
F (x) - G (x) = - 1 / (x + 1) (formula 2)
By solving the equation composed of equation 1 and equation 2, we can get
f(x)=1/(x^2-1),g(x)=x/(x^2-1）
Where x ^ 2 is the square of X