Given f (x6) = log2 (x), what is f (8) equal to? PS: the above 6 is the sixth power, 2 is the base, and the following x is the real number. It's best to write down the process, quickly! --- there is another question: what is log √ 2 - 1 (3 + 2 √ 2) equal to? PS: - 1 is not in the root sign

Given f (x6) = log2 (x), what is f (8) equal to? PS: the above 6 is the sixth power, 2 is the base, and the following x is the real number. It's best to write down the process, quickly! --- there is another question: what is log √ 2 - 1 (3 + 2 √ 2) equal to? PS: - 1 is not in the root sign

f(8)=f(√2^6)=log(2)√2=1/2
log(√2-1)(3+2√2)=log(√2-1)(√2+1)^2
=-2
（√2+1=1/(√2-1)）

The function f (x) defined on R satisfies f (x)= log2(1−x)，x≤0 If f (x − 1) − f (x − 2), x > 0, the value of F (2013) is () A. -1 B. 0 C. 1 D. 2

When x > 0, f (x) = f (x-1) - f (X-2), then f (x + 1) = f (x) - f (x-1), then f (x + 1) = - f (X-2), f (x + 3) = - f (x), so f (x + 6) = f (x)

If the definition domain of y = f (x) is R and X is not equal to 0, and for any x, there is f (- x) = - f (x). When x belongs to 0 to positive infinity, f (x) = X-1 If the definition domain of y = f (x) is R and X is not equal to 0, and for any x, there is f (- x) = - f (x). When x belongs to 0 to positive infinity, f (x) = X-1 (1) find the analytic formula of F (x) when x belongs to negative infinity to 0. (2) the solution inequality f (x-1) a holds, and find the value range of A

If the definition domain of y = f (x) is R and X is not equal to 0, and for any x, there is f (- x) = - f (x). When x > = 0, f (x) = X-1 (1). If x belongs to negative infinity to 0, the analytic formula of F (x) is obtained. (2) the inequality f (x-1) a holds. When x > = 0, f (x) = X-1, then f (x-1) = X-2 > 0 -- > x > 2 when x is

F (x) is an even function. When x is greater than or equal to 0, f (x + 2) = f (x) when x [0,2) f (x) = log2 ^ (x + 1), then f (- 2010) + F (2011)=

F (x + 2) = f (x), f (x) is even function
∴ f(-2010)=f(2010)=f(0)=log2(0+1)=0
f(2011)=f(1)=log2(1+1)=log2(2)=1
∴ f(-2010)+f(2011）=0+1=1

The function f (x) defined on R satisfies f (x) = log2(1−x)，x≤0 If f (x − 1) − f (x − 2), x > 0, then the value of F (2012) is______ .

Because the function f (x) defined on R satisfies f (x) =
log2(1−x)，x≤0
f(x−1)−f(x−2)，x＞0 ，
So f (- 1) = 1, f (0) = 0, f (1) = f (2) = - 1, f (3) = 0, f (4) = f (5) = 1, f (6) = 0,
When k ∈ Z, f (1 + 6K) = f (2 + 6K) = - 1, f (3 + 6K) = 0, f (4 + 6K) = f (5 + 6K) = 1, f (6K) = 0,
f（2012）=f（6×335+2）=-1．
So the answer is: - 1

Given that even function f (x) satisfies f (2 + x) = f (2-x)) for any x ∈ R, and f (x) = log2 (1-x) when - 2 ≤ x ≤ 0, what is the value of F (2011)?

F (2 + x) = f (2-x), so the period of the function is 2, so f (2011) = f (- 1) = log2 (2) = 1

If the definition domain of even function f (x) is [P, q], then p + Q=

The sum of 0 even functions with respect to the end points of symmetric interval of y-axis is 0

Let f (x) be an even function defined on R when 0

When x > 2, Let f = a (x-3) ^ 2 + 4, because through the point (2,2), substituting into the solution a = - 2 x 0 < = x < = 2, so f = {-...)

p: B = 0, Q: function f (x) = ax ^ 2 + BX + C is even function, then p is the () condition of Q. thank you

When B = 0, we can deduce that f (x) is even function
P can deduce Q
If f (x) is an even function
f(-x)=ax^2-bx+c=(ax^2+bx+c)
So B = 0
therefore
It is a necessary and sufficient condition

If P: ψ = π / 2 + K π, K ∈ Z, Q: F (x) = sin (ω x + ψ) (ω ≠ 0) is even function, what condition is p for Q?

When p: ψ = π / 2 + K π, K ∈ Z, f (x) = sin (ω x + ψ) = sin (ω x + π / 2 + K π) = cos (ω x + K π) = ± cos ω X. therefore, f (x) is an even function, that is, P can deduce Q. if Q: F (x) = sin (ω x + ψ) (ω ≠ 0) is an even function, then f (- x) = f (x), that is sin (- ω x + ψ) =