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. f (x) is an odd function over the definition field R and satisfies f (x + 2) = f (x). When x belongs to (0,1). F (x) = 2 ^ X-1, find f (log2 / 1 6
log(1/2) 6=-log2 6=-1-log2 3-1-log2 4=-3
Then - 32
So 0
If f (x) is an odd function whose domain is r, and f (x) = X-1 when x ∈ (0, + ∞), the solution set of inequality f (x) ≤ 0 is
∵ f (x) is an odd function whose domain is r
When x ∈ (0, + ∞), f (x) = X-1
When x = 0, f (x) = 0
When x ∈ (- ∞, 0), f (- x) = - X-1
That is, f (x) = - (- x-1) = x + 1
‡ make f (x) ≤ 0
1 ° if x0, then X-1 ≤ 0
That is, 0
Note that the solution set of the inequality log2 (x-1) about X is less than 1 is p, and the definition domain of a minus x under the function FX = radical is Q (1) Find set p (2) If P belongs to Q, find the value range of A
First question P: X is less than 3 and greater than 0, a is greater than or equal to 3 and less than or equal to 0
It is known that the definition field of function f (x) is r, and the real number a satisfies f (2a ^ 2 + 3A + 1) < f (0) It is known that the definition field of function f (x) is r, and the real number a satisfies f (2a ^ 2 + 3A + 1) < f (0). If f (x) is an increasing function on R, find the value range of real number a Wait, wait
∵ f (x) is an increasing function on R
∴2a ²+ 3a+1
Ask for a quick answer! Senior one math! It is known that the function y = f (x) defined on the real number set R satisfies the condition that for any real number x, y has f (x + y) = f (x) + F (y) 1. Find f (0) 2. Prove that f (x) is an odd function 3. If x > 0, f (x) > 0. F (- 1) = 2, find the value range of F (x) on [- 2,1] The key is the third question! If x > 0, f (x) > 0. F (- 1) = - 2, find the value range of F (x) on [- 2,1]. Sorry = a = change the title
Is there a problem with the title
Odd function f (- x) = - f (x)
f(-1)=-f(1)=2
Namo f (1) = - 2
This is in contradiction with the question: if x > 0, f (x) > 0!
Given that f (x) = loga1 + X / 1-x (a > 0 and a ≠ 1), ① find the definition domain of F (x) ② judge the parity of F (x) and prove it ③ Find the value range of X that makes f (x) > 0 Detailed steps
①(1+x)/(1-x)>0
-1②f(-x)=loga(1-x)/(1+x)=-loga(1+x)/(1-x)=-f(x)
Therefore, it is an odd function
③ If 0 (1 + x) / (1-x) < 1
x> 1, or x < 0
∵ - 1 ∵ - 1 if a > 1
(1+x)/(1-x)>1
0
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