If the odd function y = f (x) (x ≠ 0) is x ∈ (0, + ∞), f (x) = X-1, then the value range of X satisfying the inequality f (x-1) < 0 is obtained
When x > 1, f (x-1) = (x-1) - 1 = X-2
RELATED INFORMATIONS
- 1. If the odd function f (x) (x is not equal to o), X belongs to (0, + 00), f (x) = X-1, then the inequality f (x-1) is satisfied
- 2. Given that the function f (x) = x + B (a is greater than 0, a is not equal to 1) satisfies f (x + y) = f (x) f (y) and f (3) = 8 to find f (x)
- 3. The definition field of odd function f (x) is R. when x > 0, f (x) = - x 2x2, then the expression of F on R
- 4. It is known that f (x) is an odd function over R, and if x
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- 7. Find the range of function y = 10 ^ X-10 ^ (- x) / 10 ^ x + 10 ^ (- x)
- 8. Find the range of the following functions y = x + √ x + 1 f (x) = (x + 5) / (X & # 178; + 4) Don't be wrong
- 9. The range of the function f (x) = - X & # 178; - 2x + 5 is
- 10. The range of function f (x) = (X & # 178; + X + 1) / (X & # 178; + 1) is
- 11. If the odd function y = f (x) (x belongs to R and X does not = 0), when x is 0 to positive infinity, f (x) = X-1, and f (x-1)
- 12. If the odd function y = f (x), (x is not equal to 0), if x belongs to (0, + ·), f (x) = X-1, find the inequality f (x-1)
- 13. It is known that the function f (x) is defined as an increasing function f (2) = 1, f (XY) = f (x) + F (y) defined on greater than 0, which solves the inequality f (4) f (X-2) less than or equal to 3
- 14. If x is greater than 0, f (x) is greater than 1, f (a + b) = f (a). F (b) It is proved that FX is a proper function on R, a and B belong to R Increasing function
- 15. Given that f (x) is defined as R, for any x, y belongs to R, f (x + y) + F (X-Y) = 2F (x) (y), and f (0) is not equal to 0 If there is a constant C, Let f (C / 2) = 0. Proof: for any x belonging to R, f (x + C) = - f (x)
- 16. It is known that the function f (x) is defined on R, for any x, y belongs to R, f (x + y) + F (X-Y) = 2F (x) f (y), and f (0) is not equal to 0 1, prove that f (0) = 1.2, y = f (x) is even function
- 17. The function f (x) defined on R, for any x, y ∈ R, f (x + y) + F (X-Y) = 2F (x) f (y), and f (0) is not equal to 0, it is proved that f (x) is even function
- 18. Given the function FX = LG (1 + x) / (1-x), find the value range of X which makes FX > 0
- 19. Given the function f (x) = LG (x + 1); (1) if 0 < f (1-2x) - f (x) < 1, find the value range of X; (2) if G (x) is an even function with a period of 2, and if 0 ≤ x ≤ 1, G (x) = f (x), find the analytic expression of the function y = g (x) (x ∈ [- 1,1])
- 20. Given the function f (x) = x2 + (a + 1) x + LG | a + 2 | (a ∈ R, and a ≠ - 2) (I) if f (x) can be expressed as the sum of an odd function g (x) and an even function H (x), find the analytic expressions of G (x) and H (x); (II) proposition p: the function f (x) is an increasing function in the interval [(a + 1) 2, + ∞); proposition q: the function g (x) is a decreasing function. If proposition p and Q have and only one is an increasing function Under the condition of (II), compare the size of F (2) and 3-lg2