Given that the domain of definition of odd function f (x) is r, and for any real number x, f (x + 2) = - f (x), and f (1) = 4, then f [f (2015)]=
The odd function shows that - f (x) = f (- x) so f (- x + 2) = - f (- x) = f (x) = - f (x + 2) then f (x + 2) = - f (2-x) = f (X-2) so let x + 2 = t then f (T) = f (T-4) so f [f (2015)] = f [f (2015-4n)] = f [f (- 1)] = f (- 4) = - f (4) = - f (0) be symmetric about the origin, so f
RELATED INFORMATIONS
- 1. For any odd function FX on the domain R, there are a. f (x) - (F-X) > 0, B. f (x) - f (- x) ≤ 0, C. f (x) × f (- x) ≤ 0, D. f (x) × f (- x). Why choose C
- 2. If f (x + 1) and f (x-1) are both odd functions, and f (0) = 2, then what is f (4)
- 3. If f (x) = radical √ x ∧ 2 + 3x-4 and G (x) = radical √ X-1 + radical √ x + 4 are defined as a and B respectively, then the relationship between a and B is a. a The definition field of function f (x) = radical √ x ∧ 2 + 3x-4 is a, and the definition field of function g (x) = radical √ X-1 + radical √ x + 4 is B, then the relationship between a and B is A. A does not include B. B. A = B. CA includes B. Da includes B
- 4. It is known that the piecewise function f (x) is an odd function on R. when x > 0, f (x) = x2-2x + 3, the analytic expression of F (x) is obtained
- 5. The definition field of function f (x) = x + 1 is [1,16], f (x) = f (2x) + f ^ 2 (x) + 1. Find the value field of function f (x)
- 6. The definition domain of function f (x) is (- ∞, 1) ∪ (1, + ∞), and f (x + 1) is an odd function. When x > 1, f (x) = 2x2-12x + 16, then the value range of real number m with two zeros of equation f (x) = m is () A. (-6,6)B. (-2,6)C. (-6,-2)∪(2,6)D. (-∞,-6)∪(6,+∞)
- 7. If the function f (x) has f (2 + x) = f (2-x) for any real number x, and the equation f (x) = 0 has four real roots, then the sum of the four real roots?
- 8. Let f (x) be an even function defined on R, and if x ≥ 0, f (x) = x2-2x-3, the root of the equation f (x) = 2a-3 (a ∈ R) is discussed Please answer as soon as possible!
- 9. The domain of function f (x) is r, and f (2 + x) = f (2-x). If f (x) is an even function, and f (x) = 2x-1 when x is [0,2], find the expression of F (x) when x is [- 4,0]
- 10. Given the real number a > b > C and a + B + C = 0, the two different real number roots of the equation AX ^ 2 + BX + C = 0 are x1, X2 (1) proof - 1 / 2C and a + B + C = 0, and the two different real number roots of the equation AX ^ 2 + BX + C = 0 are x1, X2 (1) proof - 1 / 2
- 11. When X & gt; 0, FX = - x + 1, then when X & gt; 0, the expression of FX is?
- 12. It is known that f (x) is an odd function defined on R. when x > 0, f (x) = x ^ 2-2x + 1, then the expression of F (x) is Its function-
- 13. It is known that y = f (x) is an odd function defined on R. when x ≥ 0, f (x) = x2-2x, then the expression of F (x) on R is () A. -x(x-2)B. x(|x|-2)C. |x|(x-2)D. |x|(|x|-2)
- 14. 5. Given that f (x) is an odd function defined on R, when x ≥ 0, f (x) = x-2x, find the expression of F (x) on R 5. Given that f (x) is an odd function defined on R, when x ≥ 0, f (x) = x-2x, find the expression of F (x) on R 6. It is known that the function f (x) is an even function on R. when x ≥ 0, f (x) = x-2x-3 (1) Write the expression of function y = f (x) with piecewise function; (2) Using symmetry to draw its image; (3) The monotone interval is pointed out; (4) It is pointed out that in what interval f (x) > 0 and in what interval f (x) < 0 by using images; (5) Find the maximum value of the function 7. Find the range of function y = 1 / X (x > - 4 and X is not equal to 0) 8. Find the range and monotone interval of the function y = | x + 2 | - | X-5 | 9. It is known that the function y = f (x) is an even function defined on R. when x < 0, f (x) is monotonically increasing. The solution set of the inequality f (x + 1) > F (1-2x) is obtained 10. The definition field of function y = x-3x-4 is [0, M], the range of value is [- 25, 4, - 4], and the value range of real number m is obtained
- 15. Given that the function defined on (0, -∞) satisfies f (x, y) = f (x) + F (y), and if x > 1, f (x) < 0, if f (half) = 1, find the solution set of the inequality f (x) + F (5-x) ≥ - 2 Is f (XY) = f (x) + F (y)
- 16. It is known that the domain of F (x) is r, and its derivative satisfies 0
- 17. It is proved that if the function f (x) is a strictly increasing function on [a, b], then the equation f (x) = 0 has at most one real root on the interval [a, b]
- 18. For any differentiable function f (x) on R, if x is not equal to 1 and satisfies (x-1) f '(x) > 0, it is proved that f (0) + F (2) > 2F (1)
- 19. Let f (x) be differentiable on [0,1] and 0
- 20. The function f (x) is differentiable on [0,1] and 0