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

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• 1. 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)
• 2. 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，+∞）
• 3. 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?
• 4. 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!
• 5. 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]
• 6. 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
• 7. F (x) = ax ^ 2 + BX + C, (a > 0), two X1 and x2,0 of the equation f (x) - x = 0
• 8. If the two roots of the equation AX & # 178; + BX + C = 0 (a ≠ 0) are X1 and X2, then X1 + x2 = - (B / a), x1x2 = C / A, prove
• 9. F (x) = ax ^ 2 + BX + C, X2 > x1, f (x1) ≠ f (x2), it is proved that f (x) = 1 / 2 [f (x1) + F (x2)] equation has a real root in (x1, x2) inside
• 10. Let f (x) = ax ^ 2 + BX + C, if 6A + 2B + C = 0, f (1) * f (3) > 0, prove that the equation f (x) = 0 must have two unequal real roots, and 3 < X1 + x2 < 5
• 11. 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
• 12. If f (x + 1) and f (x-1) are both odd functions, and f (0) = 2, then what is f (4)
• 13. 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
• 14. 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)]=
• 15. When X & gt; 0, FX = - x + 1, then when X & gt; 0, the expression of FX is?
• 16. 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-
• 17. 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）
• 18. 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
• 19. 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)
• 20. It is known that the domain of F (x) is r, and its derivative satisfies 0