F (x) = ax ^ 2 + BX + C, (a > 0), two X1 and x2,0 of the equation f (x) - x = 0
1) Prove: Let f (x) = f (x) - x,
∵ x1, X2 are the roots of F (x) - x = 0,
∴ F(x)=a(x-x1)(x-x2).
When x ∈ (0, x1), because x 10,
If a > 0, then f (x) = a (x-x1) (x-x2) > 0
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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
- 4. Let f (x) = ax ^ 2 + BX + C, if 6A + 2B + C = 0, f (1) times f (3) > 0, if a = 0, find the value of (2)
- 5. How to prove that f (x) = √ (x-1) is an increasing function in the domain of definition
- 6. It is proved that y = 1-sinx + 7cos3x is a bounded function on its domain
- 7. Does bounded function mean that there are both upper and lower bounds in its domain of definition
- 8. To prove that a function is bounded, must its upper and lower bounds be opposite to each other
- 9. It is proved that the necessary and sufficient condition for f (x) to be bounded in (a, b) is that f (x) has both upper and lower bounds in (a, b)
- 10. Given the function f (x), for X ∈ R, f (4-x) = f (x). If f (x) has exactly four unequal zeros x1, X2, X3, x4, then X1 + x2 + X3 + X4=
- 11. 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
- 12. 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]
- 13. 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!
- 14. 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?
- 15. 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,+∞)
- 16. 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)
- 17. 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
- 18. 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
- 19. If f (x + 1) and f (x-1) are both odd functions, and f (0) = 2, then what is f (4)
- 20. 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