Try to find an interval of length one where the function y = (x-1) \ (3x + 2) has at least one zero point
It is easy to know that when x = 1, y = 0, so it can be (0.5,1.5)
RELATED INFORMATIONS
- 1. The interval where the zeros of the function f (x) = - 1x + lgx are located is () A. (0,1)B. (1,2)C. (2,3)D. (3,10)
- 2. The interval where the zeros of the function f (x) = - (1 / x) + lgx lie is
- 3. The interval of zero point of function f (x) = lgx + X-5 is () A. (1,2)B. (2,3)C. (3,4)D. (4,5)
- 4. X 2 ∈ (0. + ∞), if f (x) = lgx, compare the sizes of [f (x 1) + F (x 2)] / 2 and f [(x 1 + x 2) / 2]
- 5. For any x1, X2 (x1 ≠ x2) in the definition field of function f (x) = lgx, the following conclusion is obtained f((x1+x2)/2)
- 6. For any x1, X2 (x1 ≠ x2) in the domain of function f (x), the following conclusions are obtained: (1) f (x1 + x2) = f (x1) + F (x2); (2)f(x1·x2)=f(x1)+f(x2); (3)[f(x1)-f(x2)]/(x1-x2)>0; (4)f[(x1+x2)/2]
- 7. For any x1, X2 (x1 ≠ x2) in the domain of definition of function f (x), we have the following conclusion: (1) f (x1 + x2) = f (x1) * f (x2) (2) f (x1 * x2) = f (x1) + F (x) (1)f(x1+x2)=f(x1)*f(x2) (2)f(x1*x2)=f(x1)+f(x2) (3)[f(x1)-f(x2)]/(x1-x2)>0 (4) f[(x1+x2)/2]
- 8. For the function f (x) = 1 / X (x > 0) in the domain x1, X2 (x1 ≠ x2), we have the following conclusions 1.f(x1+x2)=f(x1)+f(x2);2.f(x1x2)=f(x1)f(x2);3.f(x1)-f(x2) / x1-x2; 4.f(x1+x2 / 2)<f(x1)+f(x2) / 2 The correct conclusion in the above conclusion is -- () the answers are 2 and 4, but I don't know why 4 is right,
- 9. Let y = log2 ^ (0, + ∞). If f (x) = lgx, try to judge 1 / 2 "f (x1) + F (x2) and 1 / 2" f (x1) + F (x2) according to the image of F (x) F "((x1) + (x2)) / 2"
- 10. It is known that the zeros of the function y = x ^ 2 + 2mx + 2m + 3 (M belongs to R) are x1, x2, Find the minimum value of X1 ^ 2 + x2 ^ 2
- 11. The interval where the zeros of the function f (x) = 2x + 3x lie is______ .
- 12. Find the zero point of function y = (3x ^ 2-x ^ 2) / (x ^ 2-1) + 1 / (1-x) - 2
- 13. The number of zeros of function y = x ^ 2-3x + 1
- 14. Finding the zeros of the function y = (3x-x ^ 2) / (x ^ 2-1) + 1 / (1-x) - 2
- 15. The zero point of the function f (x) = x2-3x + 2 is () A. (1,0),(2,0)B. (0,1),(0,2)C. 1,2D. -1,-2
- 16. The even function f (x) defined on R satisfies that for any x1, X2 belongs to (negative infinity, 0) (x1 ≠ x2), and (x2-x1) (f (x2) - f (x1)) > 0, then when n belongs to N +, f(n+1)
- 17. Even functions defined on R satisfy the following conditions: for any X &;, X &; ∈ [0, + ∞], there is f (x1) - f (x2) / x1-x2 < 0 The size order of F (3), f (- 2), f (1) (solving process)
- 18. Given the function f (x) = TaNx, X belongs to (0, Wu / 2), if x1, X2 belong to (0, Wu / 2), X1 is not equal to X2, try to prove: 1 / 2 [f (x1) + F (x2)] > F [(x1 Given the function f (x) = TaNx, X belongs to (0, Wu / 2), if x1, X2 belong to (0, Wu / 2), X1 is not equal to X2, try to prove: 1 / 2 [f (x1) + F (x2)] > F [(X1 + x2) / 2]
- 19. Given the function f (x) = TaNx, X belongs to (0 ~ π / 2) and X1 = X2, compare the size of 1 / 2 (FX1 + FX2) and f (x1 + x2) / 2 Can we not use derivatives
- 20. Given the function f (x) = TaNx, X belongs to 0 to 90 degrees, if X1 and X2 belong to 0 to 90 degrees, and x1 ≠ X2, the proof is: 1 / 2 {f (x1) + F (x2)} > F {(x1 + x2) / 2}