Given that the function f (x) = (A-2) x & # 178; + (A-1) x + 3 is even, then the monotone increasing interval of F (x) is?

RELATED INFORMATIONS

• 1. If the function f (x) is defined as an odd function on (- 1,1) and a decreasing function, if f (x-1) + F (1-x & # 178;) < 0, the value range of X is obtained
• 2. The function y = f (x) is a decreasing function defined on R and an odd function. Solve the equation f (X & # 179; - x-1) + F (X & # 178; - 1) = 0
• 3. It is known that the odd function f (x) is a decreasing function defined on (- 1,1), and f (1-T) + (1-T & sup2;)
• 4. The monotone increasing interval of function y = 2x Λ & #178; + x-3 is
• 5. The function f (x) = x-1-alnx (a ∈ R) is known. It is proved that f (x) ≥ 0 is constant if and only if a = 1 ② Necessity F '(x) = 1-ax = x-ax, where x > 0 (i) When a ≤ 0, f '(x) > 0 is constant, so f (x) is an increasing function on (0, + ∞) And f (1) = 0, so when x ∈ (0,1), f (x) < 0, which is contrary to f (x) ≥ 0 A ≤ 0 does not satisfy the problem (II) when a > 0, ∵ x > A, f '(x) > 0, so f (x) is an increasing function on (a, + ∞); When 0 ＜ x ＜ a, f '(x) ＜ 0, so the function f (x) is a decreasing function on (0, a); ∴f（x）≥f（a）=a-a-alna ∵ f (1) = 0, so when a ≠ 1, f (a) < f (1) = 0, which is in contradiction with F (x) ≥ 0 ∴a=1 In the process of proving the necessity above, what is the meaning of "∵ f (1) = 0, so when a ≠ 1, f (a) < f (1) = 0, which is in contradiction with F (x) ≥ 0?" why is there f (a) < f (1) when a ≠ 1?
• 6. Given the function f (x) = x-1-alnx, it is proved that f (x) ≥ 0 is constant if and only if a = 1
• 7. If the function f (x) = {(2b-1) x + B-1, x > 0, is an increasing function on R, then the value range of function B is {- X & # 178; + (2-B) x, X ≤ 0 According to the meaning of the title {2b-1 ＞ 2, ﹛2-b＞0 ﹛b-1≥f（0）, How do these three inequalities come from?
• 8. If the function f (x) = x & # 178; + 2 (A-1) x + 2 is an increasing function on [4, + ∞), then the value range of real number a is?
• 9. Let f (x) = x2 + X, x < 0-x2, X ≥ 0, if f (f (a)) ≤ 2, then the value range of real number a is___ ．
• 10. If the function y = x2-2x + 3 has a maximum value of 3 and a minimum value of 2 in the interval [0, M], then the value range of M is () A. [1，∞）B. [0，2]C. （-∞，2]D. [1，2]
• 11. If f (x) = (K-3) x & # 178; + (K-2) x + 3 is an even function, then the increasing interval of the function is________
• 12. Given that even function f (x) (x ≠ 0) is monotone on interval (0, + ∞), what is the sum of all x satisfying f (X & # 178; - 2x-1) = f (x + 1)?
• 13. The function f (x) is even and is a decreasing function on (- ∞ 0). Try to compare the size of F (- 7 / 8) and f (2a & # 178; - A + 1)
• 14. If the function f (x) = ax & # 178; + BX + 3A + B defined on [a-1,2a] is even, then a + B =?
• 15. Given the function f (x) = x cube minus 4x & # 178; 1) find the monotone interval of function f (x) (2) find the function f (x) in the closed interval 0 The known function f (x) = x cube minus 4x & # 178; 1) Finding monotone interval of function f (x) (2) Finding the maximum and minimum of function f (x) in the closed interval 0 4
• 16. Given the function f (x) = & # 188; X & # 8308; = & # 8531; ax & # 179; - A & # 178; X & # 178; + A & # 8308; (a > 0), find the monotone interval of the function
• 17. The known function f (x) = - 2A & # 178; X & # 178; + ax + 1 If f (x) = - 2A & # 178; X & # 178; + ax + 1 ≤ 0 is constant in the interval (1, + ∞), find the value range of real number a
• 18. Given the function f (x) = ax & # 178; - (2a & # 178; - 1) x-2a (a ∈ R), let the solution set of the inequality f (x) > 0 be a, and also know that B = {x | 1 ＜ x ＜ 3}, a ∩ B ≠ & # 216;, and find the value range of A
• 19. How much is the limit when x is close to zero
• 20. X divided by (1-radical x + 1) (x is greater than or equal to negative 1. X is not equal to 0)