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)?
Because f (X & # 178; - 2x-1) = f (x + 1) and f (x) is even function, so x ^ 2-2x-1 = x + 1 or x ^ 2-2x-1 = - X-1, from ① we get x ^ 2-3x-2 = 0 = = > x is irrational root, X1 + x2 = 3, from ② we get x ^ 2-x = 0 = = > x is rational root, X1 + x2 = 1, so the sum of X is 4
RELATED INFORMATIONS
- 1. If f (x) = (K-3) x & # 178; + (K-2) x + 3 is an even function, then the increasing interval of the function is________
- 2. 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?
- 3. 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
- 4. 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
- 5. It is known that the odd function f (x) is a decreasing function defined on (- 1,1), and f (1-T) + (1-T & sup2;)
- 6. The monotone increasing interval of function y = 2x Λ & #178; + x-3 is
- 7. 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?
- 8. Given the function f (x) = x-1-alnx, it is proved that f (x) ≥ 0 is constant if and only if a = 1
- 9. 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?
- 10. 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?
- 11. 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)
- 12. If the function f (x) = ax & # 178; + BX + 3A + B defined on [a-1,2a] is even, then a + B =?
- 13. 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
- 14. 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
- 15. 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
- 16. 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
- 17. How much is the limit when x is close to zero
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- 19. The limit of F (x) is a, a > 0. It is proved that f (x) is equal to a under the triple root sign Can be added!
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