Given that the function f (x) is an odd function, when x belongs to (0,1), f (x) = LG (x 1), then when x belongs to (- 1,0), what is the expression of F (x)?
Odd function f (- x) = - f (x), so if x ∈ (- 1,0) has f (x) = - f (- x) = - LG (- x 1)
RELATED INFORMATIONS
- 1. It is known that f (x) is an odd function on R. when x > 0, f (x) = x (1-x), find the expression of F (x), and draw the image of F (x) in the given coordinate system
- 2. Let f (x) be an odd function defined on R, and the image of y = f (x) is symmetric with respect to x = 1 / 2, f (1) + F (2) + F (3) + F (4) + F (5)=
- 3. Given the function f (x) = x2 + (a + 1) x + LG | a + 2 | (a ∈ R, and a ≠ - 2) (I) if f (x) can be expressed as the sum of an odd function g (x) and an even function H (x), find the analytic expressions of G (x) and H (x); (II) proposition p: the function f (x) is an increasing function in the interval [(a + 1) 2, + ∞); proposition q: the function g (x) is a decreasing function. If proposition p and Q have and only one is an increasing function Under the condition of (II), compare the size of F (2) and 3-lg2
- 4. Given the function f (x) = LG (x + 1); (1) if 0 < f (1-2x) - f (x) < 1, find the value range of X; (2) if G (x) is an even function with a period of 2, and if 0 ≤ x ≤ 1, G (x) = f (x), find the analytic expression of the function y = g (x) (x ∈ [- 1,1])
- 5. Given the function FX = LG (1 + x) / (1-x), find the value range of X which makes FX > 0
- 6. The function f (x) defined on R, for any x, y ∈ R, f (x + y) + F (X-Y) = 2F (x) f (y), and f (0) is not equal to 0, it is proved that f (x) is even function
- 7. It is known that the function f (x) is defined on R, for any x, y belongs to R, f (x + y) + F (X-Y) = 2F (x) f (y), and f (0) is not equal to 0 1, prove that f (0) = 1.2, y = f (x) is even function
- 8. Given that f (x) is defined as R, for any x, y belongs to R, f (x + y) + F (X-Y) = 2F (x) (y), and f (0) is not equal to 0 If there is a constant C, Let f (C / 2) = 0. Proof: for any x belonging to R, f (x + C) = - f (x)
- 9. If x is greater than 0, f (x) is greater than 1, f (a + b) = f (a). F (b) It is proved that FX is a proper function on R, a and B belong to R Increasing function
- 10. It is known that the function f (x) is defined as an increasing function f (2) = 1, f (XY) = f (x) + F (y) defined on greater than 0, which solves the inequality f (4) f (X-2) less than or equal to 3
- 11. Given that the function y = f (x) is an odd function on R, and when x > 0, f (x) = 1, then the expression of the function y = f (x) is______ .
- 12. It is known that the function y = f (x) is an odd function on R, and when x is greater than 0, f (x) = 1. Try to find the expression of y = f (x)
- 13. Given that the function y = f (x) is an odd function on R, and when x > 0, f (x) = 1, then the expression of the function y = f (x) is______ .
- 14. Given that function f (x) is an odd function on R, and x > 0, f (x) = 1, try to find the expression of function y = f (x)
- 15. If f (x) is an odd function defined on R, and X ∈ (0, + ∞), f (x) = LG (x + 1), find the expression of F (x) and draw the diagram
- 16. Y = (1 / 3) ^ (2 / 3x-1), find the range of function y=(1/3)^[2/(3x-1)]
- 17. Find the range of y = 3x + 1 / 2x-1, X ∈ [1,2]
- 18. Given the function FX = {X & # 178; + 1, X ≥ 0 and 1, x < 0, then the value range of X satisfies the inequality f (1-x & # 178;) > F (2x)
- 19. Given the function FX = | X-1 | + | x-3 | (1), find the value range of X, make FX a constant function (2) if the inequality f about X The known function FX = | X-1 | + | x-3| (1) Find the value range of X so that FX is a constant function (2) If the inequality f (x) - a ≤ 0 about X has a solution, find the value range of real number a
- 20. Given the function f (x) = (x2-x-1 / a) e ^ ax (a > 0), if the inequality f (x) + 3 / a > = 0 holds for X ∈ (- 3 / A, + infinity), then the value range of real number a is?