Let f (x) be an odd function defined on R. when x ≥ 0, f (x) = 3 ^ x + 3x + A, then f (- 2)=

Let f (x) be an odd function defined on R. when x ≥ 0, f (x) = 3 ^ x + 3x + A, then f (- 2)=

Because f (x) is an odd function, f (0) = 3 ^ 0 + 3 * 0 + a = 0 deduces a = - 1, f (x) = 3 ^ x + 3x-1
So f (2) = 14, so f (- 2) = - f (2) = - 14

It is known that the odd function FX defined on R is a decreasing function. If X1 plus X2 is less than 0, X2 plus X3 is less than 0, X3 plus X1 is less than 0, then the value of FX1 + FX2 + fx3 is

Classmate, is this a question to fill in the blanks? It seems that I can't find out the specific value

It is known that the function y = f (x) is an odd function and a decreasing function in the domain of definition [- 1,1]. (1) it is proved that for any x1, X2 ∈ [- 1,1], there are [f (x1) + F (x2)] (x1 + x2) ≤ 0 (2) solutions to the inequality f (1-A) + F (1-a2) < 0

(1) If X1 + x2 = 0, the inequality holds obviously; if X1 + x2 < 0, then - 1 < X1 < - x2 < 1, ∵ function y = f (x) is an odd function and a decreasing function in the domain of definition [- 1, 1], ∵ f (x1) > f (- x2) = - f (x2), f (x1) + F (x2) > 0, so the original inequality holds; similarly, it can be proved that when X1 + x2 > 0 & nbsp;, the original inequality also holds. (2) from F (1-A) + F (1-a2) < 0 And given the following inequality system − 1 ≤ 1 − A2 ≤ 1 − 1 ≤ a − 1 ≤ 11 − A2 > a − 1, the solution is 0 ≤ a < 1

If f (x) = - x-x3, x1, X2, x3 ∈ R, and X1 + x2 ＞ 0, X2 + x3 ＞ 0, X3 + x1 ＞ 0, then the value ()
A. Must be greater than zero B. must be less than zero C. equal to zero D. both positive and negative are possible

It is easy to know that the function f (x) = - x-x 3 is an odd function, a decreasing function, ∵ x 1 + x 2 > 0, x 2 + x 3 > 0, ∵ x 1 > - x 2, x 2 > - x 3 > x 1, ∵ f (x 1) < f (- x 2,) f (x 2) < f (- x 3), f (x 3) < f (- x 1) ∵ f (x 1) + F (x 2) < 0, f (x 2) + F (x 3) < 0, f (x 3) + F (x 1) < 0 Choose B

The functions f (x) = x + X3, x1.x2.x3 belong to R, X1 + x2

x1+x2

F (x + 2) is defined as the odd function (x1-x2) / (f (x1) - f (x2)) on R is less than zero
Solving inequality f (3-x)

The solution is (x1-x2) / (f (x1) - f (x2)), so f (x) is a decreasing function
Because f (x + 2) is defined as an odd function over R
So f (0 + 2) = f (2) = 0
f(3-x)

It is known that f (x) is an odd function defined on [- 1,1], and f (1) = 1. If x1, X2 ∈ [- 1,1], and X1 + x2 ≠ 0, then (f (x1) + F (x2)) / (x1 + x2) > 0. (1) prove by definition that f (x) is an increasing function on [- 1,1] (2) if the inequality f (x) ≥ 4 + log2 m holds for any x ∈ [- 1,1], find the value range of real number M

(1) Because X1 and X2 belong to [- 1,1], so - x2 also belongs to this interval, and f (x) is an odd function in this interval. Let X1 > - X2, so 0-x2 (x2 also belongs to this domain) holds, so f (x) is an increasing function in the domain. (definition) (2) from (1), we can know that the minimum value of F (x) is f (- 1) = - f (1) = - 1 (because it is an increasing function), so we should make the inequality hold, Only 4 + log2 m is less than its minimum value, that is - 1. If we solve the inequality 4 + log2 m ≤ - 1, we get 0

Y = f (x) is an odd function defined on R. when x ≥ 0, f (x) = x & sup2; - 2x, find the analytic expression of F (x) on R
What I want to ask is, can't we use an analytic expression?

When x < 0
-x > 0
So f (- x) = (- x) &# 178; + 2x
= x² + 2x
Because y = f (x) is an odd function
So f (x) = - f (- x) = - X & # 178; - 2x
To sum up, f (x) = x & # 178; - 2x x ≥ 0
= -x²-2x x 0
= 0 x = 0
= -1 x < 0

It is known that y = f (x) is an odd function whose domain is r, and x = > 0, f (x) = 2x-x & sup2;
1 for X

1. Y = f (x) is an odd function whose domain is r, and x = > 0, f (x) = 2x-x & sup2;
When x

Let y = f (x) be an odd function defined on R. when x > 0, f (x) = x & sup2; - 2x + 3,
Try to find the expression of (x) on R and write the monotone interval

When x > 0, f (x) = x & sup2; - 2x + 3,
Let x0, so f (- x) = (- x) ^ 2-2 (- x) + 3 = x ^ 2 + 2x + 3
F (x) = - f (- x) for odd functions
So, x0, f (x) = x ^ 2-2x + 3
x=0,f(0)=0
x