It is known that f x is a monotonic function on R, and f x is an odd function, if f 3 2

In this problem, the domain of the function f(x) is r, and it is an odd function, indicating that the function is defined at the origin.

The odd function is defined at the origin, which means f(0)=0 (if the odd function is not defined at the origin, then f(0)=0 is not necessarily true).

Because the function is monotonic and f(-3)=2>f(0), it means that the function is monotonically subtractive.

The inequality f((m-x) x)+f(m)<0 is equivalent to: f((m-x) x)<-f(m)=f(-m).

That is, the inequality can be reduced to: (m-x) x>-m, i.e., ((m-1)x+m) x>0 is equivalent to: x>0, (m-1)x+m>0 (1) or: x<0, (m-1)x+m<0 (2).

When m>1, the solution set of (1) is: x>0;The solution set of (2) is: x0 becomes: 1 x>0, so the solution set at this time is x>0

In summary, when m>1, x>0 or x0

I'm going to go. What is the odd function tangram guid 1356781792954???The third year of junior high school, right? It's a pity that I'm in the second year of junior high school. I'm sorry.

Because it is an odd function, f(x)=-f(-x), so f(0)=0

And because f(-3)=2, f(3)=-2 is monotonically decreasing on r.

f[(m-x) x]+f(m) 0 is transformed into f[(m-x) x] f(-m), and (m-x) x m is obtained from the monotonically decreasing property to obtain x m m+1

Because when x 0, f(x) is decreasing, and f(x) is an odd function defined by the middle ant on r.

So f(x) is a subtraction function over r.

Because x1+x2 0

So x1>-x2

Therefore, f(x1) cannot find the value of the specific Li Chang, but only the range.

f(x) is a singular function defined on r, the image is symmetrical with respect to the origin, and the function f(x) decreases monotonically on [0,1), then f(x) decreases monotonically on [-1,1), f(2-x)=f(x) then the f(x) image is symmetrical with respect to the line x=1 (2-x corresponds to x with equal values, regardless of x, x+(2-x)=2,x+(2-x)] 2=1, and the function f(x) decreases monotonically on [-1,1), then the function f(x) is in [1,3f(x)=-1 has a real root on [0,1), the straight line y=-1 and f(x) graphs have an intersection point on [0,1), and each has an intersection point on [-11) and [1,3], the straight line y=-1 and f(x) graphs have 2 intersections on [-,3], and these two intersection points are symmetrical with respect to the straight line x=1, and the sum of the abscissa of the intersection points = 2, that is, the sum of the roots of all real numbers on the interval [-1,3] f(x)=-1 = 2.

It is known that the function f(x) is monotonic on r, and satisfies that for any x r, there is f(f(x)-2 x)=3, then f(3)=?

By the title, version f(x)-2 x is a constant weight, set to m

i.e. f(x)=2 x+m, by f(f(x)-2 x)=3, i.e., f(m)=3, i.e., 2 m+m=3, so 2 m=3-m

So m=1 [1, draw the image, y=2 x and y=3-x images have one and only one intersection; 2. Make up the number - x=1 is the solution of the equation].

So f(x)=2 x+1, so f(3)=2 3+1=9.

Hope it helps.

1) Because the function f(x) is an odd function defined on r.

So f(-x) = -f(x).

and f(1)=-2

So f(-1)=2

Because the function f(x) is a monotonic function defined on r.

and f(1)-f(2 x-4 x-1)=f(4 x+1-2 x) according to the monotonicity monotonic reduction function.

2^x<-2^x+4^x+1

4^x-2*2^x+1>0

2^x-1)^2>0

2^x-1≠0x≠0

(1) Odd function So f(1)=-2 f(-1)=2 monotonic, f(1) so f(x) is a monotonic decreasing function.

2) The title is wrong.

f(x) is a singular function defined on r, and when x0 and f(x) decreases monotonically, then the function f(x) decreases monotonically on r, if x1+x2 0, then x1 -x2, f(x1) f(-x2)=-f(x2).

f（x1）+f（x2）＜0

So choose A

Let f(x)-2x=t give f(x)=t+2x f(t)=t+2t

From the properties of the function, it can be seen that the function f(t) increases monotonically on r f(1)=1+2=3

f[f（x）-2x]=3=f（1）

f（x）=1+2x

f（3）=9

Therefore choose C

Let x<0, then -x>0, f(-x)=-2(-x) 2+3(-x)+1=-2x 2-3x+1

i.e. -f(x)=-2x 2-3x+1

f(x)=2x^2+3x-1

Therefore f(x)={-2x 2+3x+1 x>0{0 x=0{2x 2+3x-1 x<0

Drawing two parabolic images, we can see that the monotonically increasing intervals are (-3 4,0) and (0,3 4); The monotonically decreasing intervals are (-3 4) and (3 4, +).

First of all, it can be seen that this problem is to be done with a piecewise function, known as x>0So when x<0, -x>0, will replace the x in the original formula with -x, f(-x)=-2x 2-3x+1, because it is an odd function, so f(x)=-(-2x 2-3x+1)=2x 2+3x-1, the monotonic interval can be found, draw a sketch to know, find the axis of symmetry, it is clear at a glance, remember to > 0Finding the intersection with x<0 is the important method, I hope it can help you

Since f(x) is a monotonically decreasing function, the inequality is constant and cosq*cosq-2t<=4sinq-3 so t>=1