It is known that the function y f x is an odd function, and it is in 0, which is an increment funct

f(x)=1 f(x) is a subtractive function on (- 0).

Prove that f(x) is an odd function

f(-x)=-f(x), f(x) at (0,+ is an increasing function, and f(x)<0, f(-x) is also an increasing function at (- 0), and f(-x) is > 0, indicating that f(x) increases with the increase of x, and because f(x)=1 f(x), x increases, f(x) increases, and 1 f(x) decreases, indicating that f(x) is a subtraction function, f(-x)=1 f(-x)=-1 f(x)=-f(x), which means that f(x) is also an odd function, so f(x)=1 f(x) is a subtractive function on (- 0).

f(x)=1 f(x) is a subtractive function on (- 0).

Proof is that according to the known condition f(-x)=-f(x), f(x) is an increasing function at (0, + is an increasing function, and f(x) is < 0, indicating that the image is an increasing function in the fourth quadrant, and the odd function is about dot symmetry, so f(-x) is also an increasing function at (- 0), and f(-x) > 0, indicating that f(x) increases with the increase of x, and because f(x)=1 f(x), x increases, f(x) increases, and 1 f(x) decreases, indicating that f(x) is a decreasing function, and f(-x)=1 f(-x)=-1 f(x)=-f(x), which means that f(x) is also an odd function, so f(x)=1 f(x) is a subtraction function on (- 0).

Proof: Let x>=0 exist, and there is an arbitrary positive number af(x+a)-f(x)>0 (1) by the title

Since f is an odd function, then f(-x-a) = -f(x+a), f(-x) = -f(x);

1 f(-x) -1 f(-x-a) (2)f(-x-a)-f(-x) f(-x)f(-x-a)-[f(x+a)-f(x)] f(x)f(x+a)(x+a)(x)f(x+a)(x)f(x)f(x)>00 (2)< 0 Therefore, at negative infinity to 0 f(x)=1 f(x) is a subtractive function.

1.is a subtraction function.

Take x1 and x2 at (- 0), and set x1 to "Brother Debate and Laugh x2 is envious of -x1>-x2>0

is an increment function on (0,+, and f(x) 0

0>f(-x1)>f(-x2)

0>-f(x1)>-f(x2)

1/f(x1)>1/f(x2)

f(x)=1 f(x) is a subtraction function on (- 0).

2.The second question lacks conditions.

1.Proof: Let x>=0 first, and there is an arbitrary positive number af(x+a)-f(x)>0 (1) from the title

Since f is an odd function, then f(-x-a) = -f(x+a), f(-x) = -f(x);

1 f(-x) -1 f(-x-a) (2)f(-x-a)-f(-x) f(-x)f(-x-a)-[f(x+a)-f(x)] f(x)f(x+a)> (x+a) is obtained from (1) and f(x)f(x+a) states Zheng 0 (2)< 0 is a subtraction function from negative infinity to 0 f(x)=1 f(x).

1.is a subtraction function. Since it is a chirp function, f(x) is also an increasing function on (- 0). 1 f(x) is to subtract the letter Zheng Shen Li Shu filial piety.

Proof omitted.

u-v>0

y=f(x) is an increment function on (0,+.

f(-u)>f(-v)

And when x>0 there is always f(x)<0

0>f(-u)>f(-v)

odd function y=f(x).

0>-f(u)>-f(v)

01/f(v)

f(u)>f(v)

f(x)=1 f(x) decreases monotonically on (- 0).

f(x) is an odd function, and in the interval (0,+ is a monotonic increasing function, and f(-2)=0, f(2)=0, and when x -2 or 0 x 2, the function image is below the x-axis, as shown in the figure

When x 2 or -2 x 0 the function image is above the x-axis, xf(x) 0 has a solution set of (-2,0) (0,2), so the answer is: (-2,0) (0,2).

Solution: y=f(x) is the function of the Shiyanqi, so the domain is symmetrical with respect to the origin, which is an increasing function at (0,+, and f(x)"Annihilation of silver 0, then the digging is an increasing function on (- 0), and f(x)>0.

Thus, f(x)=1 f(x) is a subtractive function on (- 0).

The function y=f(x)(x≠0) is an odd function, and when x (0,+ is an increasing function, f(x) is an increasing function at (- 0), and f(x-1 2)<0=f(soil 1),0< x-1 2<1, or x-1 2<-1, 1 2

f(x) is an odd function, then the graph of f(x) is symmetrical with respect to the origin.

It is an increment function on (0,+.

So on (- 0) is also an increment function.

If x2>x1>0 has -x2<-x1<0

f(x2)-f(x1)=-f(-x2)+f(-x1) is an increasing function when x<0, i.e., f(-x1)>f(-x2), so.

f(x2)-f(x1)=-f(-x2)+f(-x1)>0, that is, the value of the function increases with x.

Therefore, at x>0 the function is also an increasing function.

y=fx is an odd function and is an increasing function at x<0, so that -x1< -x2 < 0, then x1>x2 > 0

f(-x1) -f(-x2) <0, i.e.: -f(x1) -f(x2)] 0, so f(x1) -f(x2) >0

So y=fx is also an increment function at x 0.

y is an odd function, then f(x)=-f(-x), at x<0, fx increments, so that any x1-x2>0, f(-x1)-f(-x2)=-f(x1)+f(x2)>0, is proved.