It is known that f x is an odd function defined on 1,1 , and when x 1,0, f x 2 x 4 x 1 .

I just answered it for someone else yesterday, and I copied it directly and changed it slightly, but you didn't have a third question. If you look at it in general, the approach is the same, very similar, but in fact, a question has been slightly changed. If you are interested, you can click on the third question I answered to take a look.

1, when 00, 10

f(x1)-f(x2)<0, so f(x) is an additive function over the interval (-1,0).

In the same way, it can be proved that f(x) is an increasing function over the interval (0,1). (odd function) 3, in fact, is to find the range of f(x) in (-1,1). Since the odd function has a good single increment, only the minimum and maximum values are required, f(-1)=-2 5, f(0)=0 [because it is an odd function], f(1)=2 5, so a belongs to the interval (-2 5, 2 5), and f(x)=a has a solution.

1) x (-1,0), -x (0,1).

f(-x)=2^(-x)/(4^(-x)+1)=2^x/(4^x+1)

f(x)=-f(-x)

2^x/(4^x+1)

x∈(-1，0))

For odd functions, f(-x)=-

f(x)f(-0)=-

f(0)f(0)=0

In summary, when x (0,1), f(x)=2 x (4 x+1)x=0, f(0)=0

x (-1,0), f(x) =

2^x/(4^x+1)

2) 01, so f(x1)-f(x2)>0

f(x1)>f(x2)

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

Because the function is odd, f(x) is also a subtraction function on (-1,0).

Thus, f(x) is a subtractive function at (-1,1).

3) When x (0,1), f(x)=2 x (4 x+1)=1 (2 x+2 (-x)).

2 x+2 (-x) 2,0<1 (2 x+2 (-x)) 1 201 2, because the function is odd, and when x (-1,0), -1 2 f(x)<0 and f(0)=0

So the function range is [-1 2,1 2].

1. From the meaning of the question, f(0)=0

Let x (0,1), then -x (-1,0), then there is.

f(-x)=2^(-x)/(4^(-x)+1)=2^x/(4^x+1)

And because f(-x) = -f(x).

So -f(x)=2 x (4 x+1).

So the f(x)= -[2 x (4 x+1)] function f(x) is a three-segmented piecewise function on the defined domain.

when x (-1,0), f(x)=2 x (4 x+1);

When x=0, f(0)=0;

When x (0,1), f(x) = -[2 x (4 x+1)]2, the monotonic interval of the original function should be in (-1,0) as the increasing function.

On (0,1) is also an increasing function.

1) x (-1,0), -x (0,1).

f(-x)=2^(-x)/(4^(-x)+1)= 2^x/(4^x+1)

f(x)=- f(-x) =- 2^x/(4^x+1) (x∈(-1，0))

For odd functions, f(-x)=- f(x).

f(-0)=- f(0) f(0)=0 When we know that x (0,1) is in summary, f(x)=2 x (4 x+1)x=0, f(0)=0

x (-1,0), f(x) = - 2 x (4 x+1)(2)01, so f(x1)-f(x2)>0 f(x1)>f(x2)f(x) is a subtractive function on (0,1).

Because the function is odd, f(x) is also a subtraction function on (-1,0).

Thus, f(x) is a subtractive function at (-1,1).

3) When x (0,1), f(x)=2 x (4 x+1)=1 (2 x+2 (-x)).

2 x+2 (-x) 2,0<1 (2 x+2 (-x)) 1 20 Because the function is odd, when x (-1,0), -1 2 f(x)<0 and f(0)=0

So the function range is [-1 2,1 2].

1.Odd function =>f(0) =0

x ∈ 0,1], f(x)=(2^x)/(4^x+1);

x ∈[1,0), f(x) =f(-x) =2^(-x) /4^(-x) +1] =2^x / 1 + 4^x)

2.Let a, b (0,1), a < b,f(a) -f(b) =2 a ( 1 + 4 a) -2 b 1 + 4 b).

2^a (1 + 4^b) -2^b (1 + 4^a) ]1 + 4^a)(1 + 4^b)]

2^a - 2^b ] 1 - 2^(a+b) ]1 + 4^a)(1 + 4^b)]

0 ∵ 2^a < 2^b, 2^(a+b) >1

3.f(x)=x+b always has a solution to a real number on [-1,1], i.e., at least one of the following three equations has a solution.

1） x ∈ 0,1], f(x) =2^x / 1 + 4^x) =x + b ①

f(x) single decrease, and -1 < f(x) 2 5 , x+b single increase, x+b b, 1+b].

b < 1, 1+b ≥ 2/5 =>3/5 ≤ b < 1

2） x ∈[1,0), 2^x / 1 + 4^x) =x + b ②

f(x) single increase, and 2 5 f(x) <1 , x+b single increase, x+b 1+b, b].

Spike -1+b 2 5, b > 1 =>1 < b 3 5

3） x = 0, 0 = x + b ③

b = 0 so: -1 < b < 1

(1) When x (0,1), -x (-1,0), bring -x into f(x)=1 4 x-a 2 x, and get f(-x)=-a2 x+2 2x=-f(x).

f(x)=a2^x-2^2x

2) Discuss, axis of symmetry x=a2, when a2 [-1 2), f(x)max=4-2a

When a2 [1 2, ]f(x)max=1-a(3), i.e. a2 is less than 0, a0

Solution: (1) The function f(x) is an odd function defined on [-1,1], again =1-a=0

The solution gives a=1, which is the analytic formula when x [-1,0].

When x [0,1], -x [-1,0].

4x-2x=-f（x）

f（x）=2x-4x（x∈[0，1]）

2) When x [0,1] is obtained from (1), f(x)=2x-4x makes t=2x(t [1,2]).

then 2x-4x=t-t2, so y=t-t2(t [1,2]).

Then it is easy to obtain that when t=1, y has a maximum value of 0

The maximum value of f(x) on [0,1] is 0

Yes, the maximum buffer is when the acid and the conjugate base are one-to-one.

1) Since the function is an odd function on (-1,1), therefore f(0)=0 ;

When -1 therefore f(x)= -f(-x)= -2 [(x) 2-2*(-x)]= -2 (x 2+2x), so the analytic expression of the function on (-1,1) is.

2^(x^2+2x) （1f(x)= ｛0（x=0）；

2^(x^2-2x) 。This is a piecewise function, written in three lines with a curly brace in front of it).

2) When -1 is x=0, f(x)=0 ;

When 0, the function range is (-1,-1, 2)u{0}u(1, 2,1).

(1) Because of the odd function, f(-x)=-f(x). f(x) on (0,1) brings x=-x into the calculation, f(-x)=(2 x) (4 x+1) so f(x)=-(2 x) (4 x+1).

In this case, the definition field is that x belongs to (0,-1). Because it is an odd function and has a defined domain at point 0, f(0)=0. That's it.

2) On (0,1), divide the numerator-denominator colleague by 2 x to get 1 (2 x+2 -x).The derivative of the denominator yields f(x)>0If the denominator is the Zeng function, then the original function is a subtraction function.

And because the function on the defined domain is an odd function, it is a subtractive function on (-1,0).

3) Simplify the inequality first, get a simple problem like f(x 2-2) and send it out, and then use your brain more.

odd function, so f(b)=-f(-b).

f(a)+f(b)]/a+b)>0

f(a)-f(-b)/(a+b)>0

Consider a>-b

then a+b>0

f(a)-f(-b)>0

f(a)>f(-b)

The same can be proved: a<-b, a+b<0, f(a)-b, there is always f(a)>f(-b).

So f(x) is an increasing function.

2.If f(x)<=m 2-2am+1 is [-1,1] constant for all x, find the range of m values.

Because f(x) is an increasing function, so.

f(x)≤f(1)=1

f(x)<=m 2-2am+1 for all x belongs to [-1,1] constant, then.

m^2-2am+1≥1

m^2-2am≥0

m(m-2a)≥0

Add to that for all a belongs to [-1,1] is true, right?

Then start the discussion with a=0, and then comprehensively intersect.

a>0m 2a or m 0

This is also true if a=1 is included. So.

m 2 or m 0

a=0 then m is an arbitrary real number.

A<0M0 or M2A

a=-1 is also included, then.

m 0 or m -2

Take the intersection. m 2 or m 0

m 2 or m -2