Abstract function inequalities, abstract functions and function inequalities

Solution: (1) Because f(x y) = f(x)-f(y)(x,y＞0).

Therefore, if x=y 0, then there is f(1)=f(x)-f(x)=0i.e. f(1)=0.(2).

Because f(x y) = f(x)-f(y) and f(1) = 0Therefore, if x=1,y0, then there is f(1 y)=f(1)-f(y)=-f(y)That is, there is f(1 x) + f(x) = 0

x＞0).f(6)=1Therefore, the original inequality can be reduced to f(x+3)+f(x) 2f(6)

=>f(x+3)-f(6)＜f(6)-f(x).===>f[(x+3)/6]＜f(6/x).Since on r+, the function f(x) is incremented, so there is 0 (x+3) 6 6 x

=>0＜x(x+3)＜36.===>0＜x＜(-3+3√17)/2.That is, the solution set is (0,(-3+3 17) 2).

By definition: 1<=x-1 2<=1 --1

1<=1/4-x<=1 --2

From the odd function, we know that f(x)=-x

So f(x-1 2)<-f(1 4-x)=f(x-1 4), i.e., x-1 21 2<=x<=5 4 from the increasing function

The first question is to join the group.

Set the orange x>0

m+x>m

f(m+x)=f(m)+f(x)-1

When x>0, f(x)>1

So f(x)-1>0

So f(m+x) > f(m).

So f(x) is an increasing function on r.

Second question. f(3)=f(1)+f(2)-1 =f(1)+f(1)+f(1)-1-1

f(3)=4

3f(1)-2=4

f(1)=2

f(a^2+a-5)<2

f(a 2+a-5) a 2+a-5

3

Solution: (1) Because f(x y) = f(x)-f(y)(x,y＞0).

Therefore, if x=y 0, then there is f(1)=f(x)-f(x)=0i.e. f(1)=0.(2).

Because f(x y) = f(x)-f(y) and f(1) = 0Therefore, if x=1,y0, then there is f(1 y)=f(1)-f(y)=-f(y)That is, there is f(1 x) + f(x) = 0

x＞0).f(6)=1Therefore, the original inequality can be reduced to f(x+3)+f(x) 2f(6)

=>f(x+3)-f(6)＜f(6)-f(x).===>f[(x+3)/6]＜f(6/x).Since on r+, the function f(x) is incremented, so there is 0 (x+3) 6 6 x

=>0＜x(x+3)＜36.===>0＜x＜(-3+3√17)/2.That is, the solution set is (0,(-3+3 17) 2).

Let x=y bring f(x y)=f(x)-f(y), and get f(1)=0f(x+3)-f(1 x)-2 < 0

f(x+3)-f(1 x)-f(6)-f(6) <0f[(x+3)x 36] <0 = f(1) multiplication, so (x+3) x 36 <1

x 2 + 3x-36 < 0 is fine, pay attention to x>0

The so-called abstract inequality is an inequality in which a concrete algebraic formula is not explicitly given on either side of the inequality.

For example, let f(x) be an increasing function defined on r, satisfying f(1)=2, and for any real number x,y, there is f(x+y)=f(x+f(y).

Solve the inequality f(x-1)+f(2x)>4

Solution: The original inequality can be reduced by the condition f(2)=f(1+1)=f(1)+f(1)=4.

f(x-1 +2x)>f(2)

again f(x) is the increasing function, thus.

x-1 +2x >2

Solve x>1

It's just that there are some high-level multiples that you can't apply to basic functions.

In the first problem, the inequality is deformed into f(1+a)<-f(1-a) Note that f is an odd function, then -f(1-a)=f(a-1) so that the original inequality can be reduced to f(1+a)a -1, note that f defines the domain as -1<1+a<1 and -1f(2x) 1, so 1-x >0 (if not, f(1-x)=1, contradicts f(1-x)>1) and note that f is monotonically not strictly increasing, so there is 1-x >2x and thus 1- 20 knows -1, so the range of x is (1-2,1).

Think of the side of x as a whole that satisfies the domain of definitions.

0,1] means that the x square is greater than or equal to 0 and less than or equal to 1

That is, 0 x 1, the domain of f(x) is a, then the domain of f[g(x)] is g(x)=a

In the first question. Transform the inequality into f(1+a)<-f(1-a) Note that f is an odd function, then -f(1-a) = f(a -1) so that the original inequality can be reduced to f(1+a).

a -1, note that f defines the domain as -1<1+a<1 and -1f(2x) 1, so 1-x >0 (otherwise f(1-x finch) = 1, which contradicts f(1-x pinch) >1) and note that f is monotonically not strictly increasing, so there is 1-x and 2x and thus 1-2