High School Math Abstract Functions, High School Abstract Functions

Defining Domains: Solving the problem of defining domains of abstract functions - well-defined, equivalence transformations.

Summary: The definition domain of a function refers to the range of values of the independent variable, and the key to finding the definition domain of an abstract function is the equality of the formulas in parentheses (that is, the formulas in parentheses after the same correspondence rule have the same range of values).

Value Range: Solve the problem of the value range of abstract functions - define the domain, correspond to the law decision.

Summary: When the definition domain of the function and the corresponding law do not change, the value range of the function does not change.

Clarify what is called at standard x.

For example, if you know that f(x) defines the domain, find f(x+1) to define the domain. Finding the f(x+1) definition domain is finding the definition domain of x. x+1 occupies the standard x, so x+1 conforms to the domain defined by f(x), which is the range of x in f(x).

For example, given that f(x) defines the domain as (-1,1), find f(x+1) to define the domain? Answer: -1 x+1 1, we can find x (-2,0), that is, f(x+1) defines the domain as (-1,0).

The definition domain refers to all the values that can be taken from the independent variable (usually x) in the function, such as the function: y=1 x, when x is 0, the function is meaningless, and other values can be taken, so its definition domain is (- 0)u(0,+

The value range refers to all the values that can be taken by the dependent variable (usually y) in the defined domain, such as the function: y=1 x, when x infinity approaches 0, y tends to negative infinity or positive infinity, and when x is infinite, y approaches 0 but does not equal to 0, so the value range is also (- 0)u(0,+

logax+logay=3 <==> loga(xy)=3 <==> y=a 3 x (a>1,a<=x<=a 2), so it is part of the upper branch of the hyperbola, for this part there is a maximum value when x takes the minimum value, and when x takes the maximum value, y gets the minimum value (monotonically decreased), when x=a = a 2, when x=a 2 y=a 2, that is, a 2 2<=y<=a 2, and a<=y<=a 2, so a 2 2>=a ==>a>=2. Choose B

Solution: It is easy to know that a(x)+ a(y)=3===> a(xy)=3===>xy=a===>y=a³/x.

again a x 2a(a＞1)===>1/a≥1/x≥1/(2a)===>a²/2≤a³/x≤a².i.e. a 2 y a

From the title, a y a∴a≤a²/2≤y≤a².===>a≥2.I choose B

The argument of f(x+1) is x, not x+1, so.

It should be f(-x+1)=-f(x+1).

Instead of f(-x-1) = -f(x+1).

If f(-x-1)=-f(x+1), it only means that f(x) is an odd function, but not that f(+1) is an odd function.

If for g(x) any value x within the defined field, it is satisfied.

g(x)=-g(2a-x) (Of course, writing g(x+a)=-g(-x+a) is the same.) ）

Then, g(x) is symmetrical with respect to the center of (a,0), which has nothing to do with what "left plus right subtraction" is not dismantling high.

So raise the ruler of the imperial stool.

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

f(-x-1)=-f(x-1) (2)

For equation (1), let x=x+2, get.

f(-x-1)=-f(x+3), compared with equation (2).

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

So the period t=4 of f(x) is nothing else, it is f(x) so, for equation (2), +4 in parentheses

f(-x+3)=-f(x+3)

So, f(x+3) is an odd function.

Abstract functions can be done with some typical examples, and then memorize and understand some special abstract function solving techniques.

It is nothing more than defining the domain value range (i.e., boundedness) around monotonicity, parity, periodicity

Solution: Take x1 x2

f(x)+f(y)=2+f(x+y) (You're wrong).

f(x+y)= f(x)+f(y)-2

f(x2)= f(x2+x1-x1) =f(x2-x1) +f(x1)-2=+f(x1)+f(x2-x1) -2

where x1 x2, x2-x1 0, so f(x2-x1) 2 is f(x2-x1) -2 0

So f(x2) f(x1).

So the function f(x) is a monotonic increasing function in r.

f(1)+f(1)=2+f(2)

f(3)+2=f(1)+f(2)

f(3)=5

The solution yields f(1)=3

For inequalities f(x -2x-1)<3=f(1) (for special functions, note the definition of x -2x-1 in the function).

It is obtained from the monotonicity of f(x).

x²-2x-1<1

af(x)+bf(1/x)=cx (1)

Let p=1 x, then x=1 p

So af(1 p) + bf(p) = c p

So af(1 x) + bf(x) = c x (2) (1) a-(2) b

a -b )f(x)=acx-bc x=(acx -bc) x, so f(x)=(acx -bc) [(a -b)x].

Specific ideas:

This problem should be done using the system of equations.

That is, f(1 x) and f(x) are treated as two unknowns to solve the system of equations

af(x) +bf(1 x) = cx (1) substituting x=1 x, where 1 x=x

af(1 x) + bf(1 (1 x)) = c x (2) synapsis (1) (2).

f(x)=(bc x-acx) (b 2-a 2);

In fact, the idea of this kind of topic is far more important, as long as there is an idea, in a slightly more flexible application, this kind of topic will not be a problem in the future!