High School Mathematics Defining Domains and Domains .

f(x)=(a-3)(a+1)x +(a+3)x+1a=3 or -1, f(x)=6x+1 or 2x+1, obviously both the domain and the domain are r

When a=-3, f(x)=12x +1, the range is [1, not r.

So a≠-3

When a≠3, -1 or 3, f(x) is a bisecond function, and the range is constant r

The value range of a is as follows:

If the range is r, it cannot be a quadratic function. Therefore, the coefficient of the quadratic term is zero.

i.e. (a 2 - 2 a - 3) = 0

Cross multiplication. (a+1)(a-3)=0

a=-1 or a=3

When a=3, the original formula = 1, contradicts the range of r and rounds.

In summary, a=-1

If it is a quadratic function, then the range cannot be r, so it is a primary function.

Let me still say it.

Categorical discussions. If a2-2a-3≠0, then f(x) is a quadratic function, and it is impossible to define both the domain and the range of values as r

If a2-2a-3=0, i.e. a=-1 or 3;

When a=3, f(x)=6x+1 conforms to the topic;

When a=-1, f(x)=-4x+1 conforms to the meaning of the question, so the value range of a is .

f(x)=(a-3)(a+1)x²+(a+3)x+1

When a=-3, f(x)=12x +1, the range is [1, not r.

The value range of a is as follows:

Molecules: 3-x>0, x<3.

Denominator: |x|-1>0，|x|>1, x>1 or x<-1.

A combination of the two: <>

So the domain is defined as: 1 is represented by a set: x.

Since the definition domain of lnx is (0, ).

Therefore, 3 x 0, the denominator cannot be zero.

And the root number should be greater than or equal to zero. Combining these three conditions results in a defined domain.

The true number of the numerator is greater than 0, the denominator is not 0 and the root number is greater than or equal to 0, so the denominator is greater than 0, you beg to see??

sinx≧0 y=|sinx|+sinx =sinx+sinx=2sinx

sinx<0 y=-sinx+sinx=0 domain [0,2] The meaning of the absolute value.

There is the formula y=tan(wx+b) for period: period t= wy=tan( x 2+ 3) in w= 2 t= ( 2)=2

The value range of sinx is -1 to +1, and when sinx 0, 0 2sinx 2, so it is natural, this problem is divided into sinx 0 and <0 two cases to discuss, and the final result is 0 y 2

f(x)=tg( x 2+ 3+ )=tan【 2(x+2)+ 3】=f(x+2)There is a word limit, so I won't go into details.

1. When sinx 0, sinx [0,1], then 2sinx [0,2].

2. The periodicity of the tangent function is divided by the coefficient of tan(ax): a; For this question: (2)=2

The first one can't read the question clearly, and the second, you think, is f(x)=f(x+2) not a formula? It's the periodic formula of the function.

The definition domain is the range of x (generally speaking), the value range is the range of y, and the high school generally requires the definition of the domain of the positive many, which is generally evaluated according to the definition domain, and there are also requirements for the value range related to the actual problem. For all function problems, first look at the definition domain of the problem, no matter what it is looking for, first figure out the definition field, absolutely only the right and not wrong. Everything that follows is discussed within the definition domain.

Defining the domain is within that scope of the problem has a beam kernel meaning. The value range is the value that x can get when the domain is defined, and the value that y can get is a prerequisite for a problem, and it is necessary to give priority to rubber, in high school mathematics, a function problem, if the problem of defining the domain is not considered, the score will be very low, and as long as you pay attention to the definition domain, there will be a score.

There is no specific theory to understand this, listen to the lecture, and then try to do the problem, all high school students learned to define the domain in the first year of high school, but they still often make this mistake in the third year of high school. Don't be too anxious about what to do, this is just accumulated with certain mistakes, and you will experience it after a period of time. The last sentence:

Consider defining the domain to form a conditioned reflection to see the function.

To define the domain, you only need to keep in mind a few special cases: the denominator of the fraction is not 0, and the number of open squares of the even square root is greater than or equal to 0, which is analyzed according to the specific situation;

The value range can be determined by looking at the relationship between the defined domain and the function and the collapse of the function. Also filial piety can be found according to the relationship of the inverse function into the orange line, or according to the image.

Let a and b be two sets of non-empty numbers, and a mapping from set a to set b is called a function from set a to set b. Denoted as f:x y=f(x),x a.

where a is called the definition domain. Usually, it is denoted by the letter D. Usually, the defined field is the range of x values in f(x).

1. Given a defined domain: For example, if the function y=2x-1, the defined domain of x is a given set.

2. The definition domain of a general function: a real number that makes a function meaningful. For example, the function y=1 x is defined in the domain of . r is any real number. It can also be written as x (—0) (0,+.)

3. Practical problem: Define the domain according to the specific situation.

This function is in the image.

The domain of definition is one of the three elements of the function (definition domain, value range, and corresponding law), which corresponds to the object of the law. There are three main types of questions in the function definition domain: abstract functions, general functions, and function application problems.

Pay attention to the following points when seeking to define a domain:1: Negative numbers cannot be turned to the even power

The denominator is not 03: The true number in the logarithm should be greater than 0, and the envy key should be buried4: Contains radicals, fractions, and logarithms to take the intersection of each defined domain

5: If the variable has practical significance, it should be consistent with the actual meaning.

The definition domain is like the security gate of an airport station"Width".

f(x) is defined by [-2,4], for security gates"Width"Yes [-2,4].In the function y=f(x)+f(-x), x, -x must pass the security check, i.e.

2<=x<=4, -2<=-x<=4, so -2<=x<=2

In this question, 0<=x<=1, 2x+1 represents the width of the security gate. i.e. the domain of f(x). And 1< = 2x + 1< = 3So the domain of the function y=f(x) is [1,3].

The domain of the function y=f(x) is [0,2], i.e. the width of the security gate is [0,2].2x need to go through security: 0<=2x<=2,0<=x<=1, again.

x is not equal to 1, so the domain is defined as [0,1).

x , lim[(ax 2+8x+b) (x 2+1)] = a , 1 y " 9 , a = 9.

g(x) = (9x^2 + 8x + b)/(x^2 + 1)= /(x^2 + 1)

g(x)min = 1 ， b - 16 9)] (x 2 + 1) = 1, and at this point 3x + 4 3) = 0, i.e. x = -4 9 b - 16 9) = 97 81 , b = 241 81 f(x) = [9x 2 + 8x + 241 81)]= = 64 - 964 9)< 0 , x r f(x)" 97 81) = ( 97) 9 i.e. f(x) range is: [(97) 9 ,+

First turn the range of g(x) back into the range of the defined domain (i.e., the group of inequalities) and then put it in f(x).

I won't forget it.

I didn't understand the one below.