Find the definition domain and value range of the function, and how to find the definition domain an

Knowing the analytic formula to define the domain: just make sure that the formula is meaningful, for example, the denominator is not 0, the base is not 0 under the even root number, the base of the 0 power is not 0, the true number of the logarithmic formula is greater than 0, the base is greater than 0 and not 1, etc.

Therefore, the above two functions define the domain as r

1) When the domain of the primary function is r, the range is also r (the image is a straight line, and y can be taken as a real number) 2) The range of the quadratic function is matched.

y=x²-6x+7=（x-3)²-2

Since (x-3) 0, the range is [-2,+

1) Define the domain (negative infinity, positive infinity).

Since the function is a monotonically decreasing function, it is easy to know that the range is (negative infinity, positive infinity)2) and defines the domain (negative infinity, positive infinity).

The know-it-all function is a parabola with an opening pointing upwards, so the range should be from nadir to positive infinity. Its lowest point is at the axis of symmetry x=-b (2a), that is, there is a minimum value of y=-2 when x=3So the final range is: [-2, positive infinity).

y=-4x+5 is the same as f(x)=-4x+5, and the domain of function (1) is negative infinity to positive infinity as well as the value range.

Function (2) can be written as:

y=（x-3）²-2

Because (x-3) 0, y -2;

So its definition range is negative infinity to positive infinity, and the value range is.

4x+5=x^2-6x+7 x^-2x+2=0 (x-1)^2=0

x=1y=1

There is only one number in the defined range of values.

Finding the definition domain of a function requires the following aspects:

1) The denominator is not zero positive (2) The number of open squares of the even root formula is not negative.

3) logarithmic.

The true part of the number in is greater than 0

4) The base of exponential and logarithmic is greater than 0 and not equal to 1

5）.x≠k + 2 in y=tanx, x≠k in y=cotx, etc.

The range is the range of y in the function y=f(x).

Commonly used methods for evaluating ranges:

1) Slag annihilation and attribution; (2) image method (combination of numbers and shapes), 3) function monotonicity method, 4) matching method.

5) Substitution method, (6) Inverse function method (inverse method), (7) Discriminant method, (8) Composite function method, (9) Trigonometric substitution method, (10) Fundamental inequality method, etc.

There are several ways to determine the definition of a function:

1) If f(x) is an integer, then the domain is defined as r;

2) If f(x) is a fraction, then its defining domain is the set of real numbers whose denominator is not 0;

3) If f(x) is an even radical, then its definition domain is the set of real numbers such that the sub-formula under the root number is not less than 0;

4) If f(x) is composed of several parts, its defining domain is the set of real numbers that make each part meaningful;

5) In practical problems, the practical significance should be considered in determining the definition domain.

Finding the range of values of a function is a complex problem, although the domain of the function is completely determined once the domain of the function and its corresponding rules are given.

When evaluating a range, the common methods are:

1) Observational method.

2) Matching method.

3) Discriminative method.

4) Substitution method.

In addition, there is also the maximum value method, the combination of numbers and shapes, etc.

How to find the definition domain of a function, mathematical knowledge.

Finding the Definition Domain: When there are no special requirements for the problem, the definition domain of the function is the range of x values that make the function expression meaningful. In order to ensure that the expression is meaningful, there are a few things to keep in mind:

1. The denominator is not 0;

2. The number of square roots to be opened is greater than or equal to 0;

3. The true number of logarithmic formula is greater than 0;

4. The base of the zero power and the negative power is not 0;

5. The tangent corresponding angle is not equal to 2+2k. Common methods for evaluating ranges:

1. The matching method is mainly for quadratic functions.

2. The separation constant method, mainly for fractional functions.

3. The commutation method is mainly aimed at functions that appear multiple times in a certain algebraic formula.

4. The monotonicity method can evaluate the monotonicity in the domain by the function in the definition domain.

5. Discriminant method, not commonly used.

6. Image method, combination of numbers and shapes and evaluation range.

The function definition domain is the range of values of the arguments that make the analytic expression of the function meaningful, and the actual meaning of the arguments.

The common analytic formulas of functions in high school generally include: fraction (ensure that the denominator is not 0), quadratic radical (ensure that the open method is greater than or equal to 0), logarithmic (ensure that the true number is greater than 0), exponential note: 0 power of 0 is meaningless, trigonometric function, pay attention to the tangent function, the terminal edge of the angle cannot be on the y-axis, and the combination of the above forms (ensure that all kinds of expressions are meaningful at the same time).

The general method of the range is to find it according to the monotonicity of the function.

The definition domain is the range of the independent variable x in the function y=f(x).

Finding the definition domain of a function requires the following aspects:

1) The denominator is not zero.

2) The number of open squares of the even root formula is not negative.

3) The true part of the logarithm is greater than 0.

4), the base of the exponent and logarithm is greater than 0 and is not equal to 1 (5). x≠k + 2 in y=tanx, x≠k in y=cotx, etc.

The range is the range of y in the function y=f(x).

Commonly used methods for evaluating ranges:

1) Naturalization; (2) image method (number combination), 3) function monotonicity method, 4) matching method, (5) commutation method, (6) inverse function method (inverse method), (7) discriminant method, (8) composite function method, (9) trigonometric substitution method, (10) basic inequality method, etc.

First of all, it must be clear that a function is composed of independent variables, correspondence laws, and definition domains, and as long as these 3 are determined, the value of the function is also determined. For example, if the independent variable is the denominator, it cannot be zero, and under the even power root, the number of squares should be greater than zero, so the first thing to be satisfied by defining the domain should be the objective existence of the independent variable, and the first thing to consider is those special forms, such as fractions, radicals, etc., which rely on accumulation; There is another category, which is to ensure the objective existence of graphs, such as ellipses and hyperbolas, the definition domain of these two functions depends on the graph, according to the graph, this mostly depends on memory. Therefore, the method of finding the definition domain is, first, to look at the objective existence of the independent variable, secondly, to draw a diagram to ensure the objective existence of the graph, and finally to find the intersection of the two, we can get the definition domain.

As for the function value, it depends on the definition domain and the corresponding law, and with the constraints of the two, it is possible to find the correct function value.

In addition, when solving the problem of functions, it is necessary to draw a diagram, and the combination of numbers and lines is one of the four major mathematical methods, and its application is very wide.

The function has two variables x, y. Generally y changes with x. x is called the independent variable and y is called the dependent variable. y is a function of x.

denote as y=(x). The range of the value of the argument variable x is the domain of definition, and the range of the value of function y is the range of values.

For example: y=2x This is a quadratic function. The value range of the argument x is arbitrary real, which means that the definition domain is arbitrary real.

The value range of the function y is a real number greater than 0, that is, the value range is a real number greater than or equal to 0.

Good luck with your studies!

There seem to be many ways to evaluate the domain of the function and find the domain of the definition. I've only mastered 2 types, so who can help me list the law? Defining Domains: The first thing to do is to understand the definition domains for each basic function. composite functions, to take into account.

Dizzy Read a good book Ask classmates and teachers A dead skin is not afraid of boiling water, and I will ask until I will meet.

How to find the definition domain of a function, mathematical knowledge.

The definition range refers to the range of values of the independent variable, and the range of values refers to the range of values of the entire function. Generally, the range is found according to the defined domain, or vice versa.

1+2sin(2x+ 3)≠0, 2x+ 3≠2k +7 6 and 2x+ 3≠2k +11 6

x≠k - 4 and x≠k +5 12

f(x)=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))

Let 2x+3=+6

f(x)=(√3-2cos(θ+/6))/(1+2sin(θ+/6))

√3-√3cosθ+sinθ)/(1+√3sinθ+cosθ)

sinθ/(1+cosθ)=(1-cosθ)/sinθ=tan(θ/2)

From the sum ratio theorem, we get f(x)=tan(2)=tan(x+12).

x≠k - 4 and x≠k +5 12

f(x)≠-3/3

Solution: To make a function meaningful, then.

1+2sin(2x+π/3)≠0

sin(2x+π/3)≠1/2

2x+3≠2k+6 and 2x+3≠2k+5 6

Solution; x≠k - 12 and x≠k + 4 (k belongs to z).

3-2cos(2x+π/3)=√3-2[cos²(x+π/6)-sin²(x+π/6)]

3[(cos²(x+π/6)+sin²(x+π/6)]-2[cos²(x+π/6)-sin²(x+π/6)]

√3-2)cos²(x+π/6)+(3+2)sin²(x+π/6)

1+2sin(2x+π/3)=1+4sin(x+π/6)cos(x+π/6)

cos²(x+π/6)+sin²(x+π/6)+4sin(x+π/6)cos(x+π/6)

y=(√3-2cos(2x+π/3))/(1+2sin(2x+π/3))

(√3-2)cos²(x+π/6)+(3+2)sin²(x+π/6)]/[cos²(x+π/6)+sin²(x+π/6)+4sin(x+π/6)cos(x+π/6)]

√3-2+(√3+2)tan²(x+π/6)]/[tan²(x+π/6)+4tan(x+π/6)+1]

So, [tan (x+ 6)+4tan (x+ 6)+1]y=[ 3-2+( 3+2)tan (x+ 6)]

Shift: (2+ 3-y)tan (x+ 6)-4ytan(x+ 6)-y-2+ 3=0

b²-4ac>=0

That is: (-4y) -4(2 + 3-y)(-y-2+ 3)>=0

Solution: 12y +8 3y + 4>=0

4(√3y+1)²>=0

The equation is constant, so y belongs to r

1+2sin(2x+ 3)≠0, sin(2x+ 3)≠-1 22x+ 3≠2k - 6 and 2x+ 3≠2k -5 6; x≠k - 4 and x≠k -7 12 (k belongs to z)y=( 3-2cos(2x+ 3)) (1+2sin(2x+ 3))=( 3 2-cost) (1 2+sint) represents the slope of the line connecting two points a(-sint, coct) and b(1 2, 3 2) on the circle (a, b do not coincide), a, b coincide indicates the slope of the tangent, so y≠- 3 3

Solution: The analytic formula of the function contains a fraction, and the independent variable only needs to satisfy that the denominator is not equal to 0, that is, 1+2sin(2x+3)<>0, sin(2x+3)<>1 2

2x+ 3<>2k +7 6 and 2x+ 3<>2k - 6

The solution is x<>k +5 12 and x<>k - 4