Three Elements of Fractions
Three Elements of Fractions:
Three Elements of Functions (Domain of Definition, Domain of Values, and Correspondence)
1. Domain of Definition; it is the range of the function’s independent variable x. Usually need to consider the following seven cases
Of these seven cases, only the 6th compound function definition domain problem is a bit difficult, the others are very simple. The key to solving problems on the domain of a composite function is to really understand what a composite function is.
Compound function: simpler to understand, a function of another function to take up the position of the independent variable and the composition of the new function.
Shape: f[g(x)]
2. function of the domain
Function of the domain of the function is the range of function y, the domain of the problem can be difficult to be simple, the method can be flexible, but still can summarize some of the method of law out.
For the seven basic elementary functions, as well as their simple deformations, can be directly observed or function images to solve
For the composite function can be solved by the method of commutation
For the fractional function can be considered to use the separation of constants to solve for the domain of the value
Use of monotonicity can be used to find the domain of the value
Use of geometric models or Boundedness, etc. to find the value domain
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3. Correspondence (function analytic)
Function analytic is also a class of test questions, the overall difficulty is not low, the common methods:
For the known type of function, you can be set up, and then find out which the unknown letter
For known Related composite function of the analytic formula, the available permutation method or matching method
For similar equations can be replaced, the available system of equations method
Using the method of assigning a special value to find the analytic formula of the function
Definition domains of common functions
Definition domains of common functions
1, Fractional function 1/f(x) type. It is sufficient to solve for the denominator f(x)≠0;
2. Irrational functions √f(x)-type. Solve f(x) ≥ 0;
3, logarithmic function type, solve the true number formula >0, the base formula >0 and not 1;
4, tangent function tanf(x) type. Solution f(x)≠kπ+π/2,k is an integer.
Generally, the actual solution is a synthesis of several types of problems, therefore, it should be applied comprehensively.
Recognition of the definition domain of a function
We can recognize f(x) in the following ways.
First: the knowledge of algebraic formulas. Every algebraic formula it is essentially a function. Like the algebraic formula x2-1, it is a function whose independent variable is x. For every value of x x2-1 there is a unique value that corresponds to it, so the set of all values of x2-1 is the domain of values of this function.
Second: the knowledge of abstract numbers, for an abstract function without a specific analytic formula, because we do not know its specific correspondence law is also difficult to know its independent variable, the definition of the domain, the value of the domain, it is difficult to understand its symbol and its meaning.
For example, what is the `autonomous variable of f(x+1)? Is its corresponding law still f?The independent variable of f(x+1) is x,its corresponding law is not f.
We may make the following assumptions,if f(x)=x+1,then f(x+1)=(x+1)+1,f(x+1) is algebraically equivalent to (x+1)+1,i.e.,the independent variable of (x+1)+1 is the independent variable of f(x+1). The corresponding rule for (x+1)+1 is to add 1 to the independent variable before squaring it and then adding 1.
Again, is f(x) the same function as f(t)?
Only one particular function must be cited to illustrate this.
Obviously, f(x) and f(t) have the same correspondence law, if the range of values of x is the same as the range of values of t, then f(x) and f(t) are the same function, otherwise, they are functions with the same correspondence law but different domains.
Example: it is known that f(x+1)=x+1, the definition domain of f(x+1) is [0, 2], find f(x) analytic formula and the definition domain
Set x+1=t, then; x=t-1, then denote the function of the independent variable f in terms of t as follows: (that is, substituting x=t-1 into f(x+1)=x+1)
f(t)=f(x+1) = (t-1)+1
=t-2t+1+1
=t-2t+2
So, f(t)=t-2t+2, which gives f(x)=x-2x+2
Or in this way – more intuitively:
Let f(x+1 ) = x-1 in x+1, which is more intuitive, and substituting x = x-1 for f(x+1) = x+1, then:
f(x)=f[(x-1)+1]=(x-1)+1
=x-2x+1+1
=x-2x+2
So f(x)=x-2x+2
So f(x)=x-2x+2
And f(x) and f(t) must have the same range of values for x and t to be the same function,
From t=x+1, the domain of definition of f(x+1) is [0, 2], and it is known that: t ∈ [1, 3]
The domain of definition of f(x) = x-2x+2 is that: x ∈ [1, 3]
In summary, f(x) = x-2x+2 (x ∈ [1, 3]
Concepts of Functions (1) problem for freshman math for ideas and process! Rush rush rush rush! Thanks!
For the fractional function, the ideas are as follows:
1, Df (i.e., the domain of definition) is R, first of all, you can confirm that the denominator can not be 0.
2, consider k>0, at this time the denominator opens upward, and the function image must not be intersected with the X-axis, by the judgment formula b*b-4ac<0 to find
0<k<3/ 4;
3. Consider k=0, when the function is Y=2/3, which satisfies the question.
4, consider k<0, at this time the denominator opening downward, the same function image shall not have an intersection with the X axis, by the judgment formula b*b-4ac<0 to find
0<k<3/4.
5, in summary, the range of values of k [0, 3/4).
If you are not sure, you can follow up.
分式函数的定义域和值域
1.y=(2x+2)/(5x-3)Domainoffunction=R\{3/5}Makexthesubject:x=(3y+2)/(5y-2).Rangeoffunction=R\{2/5}2.y=(2x²-3x+3)/(3x²+x-1)=(2x²-3x+3)/[3(x-(-1+√13)/6)(x-(-1-√13)/6)]Domainoffunction=R\{(-1+√13)/6
(-1-√13)/6}y=(2x²-3x+3)/(3x²+x-1)⇒(3y-2)x²+(y+3)x+(y+3)=0Asxisreal
Δ=(y+3)²-4(3y-2)(y+3)≥0.y²+2y-3≤0⇒(y+3)(y-1)≤0⇒-3≤y≤1.Rangeoffunction={y∈R:-3≤y≤1}3.y=(x+1)/(2x³-x²-8x+4)=(x+1)/[(2x-1)(x-2)(x+2)]Domainoffunction=R\{1/2
2
-2}Whenx=-1
y=0.∴0∈rangeoffunction.Foranyy∈R\{0}
rewritethefunctioninto2x³-x²-(8+1/y)x+(4-1/y)=0….(*)(*)isacubicequationwithrealcoefficients
theremustbearealsolution.Thusy∈rangeoffunction.Inconclusion
rangeoffunction=R.2012-08-2111:37:11补充：多谢指正。现改正值域如下：2.(值域部份)y=(2x²-3x+3)/(3x²+x-1)⇒(3y-2)x²+(y+3)x-(y+3)=0Asxisreal
Δ=(y+3)²+4(3y-2)(y+3)≥0.13y²+34y-15≥0⇒(13y-5)(y+3)≥0⇒y≤-3ory≥5/13.Rangeoffunction={y∈R：y≤-3ory≥5/13}
Segmented function definition domain how to find segmented function definition domain and value domain how to find
1, how to find the definition domain
The basis for finding the definition domain of the function is to make the function’s analytic formula meaningful range of values of the independent variable. The basis for its solution is generally: the denominator is not zero; even root formula, the number of non-negative square; logarithm of the truth is greater than 0.
2, how to find the domain
Seeking segmented function of the domain of the value of the segmented function to be segmented, that is, the segmented function of the function of the individual segments of the function as a function of an independent, and respectively, their domain, then the individual segment on the function of the value of the domain of the concatenation is the value of the segmented function.
3, segmented function definition
Segmented function of the independent variable x for different ranges of values, has a different corresponding law, such a function is usually called segmented function. It is a function, not several functions: the definition domain of the segmented function is the concatenation of the definition domains of the segmented function, and the value domain is also the concatenation of the value domains of the segmented function.
The domain of definition of a general function, to be full
Domain of definition of a general function: all real numbers that make sense of the function. For example, the function y=1/x has a domain of definition of
R for any real number. Can also be written as
1, the natural domain of definition, if the correspondence of the function has an analytic expression to represent, the range of values of the independent variables that make sense of the analytic formula is called the natural domain of definition. For example, the function
To make sense of the analytic expression of a function, the
The natural domain of definition of a function is therefore
2. Functions have a practical context of specific applications. For example, the function v=f(t) represents the relationship between speed and time, and to make sense of the physical problem, the time
Therefore the function has a domain of definition of
3. Artificially-defined domain of definition. For example, when studying a particular function, one is only concerned with a section of the function’s independent variable x in the range [0,10], and therefore defines the function’s domain of definition as [0,10].
Steps to find the definition domain of a function: the definition domain of a function is generally solved by first looking for the constraints in the analytic formula to establish an inequality, and then solving the inequality to obtain the definition domain of the function, when the function y=f(x) is given by the actual problem, pay attention to the practical significance of the independent variable x.
Extended information