The so-called application of function thought is to construct the corresponding function for a practical or mathematical problem, so as to solve the problem faster and better. Constructor is an important embodiment of function thought, and the application of function thought should be good at grasping the unchangeable laws and properties of things in the process of motion.
Let's briefly introduce how to solve problems such as equations, inequalities, series, and parameter ranges by using function ideas.
First, solve the equation problem with the idea of function.
Function and equation are not only two different concepts, but also closely related. If a function can be expressed by an analytical expression, then this expression can be regarded as an equation; There is a corresponding relationship between the two unknowns of the binary equation. If this correspondence is single-valued, then this equation can also be regarded as a function. The two ends of an equation can be regarded as functions respectively, and the solution of the equation is the abscissa of the intersection of two function images. Therefore, many problems related to equations can be solved by functional ideas.
Example 1 Prove that no matter what real number A takes, the equation x2-(a2+a) x+a-2=0 must have two unequal real roots.
Analysis: If this problem is solved by conventional methods, the discriminant △ is a univariate quartic polynomial about A, and the sign is difficult to judge. If we analyze the meaning of the problem with the idea of function, let the function f(x)=x2-(a2+a)x+a-2. In order to prove the proposition, we only need to prove that the image of the function y=f(x) has two intersections with the X axis. Because its opening is upward, we only need to find a real number X0 to make f (x0).
Example 2 It is known that the real coefficient quadratic equation x2+ax+b=0 about X has two real roots α, β. Facts have proved that:
(i) If | α|
(II) If 2 | a |
|β| & lt； 2;
Analysis: On the surface, this problem is an equation problem. If it is treated by pure equation theory, the relationship between the distribution of equation roots and parameters A and B is very complicated. If we use the thought of function to analyze and transform the distribution problem of equation roots into the intersection of function image and X axis, we can grasp the essence.
Answer: The results of (1) and (2) of this question are
2 | a | & lt4+b
{ & lt= = & gtα,β ∈(-2,2)
| b |< Four
Let f(x)=x2+ax+b( I) be known from the image of quadratic function.
f(2)>0
α，β∈(-2，2)= = & gt； { f(-2)>0
|b|=|α？ 6? 1β| & lt； 44+2a+b & gt； 02a & gt； - (4+b)
= = & gt{ = = & gt{
4-2a+b & gt； 02a & lt； 4+b== >2 | a | & lt4+b and | b |
(ii) If {= > {= >; {then
| b | & lt4 4-2a+b & gt； 0 f(-2)>0α, β is within (-2,2) or outside (-2,2). If α and β are outside (-2,2), then | α? 6? 1β| = b & gt； 4, same as | b |
Second, prove the inequality with the thought of function.
Example 3 assumes that a, b and c are all positive numbers, and a+b >; c，
B.C.
Verification:-+->-
1+a 1+b 1+c
B.C.
Analysis: the left and right sides of inequality are similar in structure:-,-,-,because.
1+a 1+b 1+c This can be related to the function f (x) = x/( 1+x) (x >); 0) Monotonicity.
Prove that the prior function f (x) = x/(1+x) (x >; 0) Monotonicity.
x 1 & gt； 0, x2>0, we might as well set x 1
Then f (x 1)-f (x2) =-.
1+x 1 1 x2( 1+x 1)( 1+x2)∵x 1 & gt； 0，x2 & gt0∴ 1+x 1 & gt； 0， 1+x2 & gt； 0
And \x 1
x 1- x2
∴-& lt； 0
( 1+ x 1)( 1+ x2)
That is, f (x 1)
∴ The function f(x) monotonically increases at (0, +∞).
∵a+b & gt； c & gt0 ∴f(a+b)>； f(c)a+b c
That is,->; -
1+a+b 1+ca b a b
∵-+-& gt； - + - = -
1+a 1+b 1+a+b 65438+a+b 1+a+b a b c
∴ - + ->-
1+a 1+b 1+c Example 4 It is known that A, B, X and Y are all real numbers, a2+b2= 1, x2+y2= 1. Verification: ax+by≤ 1.
Analysis: Under the known conditions, if the sum of squares is equal to 1, we can associate the square relation between sine and cosine, and then use the boundedness of the function to prove it.
Proof: ∫a2+B2 = 1, x2+y2 = 1.
∴ let a=sinα, b=cosα, x=sinβ, y=cosβ.
Then ax+by=sinαsinβ+cosαcosβ.
=cos(α-β)≤ 1
∴ax+by≤ 1
Third, use the function thought to solve the sequence problem.
A sequence can be regarded as a positive integer set N* (or its finite set {1, 2...n}). When the independent variable takes the value from small to large, the general term formula of the sequence is also the analytical formula of the corresponding function. So some problems of sequence can be solved by function thought.
In example 5 arithmetic progression, the first n terms are Sn, and SP = Q and SQ = P are known.
(p, q∈ N* and p≠q), find sp+q.
Analysis: The conventional solution to this problem is to establish equations with summation formula, then find a 1 and D, and then find Sp+q, but the calculation is very complicated. If we consider that the sum of the first n terms of arithmetic progression is a quadratic function about n, and there are no constant terms. Therefore, the objective function Sn = AN2+BN (A and B are undetermined coefficients) can be considered to optimize the problem solving process.
Solution: Let Sn = AN2+BN (A and B are undetermined coefficients).
Then Sp=ap2+bp ∴ap2+bp=q (1).
Sq=aq2+bq ∴aq2+bq=p (2)
(1)-(2) sorting (P-Q) holding, and (d)
f(-2)& lt； 0 2 x2+2x-3 & gt； 0 √7 - 1 √3 + 1
{ = & gt{ = & gt-& lt； x & lt-
f(2)& lt； 0 2 x2-2x- 1 & lt； 0 2 2√7 - 1 √3 + 1
The value range of ∴x is (-).
2 2
(2) Construct a quadratic function and find out the range of variables.
Example 7 Real numbers A, B, C, D are known, and satisfy a+b+c+d=5.
A2+b2+c2+d2=7, find the range of A.
Solution: construct a quadratic function about x.
f(x)=(x - b)2+(x - c)2+(x - d)2
=3 x2 - 2(b + c + d) x+(b2 + c2 + d2)
∵f(x)≥0 ∴△≤0
That is, 4 (b+c+d) 2-12 (b 2+C2+D2) ≤ 0.
That is 4( 5-a)2- 12(7-a2)≤0.
∴2a2-5a+2≤0
∴ 1/2≤a≤2
The value range of ∴a is [1/2,2], and the words at the beginning and some in the middle are good. Specifically, the points whose coordinates satisfy the resolution function must be on the function image, and the coordinates of the points on the function image must satisfy the resolution function. Therefore, in order to judge whether a point in the plane rectangular coordinate system is on the function image, it is only necessary to substitute the coordinates of the point into the resolution function for testing. 2. Find the coordinates of the intersection of two functions, that is, find the solution of the binary equations composed of two resolution functions. 3. When solving problems related to functions, we should pay attention to the use of plane geometry knowledge such as the right angle between X axis and Y axis in plane rectangular coordinate system and Pythagorean theorem, and we should be able to skillfully find out the intersection coordinates of functions and coordinate axes. 5. According to the concept, properties and images of functions, it is an important method to transform shapes and numbers, shapes and equations, and shapes and inequalities. The concept of function occupies an important position in mathematics. It plays a key role in the whole middle school function teaching. The concept of function and its thinking method have become one of the main lines of mathematics teaching in middle schools. The study of function concept is a leap from students' understanding of concrete quantitative relations in the real world to their understanding of abstract quantitative relations. However, due to the complexity of the concept of function, it has become a difficult point in junior high school teaching. On the basis of previous studies, starting from the concept of function, this paper investigates junior high school students' understanding of the concept of function from three aspects: definition, expression and application, and makes a comparative analysis of the results, and draws the following conclusions: 1 Junior high school students don't have a deep understanding of the essence of the concept of function and can't fully understand the relationship between independent variable X and dependent variable Y, which is related to cultivating students under the requirements of the new curriculum standard. 2. The development of students' recognition of functions represented by graphs and charts obviously lags behind the recognition of functions represented by analytical expressions. 3. Junior high school students' ability to apply the concept of function is low. 4. There are differences in junior high school students' cognitive development level of function, but there is no obvious difference on the whole: (1) Junior high school students are better than junior high school students in using analytical expressions to describe the concept of function; (2) Senior two students are superior to senior three students in the application of charts and images. In this paper, the research results are deeply analyzed, and combined with the teaching practice, the following improvement measures are put forward for the function concept teaching in junior middle schools at this stage: (1) Strengthen the understanding of the essence of the function concept; (2) Strengthen the transformation between function representations; (3) Pay attention to the function model in daily life. These can also be used ~