2. 1. 1 Function concept and image difficulty: Understand the concept of function and the meaning of the symbol "y=f(x)" on the corresponding basis, and master the solution of function domain and value domain; The mutual transformation of three different representations of functions, the expression of resolution function, and the understanding and expression of piecewise functions; The drawing of function and how to choose points to draw and the understanding of drawing concept. Examination requirements: ① Knowing the elements that make up a function, you can find the definition and value range of some simple functions; ② In actual situations, appropriate methods (such as image method, list method and analysis method) will be selected according to different needs to express functions; ③ Understand the simple piecewise function and apply it simply; Classic example: let the domain of function f(x) be [0, 1], and find the domain of the following function: (1) h (x) = f (x2+1); (2) g (x) = f (x+m)+f (x-m) (m > 0)。 Classroom exercise: 1. In the following four groups of functions, the number of intersections between an image representing the same function and a straight line is () A.B.C.D.2 The function is () A. There must be one B. 1 or two C. There may be at most one D. There may be more than two 3. If the function is known, the domain of the function is () A.B.C.D. Represents the sales of products in each year. The following statement: () (1) The output and sales volume of the products have increased linearly and can still be carried out according to the original production plan; (2) The supply of products has exceeded the demand, and the price will tend to fall; (3) The backlog of products will become more and more serious, and it is necessary to reduce production or expand sales; (4) The production and sales of products are increasing at a certain annual growth rate. What do you think is reasonable? ()A. ( 1)、(2)、(3) B. ( 1)、(3)、(4) C. (2)、(4) D .)。 Then. 8. Specify that the symbol ""represents an operation, that is, if, the value range of the function is _ _ _ _ _ _ _ _. 9. It is known that the quadratic function f(x) satisfies the condition at the same time: (1) the symmetry axis is x =1; (2) The maximum value of f (x) is15; (3) The sum of two cubes of f (x) is equal to 17. Then the analytical formula of f (x) is. 10. The range of the function is. 1 1. Find the definition range of the following function: (13) f (x 2)12. Find the range of function. 13. Given f(x)=x2+4x+3, find the minimum value of f(x) in the interval [t, t+ 1] and the maximum value of g(t) ABCD (t) .5438+04 (2) and find f [f (3). Reference answer: classic example: solution: the domain of (1)∵f(x) is [0, 1], ∴f(x2+ 1).② When 1-m= m, ③ When 1-m > m > 0, that is, 0 g (a)-g (-b) 2f (b)-f (-a) < g (a)-g (-b) 3f (a)-f (-b) > g (b)-g (. ③ D. ② ④ Exercise in class: 1. It is known that the function f(x)=2x2-mx+3, which was increasing function at that time, was a subtraction function. Then f( 1) is equal to () a.-3b.13c.7d. A variable containing m.2. Functions include () A. Non-odd non-even function B. Even function odd function C. Even function D. odd function 3. Known functions (1), (2) and (3) (4), among which () A.1b.2c.3d.44. The mapping f:AB is known, where the set A = {-3, -2,-1, 1, 2, 3, 4}, and the elements in the set b are all images of the elements in the set a under the mapping f, and the number of elements in the set b is () A.4B.5C.6D.76 for any one. It is known that the function f(x) is a decreasing function in the interval, so the relationship with the size of is 0.8. It is known that f(x) is a domain. 0, f(x) is increasing function, if x 1 b > 0, then f (a) > f (b) > f (0) = 0, thus ① and ③ in the above inequality hold. So we choose C. Method 2: Combine function mirroring. F can be obtained from the following figure, and the results show that ① and ③ are correct, so C. Method 3: Use indirect method, that is, construct two functional models that satisfy the meaning of the question, f(x)=x and g(x)=|x|, and take special values, such as a=2 and b= 1. It can be verified that ① and ③ are correct, so C. Answer: 2. d； 3.b； 4.d； 5.a； 6.； 7.； 8.& gt； 9.x =- 1； 10.(); 1 1. solution: (1) function, set the time, so it increases monotonously in the interval; (2) Therefore, when x= 1, there is a minimum value. 12. The solution: (1) is arbitrary, and, because,,, therefore, it is monotonically increasing in the world. (2) Because it is monotonically increasing in the world, the definition domain and the value domain, that is, the equation. (2) From (1) =. 14. Solution: (1); (2) At that time, f 1(x) decreased monotonically, while f 1(x) increased monotonically; At that time, f2(z) decreased monotonically, and at that time, f 1(x) increased monotonically. (3) When summing, f(x) decreases monotonically. When and respectively increase monotonically.