Example 1 which of the following can determine that y is a function of x? Why? (1) X2+y =1(2) x+y2 =1solution (1)Y = 1- X2+Y = 1 from x2, it can be determined that y is a function of x. The definitions of the two formulas in (2) and (3) are different, so they represent different functions. In (4), the domain of the two formulas is-1 ≤ x ≤ 1. Therefore, the two formulas are the same function. Example 3 Find the domain of the following function: Example 4 It is known that the domain of function f(x) is (9) y = | x-2 |-x+ 1 | solution (1) ∵ x ∈ r, ∴-5. The range y is less than or equal to/. (6) the definition range is r, the definition range is x≠ 1 and x ≠ 2 (y-4) x2-3 (y-4) x+(2y-5) = 0 ① When y-4 ≠ 0, the formula ① is invalid. So the range is y 0) is found, the range (or maximum value) within a given interval [m, n] can be considered in three cases: (Example 5) The range of y can be found by the formula (Example 6-7). It is the evaluation domain of quadratic function, but you should pay attention to the range of the intermediate quantity t (Example 6-8). 6 separation of bounded variables: solving bounded variables by known functions. Use the range of bounded variables to find the range of function y (as in Example 6-6). 7 Mirror Method (as in Example 6-9): Since there are no rules and procedures for finding the range of function values as there are for finding the range defined by functions, various methods should be adopted to flexibly solve the problem according to the different characteristics of the resolution function. Solution (2)∫f(-7)= 10, ∴. But the main intention is to deeply understand the meaning of the function symbol f(x). When calculating the piecewise function value, we should pay attention to the definition domain. Example 8 Find the function expression according to the known conditions. (1) It is known that f (x) = 3x2- 1, and find 1f (x- 1) and 2f (x2). Find f [g (x)]. Find f(x). (4) it is known that f (x) is a quadratic function and f a(a>0) = 2, and f (x+ 1)-f (x) = x- 1. (5). And find its domain and value domain. (1) Analysis: This question is equivalent to the function value when x = x- 1 Function expressions can be obtained by substitution. The solution ∫ f (x) = 3x2-1∴ f (x-1) = 3 (x-1) 2-1= 3x2-6x+2f (x2). The analytical formula obtained by replacing it is still solved by replacement method. The solution is obtained by the known method f [g (x)] = 3 (2x-1) 2+1=12x2-12x+4 (or observation method). ∴ x = (t)-7 = T2-4t-12 (t ≥-1), that is, f (x) = x2-4x-12 (x ≥-1) shows that it is used in solution 2. Note that both methods involve intermediate quantity, and the range of intermediate quantity must be determined. It can be solved by undetermined coefficient method. Let f (x) = AX2+BX+C (A ≠ 0) be f (0) = 2, and c = 2 is obtained. Let f (x+ 1)-f (x) = x- 1, and the identity 2ax+ is obtained. ∴ 2x+2x > A, and \y > 0, which shows that the key to finding the functional expression of practical problems is to analyze the quantitative relationship in practical problems and establish the resolution function, and the definition and scope of the resolution function should consider the significance of practical problems.