1. abstract function of proportional function type
Example 1 It is known that the function f(x) has f (x+y) = f (x)+f (y) for any real numbers x and y, and when x >; 0, f(x)>0, f (- 1) =-2 Find the range of f(x) in the interval [-2, 1].
Analysis: First, it is proved that the function f(x) is a increasing function on R (note that f (x2) = f [(x2-x1)+x1] = f (x2-x1)+f (x1)); Then find its range according to the interval.
Example 2 It is known that the function f(x) has f (x+y)+2 = f (x)+f (y) for any real numbers x and y, and when x >; 0, f(x)>2, f (3) = 5, find the inequality f (A2-2A-2).
Analysis: First, it is proved that the function f(x) is a increasing function on R (parody1); Then find f (1) = 3; Finally, remove the function symbol.
2. Power function abstract function
Example 3 it is known that the function f(x) has f (xy) = f (x) f (y), f (- 1) = 1, and f (27) = 9 for any real numbers x and y. When 0 ≤ x < 1, f (x)
(1) to judge the parity of f(x);
(2) The monotonicity of f(x) on [0, +∞] is judged and proved;
(3) If a≥0 and F (A+ 1) ≤, find the value range of a. 。
Analysis: (1) makes y =-1;
(2) use f (x1) = f (x2) = f () f (x2);
(3)0≤a≤2。
3. Exponential abstract function
Example 4 Let the domain of function f(x) be (-∞, +∞), and satisfy the following conditions: there is x 1≠x2, so that f (x1) ≠ f (x2); For any x and y, f (x+y) = f (x) f (y) holds.
( 1)f(0)；
(2) For any value x, judge the sign of the value f(x).
Analysis: (1) Make y = 0；; (2) Let y = x ≠ 0.
Example 5 Whether the function f(x) exists or not depends on the following three conditions: ①f(x)>0, x ∈ n; ②f(a+b)= f(a)f(b)，a、b∈N； ③ f (2) = 4。 At the same time If it exists, find the analytical formula of f(x); if it does not exist, explain the reason.
Analysis: first guess f (x) = 2x; And then prove it by mathematical induction
4. Logarithmic function type abstract function
Example 6 let f(x) be a monotone increasing function defined on (0, +∞), and satisfy f (x y) = f (x)+f (y), f (3) = 1, and find:
( 1)f( 1)；
(2) If f (x)+f (x-8) ≤ 2, find the range of x 。
Analysis: (1) Use 3 =1× 3;
(2) Using monotonicity and known relation of functions.
Example 7 Let the inverse function of the function y = f (x) be y = g (x). If f (ab) = f (a)+f (b), then whether g (a+b) = g (a) g (b) is correct, please explain the reasons.
Analysis: let f (a) = m and f (b) = n, then g (m) = a and g (n) = b,
Then m+n = f (a)+f (b) = f (ab) = f [g (m) g (n)].
5. Trigonometric abstract function
Example 8 It is known that the domain of the function f(x) is symmetrical about the origin and satisfies the following three conditions:
(1) When x 1 and x2 are numbers in the domain, there is f (x1-x2) =;
② f (a) =- 1 (a > 0, where a is the number in the definition domain);
③ when 0 < x < 2a and f (x) < 0.
Q:
What is the parity of (1) f(x)? Explain the reasons;
(2) What is the monotonicity of f (x) on (0,4a)? Explain why.
Analysis: (1) f [-(x1-x2)] =-f [(x1-x2)] is used to determine that f(x) is odd function;
(3) First prove that f(x) is increasing function in (0,2a), and then prove that F (x) is also increasing function in (2a,4a).
For solving abstract function problems, although we can't use special models instead of solving them, we can use special models to understand the meaning of the questions. For some abstract function problems, the corresponding special model is not the basic elementary function we are familiar with. Therefore, it is necessary to make appropriate modifications for different functions, seek special models, and better solve the problem of abstract functions.
Example 9 it is known that the function f(x)(x≠0) satisfies f (xy) = f (x)+f (y),
(1) Verification: f (1) = f (-1) = 0;
(2) Prove that f(x) is an even function;
(3) If f(x) is a increasing function on (0, +∞), solve the inequality f (x)+f (x-) ≤ 0.
Analysis: the function model is: f (x) = loga | x | (a > 0)
(1) shilling x = y = 1, and then make x = y =-1;
(2) Let y =-1;
(3) If f(x) is an even function, then f (x) = f (| x |).
Example 10 it is known that for all real numbers x and y, the function f(x) satisfies f(0)≠0, f (x+y) = f (x) f (y), and when x < 0, f (x) > 1. Prove:
(1) when x > 0, 0 < f (x) <1;
(2) f(x) is a decreasing function on x∈R 。
Analysis: (1) shilling x = y = 0 to get f (0) = 1, and then make y =-x;
(3) Inspired by monotonicity of exponential function:
F (x+y) = f (x) f (y) gives f (x-y) =,
Furthermore, from x 1 < x2, there is = f (x 1-x2) > 1.
In a word, abstract function is closely related to monotonicity, parity and many other properties of function, coupled with its abstraction and variability, so there are many kinds of questions and complex and changeable methods to solve problems. However, replacing abstract function with special model is an effective teaching method, which can solve most abstract function problems in middle school mathematics. This is in line with students' age characteristics and cognitive level. Students are not only easy to understand and accept, but also feel really reliable.