[Question 1] Matching method
Example 1. Given f(■+ 1)=x+2■, find f(x).
Analysis: The analytic formula of function y=f(x) is the relationship between the values of independent variable X and Y, and its essence is the corresponding law f: x → y, so the key to solve this kind of problem is to find out what the law of "Y" to "X" is.
Solution: ∫ f (■+1) = x+2 ■ = (■+1) 2-1.
(■+ 1 1)
∴f(x)=x2- 1(x 1)
Summary: This solution is a collocation method. Through observation and analysis, the right end "x+2■" becomes the expression "■+ 1", that is, the expression containing (■+ 1). This scheme requires a certain deformation ability and observation ability.
[Question 2] Alternatives
Example 2. Given f( 1-cosx)=sin2x, find f(x).
Analysis: Take 1-cosx as a whole, apply the idea of mathematical integration, and get it by changing elements.
Solution: let t= 1-cosx.
∫-1cosx1∴ 01-cosx2 is 0t2.
∴cosx= 1-t
∴sin2x= 1-cos2x= 1-( 1-t)2=-t2+2t
∴f(t)=-t2+2t(0t2)
That is f(x)=-x2+2x(0x2)
Summary: ① It is known that f[g(x)] is a function of x, that is, f[g(x)]=F(x). To find the analytical formula of f(x), usually let g(x)=t, from which x=(t) can be solved and substituted into f[g(x].
Note: the value range of the new t should be determined after the exchange.
(2) The substitution method is a problem-solving method that replaces some original variables by introducing one or several new variables. Its basic functions are: changing the difficult to the easy, simplifying the complex, and quickly transforming the unknown into the known, so as to achieve the goal of solving the problem smoothly. There are many commonly used substitution methods, such as partial substitution, whole substitution, triangle substitution, denominator substitution, etc., which are widely used.
[Question 3] Method of undetermined coefficient
Example 3. Let the quadratic function f(x) satisfy f(x+2)=f(2-x), and the sum of squares of two real roots of f(x) = 0 is 10, and the image passes through point (0,3), so as to obtain the analytical formula of f(x).
Analysis: Because f(x) is a quadratic function, the basic structure of its analytical formula has been fixed, which can be treated by the method of undetermined coefficient.
Solution: Let f (x) = AX2+BX+C (A ≠ 0).
It can be seen from f(x+2)=f(2-x) that the function image is symmetrical about the straight line x=2.
∴-■=2, which means b =-4a...( 1)
Image passed (0,3) ∴ c = 3 ... ②.
From the equation f(x)=0 and the sum of squares of two real roots is 10, we get (-■)2-■=0.
That is B2-2ac = 10a2...③.
A= 1, b=-4 and c=3 are obtained from ① ② ③ solution.
∴f(x)=x2-4x+3
Summary: As long as the type of resolution function is clear, you can set its resolution function, try to find its coefficient and get the result. Similarly, when f(x) is known as a linear function, it can be set as f (x) = ax+b (a ≠ 0); When f(x) is an inverse proportional function, f (x) = ■ (k ≠ 0); When f(x) is a quadratic function, it can be set according to the conditions.
① general formula: f(x)=ax2+bx+c(a≠0)
② Vertex type: f(x)=a(x-h)2+k(a≠0)
③ binomial formula: f(x)=a(x-x 1)(x-x2)(a≠0).
[Question 4] Exclusion method
Example 4. It is known that the function y=f(x) satisfies af(x)+bf(■)=cx, where a, b and c are nonzero constants, and A ≠ B, find the analytical expression of the function y=f(x).
Analysis: To find the analytical expression of the function y=f(x), we know that f(■) must be eliminated from the known conditions. It is not difficult to find another equation to form a set of equations and get f(x) by eliminating f(■). How to form? Make full use of the reciprocal relationship between x and ■, and replace the known x with ■ to get another equation.
Solution: In the known equation, replace X with ■ to get af(■)+bf(x)=■, and combine with the original conditional expression to get AF (x)+BF (■) = CX ... ① AF (■)+BF (x) =■ ... ②.
(a2-b2)f(x)=c(ax-■) of ①× A-②× B.
∵a≠ b ∴f(x)=■(ax-■)(x≠0)
(Continue on Saturday)
Some students asked the following math questions through QQ, and we invited Teacher Meng from Tianjin No.4 Middle School to answer them.
Q 1。 It is known that two of the equations: x2+ax+a+ 1=0 satisfy one condition: one is greater than k, and the other is less than k (where k is a real number). One method of this problem is the mirror image method, and there is another method. Can you tell me these two methods? )
Answer: Method 1: ∵f(x)=x2+ax+a+ 1 The image is a parabola with an upward opening, so only f (k) < 0 is required.
∴ k2+AK+a+ 1 < 0, that is, a (k+ 1)
∴ when k >- 1, a
Method 2: (x 1-k) (x2-k) < 0 △ > 0.
Only (x 1-k) (x2-k) < 0, while x1x2-k (x1+x2)+k2 < 0.
That is, a+ 1+ka+k2 < 0, and the following method is the same as 1.
Question 2. Why do you only need (x 1-k) (x2-k) < 0 when solving, and you don't need to ask whether the discriminant of the root is greater than 0?
Answer: Method 2 does not require discriminant. The reason can be illustrated by a simple example. For example, if we study that x2+ax+b=0 satisfies that one root is greater than 0 and the other root is less than 0, we only need x 1x2 < 0, that is, b 0 holds.