( 1)x2+y= 1
(2)x+y2= 1
The solution (1) gets Y = 1 from X2+Y = 1-X2, and it can be determined that y is a function of X. 。
At any x∈{x|x≤ 1}, the function value is not unique.
Example 2 Does the following formula represent the same function, and why?
In the solution (1), the domains of the two formulas are both R, and the corresponding rules are the same, so the two formulas are the same function.
The two formulas in (2) and (3) have different definitions, so they represent different functions.
(4) The definition domain of the two formulas is-1 ≤ x ≤ 1, and the corresponding rules are the same, so the two formulas are the same function.
Example 3 Find the domain of the following function:
Example 4 The domain of the known function f(x) is
(9)y=|x-2|-|x+ 1|
Solution (1)∫x∈r, ∴-5x2+ 1 ≤ 1, interval y ≤ 1.
(6) the domain is R.
The domain x≠ 1 and x≠2.
(y-4)x2-3(y-4)x+(2y-5)=0 ①
When y-4 ≠ 0, ∵ equation ① has a real root, ∴ δ≥ 0.
That is, 9 (y-4) 2-4 (y-4) (2y-5) ≥ 0.
Simplify it to y2-20y+64 ≥ 0, and get
Y < 4 or y≥ 16
When y = 4, ① does not hold.
So the range is y < 4 or y ≥ 16.
When t≥0, the function y is increasing function (see Figure 2.2-3).
Remove the absolute value sign,
The image is shown in Figure 2.2-4.
The range y ∈ can be obtained from Figure 2.2-4.
Explain how to find the range of functions:
1 observation method: non-negative number: square number, arithmetic root, absolute value, etc. Often used (for example, 1, 2).
The problem of finding the value range (maximum value) of quadratic function in the specified 2-degree interval is commonly used, and it is handled in combination with the image properties of quadratic function and the position of symmetry axis. If the function f (x) = ax2+bx+c (a > 0) is found, the range (or maximum value) of a given interval [m, n] can be considered in three cases:
Example 5 can be used as a formula.
Find the range of y (as in Example 6-7).
Find the domain of quadratic function. But pay attention to the range of intermediate quantity t (as in Example 6-8).
6 separation of bounded variables: solving bounded variables by known function formulas. Find the range of function y with the range of bounded variables (for example, 6-6).
7 Imaging method (as in Examples 6-9):
Because there are no certain rules and procedures for finding the function value domain, we should use various methods to solve it flexibly according to the different characteristics of the resolution function.
Solution (2)∫f(-7)= 10, ∴ f [f (-7)] = f (10) =100.
It is simple to illustrate this example, 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) Given f (x) = 3x2- 1, find 1f (x- 1) and 2f (x2).
(2) Given f (x) = 3x2+ 1 and g (x) = 2x- 1, find f [g (x)].
Find f (x).
(4) Given that f (x) is a quadratic function, f (0) = 2 and f (x+ 1)-f (x) = x- 1, find f(x).
(5) Let an isosceles triangle with a perimeter of a (a(a>0), a waist length of x and a base length of y, try to express y as a function of x, and find its domain and value domain.
(1) Analysis: This question is equivalent to the function value when x = x- 1, and the function expression can be obtained by substitution.
Solution \f(x)= 3 x2- 1
∴f(x- 1)=3(x- 1)2- 1=3x2-6x+2
f(x2)= 3(x2)2- 1 = 3 x4- 1
(2) Analysis: the function f[g(x)] represents an analytical formula obtained by replacing x in the function f(x) with g(x), which is still solved by substitution method.
The solution is given by f [g (x)] = 3 (2x-1) 2+1=12x2-12x+4.
Method (or observation).
∴ x = (t+ 1) 2 is substituted into the original formula f (t) = (t+1) 2-6 (t+1)-7.
= T2-4t- 12(t≥ 1)
That is, f (x) = x2-4x-12 (x ≥-1)
Explain the alternatives used in the second solution. Note that both methods involve intermediate quantity, so we must determine the range of intermediate quantity and master method of substitution skillfully.
(4) Analysis: This question gives the basic characteristics of the function, that is, the quadratic function, which can be solved by the undetermined coefficient method.
Let f (x) = ax2+bx+c (a ≠ 0)
If f (0) = 2, c = 2. If f (x+ 1)-f (x) = x- 1, the identity 2ax++ is obtained.
It shows that the undetermined coefficient is an important mathematical method and should be mastered skillfully.
∫2x+y = a, ∴ y = a-2x is the expected function.
The sum of any two sides of a triangle is greater than the third side.
∴ 2x+2x > A, and ∵ y > 0,
This paper explains the function expression of finding practical problems, focusing on analyzing the quantitative relationship in practical problems and establishing the resolution function, its definition range and value range, and considering the significance of practical problems.