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If the domain of function y = f (x) is [0,3], then the domain of function g (x) = (f (x + 1)) / (X-2) is
X + 1 in brackets is a variable, and its range is [0,3]
0≤x+1≤3,x≠2
So - 1 ≤ x ≤ 2, and X ≠ 2
That is - 1 ≤ x < 2
Definition domain of abstract function (senior one mathematics) Let's take a look at the following example (1) Given that the definition domain of function f (x) is [0,1], find the definition domain of F (x2 + 1) (where x2 represents the square of x) (2) Given that the domain of function f (2x-1) is [0,1], find the domain of F (1-3x) (1) ∵ x2 + 1 in function f (x2 + 1) is equivalent to X in function f (x) The definition field of - 1 ≤ x2 ≤ 0 ‡ x = 0 ‡ f (x2 + 1) is {0} (2) ∵ the definition domain of function f (2x-1) is [0,1), that is, 0 ≤ x < 1 ∴-1≤2x-1＜1 The definition domain of F (x) is [- 1,1), i.e. - 1 ≤ 1-3x < 1 The definition field of 0 ＜ x ≤ 2 / 3 ‡ f (1-3x) is (0,2 / 3] Now my question is: why is x2 + 1 in function f (x2 + 1) equivalent to function f (x) In my reference book, it is said that the key to solving such problems is to pay attention to the correspondence law. Under the same correspondence law, no matter what letter or algebraic formula the object of the acceptance law is, its constraints are consistent, that is, they are all within the same planting range. Then, what is this correspondence law and how does it produce this correspondence law? Who can have a good explanation, Zxj_123 thank you for your answer, but I still don't understand the second question: I use the definition field of function f (2x-1) as [0,1) to calculate the range of 2x-1 - 1 ≤ 2x-1 < 1, but for function f (1-3x) Why is there - 1 ≤ 1-3x < 1? Aren't these two different functions? Why can the range of the former be replaced by the second? I hope you can explain it for me again,
Abstract function means that the correspondence rule is not given. What you pay attention to is the definition domain and value domain of the function. For example, X2 + 1 in function f (x2 + 1) is equivalent to X in function f (x). This is because the object applied by the correspondence rule is x2 + 1 instead of X! Therefore, at this time, X2 + 1 can be regarded as a whole, so that x2 + 1 =
Correct format of function definition field in senior one mathematics Example: F (x) = 1 / X-1 write the problem-solving process in the correct format
F (x) = 1 / X-1 domain x ≠ 0
f(x)=1/（x-1）
From X-1 ≠ 0, X ≠ 1 is obtained
Then the definition field is {x|x ≠ 1}
Find the answers to the following two questions, Given the square of F (1 / x) = x / 1-x, then f (x) =? F (root 2-1) =? It is known that f (x) = MX / 4x-3 (x is not equal to 3 / 4) has exactly f (f (x)) = x in the definition field to find the real number M Among the squares of X / 1-x in question 1, the squares of 1-x are connected together, that is, X of the square of 1-x. the same is true for 4x-3 in question 2. Again, 4x-3 is the bottom of the connected fractions.
Let 1 / x = t, then x = 1 / T, so f (T) = (1 / T) / (1-1 / T ^ 2) = t / (T ^ 2-1), so f (x) = x / (x ^ 2-1) f (sqrt (2) - 1) = [sqrt (2) - 1] / [(3-2 * sqrt (2)) - 1] = - 1 / 2. The second one: F (1) = MF (f (1)) = f (m) = m ^ 2 / (4m-3) = 1, the solution is M1 = 1, M2 = 3. That is, M = 1 or 3, that is, f (x) = x / (4x-3) or F (
2F (x) + F (x-1) = x ² Find the analytical formula of F (x)?
F (x) is a quadratic function. Let f (x) = ax ^ 2 + BX + CF (x-1) = a (x-1) ^ 2 + B (x-1) + C = ax ^ 2-2ax + A + bx-b + C = ax ^ 2 + (b-2a) x + A-B + C2F (x) + F (x-1) = 2aX ^ 2 + 2bx + 2C + ax ^ 2 + (b-2a) x + A-B + C = 3ax ^ 2 + (3b-2a) x + A-B + 3c3a = 13b-2a = 0a-b + 3C = 0A = 1 / 3 B = 2 / 9 C = - 1
It is known that the domain of function f (x) is (0, + ∞). For any x, y ∈ (0, + ∞), f (XY) = f (x) + F (y). If and only if x > 1, f (x) < 0 holds, (1) Let x, y ∈ (0, + ∞), and prove that f (Y / x) = f (y) - f (x); (2) Let x1, X2 ∈ (0, + ∞), if f (x1) ＜ f (x2), try to compare the sizes of X1 and X2; (3) Solving inequalities f[x about X ²- (a+1)x+a+1]＞0. Seeking ideas
F (XY) = f (x) + F (y) let x = A / B, y = B substitute f (a) = f (A / b) + F (b) then replace a with X, B with YF (Y / x) = f (y) - f (x) let x = x2 y = X1 substitute f (Y / x) = f (y) - f (x) f (x1 / x2) = f (x1) - f (x2) 1, f (x) < 0 holds, so X1 / x2 > 1 because x1, X2 ∈ (0, + ∞), X1 > x2 if and only if
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