Given that the range of square + 4 of y = 3x is [7 19], find the number of functions with the same range but different domain

Given that the range of square + 4 of y = 3x is [7 19], find the number of functions with the same range but different domain

Seven
M & sup2; + 58m-2007 = 0, n & sup2; + 58n-2007 = 0 and m ≠ n
So m and N are the roots of the equation x & sup2; + 58m-2007 = 0
By Weida theorem
m+n=-58
mn=-2007
1/m+1/n
=(m+n)/mn
=-58/(-2007)
=58/2007
How many twin functions are there with y = 2x & sup2; + 1 and range {5,19}?
If a series of functions have the same analytic formula, the same range of values, but different domains of definitions, are these "twin functions" 8 or 9? Do different domains mean that four numbers cannot be taken at the same time?
9. "Different domains mean you can't take 4 numbers at the same time." no, you can take 4 numbers at the same time
Find the set s composed of positive integers so that the sum of the elements in S is equal to the product of the elements
If there is only one element in the set, it is obvious
If there are two elements in the set, a and B, then AB = a + B, (A-1) (B-1) = 1, a = b = 2, which is contradictory to a! = B, so there are at least three elements in the set
Let three elements be a, B, C, and let 1 = 1, so ABC - (a + B + C) = a ^ 3 + (K + m) a ^ 2 + kma-3a - (K + m) > = 1 + (K + m) a + km-3a - (K + M) = (k-1) (m-1) + (K + M-3) a > = (1-1) (2-1) + (1 + 2-3) a = 0, so ABC > = a + B + C, when the equal sign holds, a = 1, B = 2, C = 3
When there are at least four elements in s, it is easy to know that the product of any two numbers greater than 1 is greater than the sum of them. Let s = {A1, A2 , an}, n > = 4, and a1a3 + a4a5a6 + +a(n-4)a(n-3)a(n-2)+a(n-1)an>a1+a2+a3+a4+a5+a6+…… +A (n-4) + a (n-3) + a (n-2) + a (n-1) + an, if n is divided by 3, then a1a2a3 an>a1a2a3+a4a5…… an>…… >a1a2a3+a4a5a6+…… +a(n-3)a(n-2)a(n-1)+an>a1+a2+a3+…… +If n is a multiple of 3, then a1a2a3 an>a1a2a3+a4a5a6+…… +a(n-2)a(n-1)an>a1+a2+…… +So s does not satisfy the condition
So only s = {1,2,3}
Given that the function y = f (x) satisfies f (x) = 2F (1 / x) + X, the analytic expression of F (x) can be obtained
F (x) = 2F (1 / x) + X (1) f (1 / x) = 2F (x) + 1 / X (2) by substituting formula (2) into (1), we get f (x) = 4f (x) + 2 / x + X, and the solution is f (x) = - (1 / 3) * (2 / x + x)
Description: a set of positive integers divided by 4 and 1
The answer is {x | x = 4K + 1, K ∈ n} but I want to ask, if k = 0, isn't it equal to 1? Can 1 / 4 get a positive integer?
Your understanding is wrong. The set of positive integers divided by 4 and 1 refers to the set of all positive integers that satisfy 4 and 1
It is not a set of numbers that are divided by 4 and the result is a positive integer,
1 / 4 = 0... 1, we don't care if the result 0 is a positive integer,
If the divisor 1 is a positive integer and the remainder is 1, then 1 is in the number set
1. Given that the function f (x) satisfies 2F (1 / x) + F (x) = x (x ≠ 0), find the analytic expression of F (x)
2. Given that the function f (x) satisfies f (x) + 3f (- x) = 4x, find the analytic expression of function f (x)
1、
2f（1/x）+f（x）=x ①
Let t = 1 / x, then x = 1 / T
2f（t）+f（1/t）=1/t
Namely
2f（x）+f（1/x）=1/x ②
① (2) it can be obtained by solving the equation
f（x）=2/3x -x/3
2、
This question is similar to question 1
f（x）+3f（-x）=4x
f（-x）+3f（x）=-4x
Solution
f（x）=-2x
A set of positive integers divided by 5 and 1
Express by description
{x|x=5k+1,k=0,1,2...n}
Given that the function y = f (x) satisfies f (x) = 2F (1x) + X (x ≠ 0), then the analytic expression of F (x) is______ .
∵ f (x) = 2F (LX) + X, ∵ f (LX) = 2F (x) + 1X, if f (LX) is eliminated by two simultaneous formulas, f (x) = - 23x − X3 (x ≠ 0) can be obtained, so the answer is: F (x) = - 23x − X3 (x ≠ 0)
Let a = {(x, y) | x ^ 2 / 4 + y ^ 2 / 16 = 1}, B = {(x, y) | y = 3 ^ x}, then the number of subsets of a ∩ B is that the answer has two elements, but the number is four
Even if there are two intersections. Calculate a and B. that subset should not be (a) (b) (a, b) (a, B, empty set) (a, empty set) (B, empty set). -- I'm messing with it. I don't know if it's like this. I want to find a positive solution!
It's right to be able to get its two intersections, but the three sets (a, B, empty set) (a, empty set) (B, empty set) proposed by you and (a, B,) (a) (b) are three identical sets. At the same time, (a, B, empty set) you are wrong to write like this. A and B are two elements, and empty set is a set
There are two points of intersection, as you wish, a and B.
The four subsets are {a}, {B}, {a, B}, and empty set.
P。 S. You're so cute
It is known that the function y = f (x) satisfies the analytic expression of F (x) = 2F (1 △ x) + X
f(x)=2f(1/x)+x①
The independent variable on the left of the above formula is 1 / X,
The results show that f (1 / x) = 2F (1 / (1 / x)) + (1 / x),
That is: F (1 / x) = 2F (x) + 1 / X (2)
② Substituting into (1), we get: F (x) = 2 [2F (x) + 1 / x] + X
That is: - 3f (x) = 2 / x + X
So f (x) = - 2 / (3x) - 1 / (3x)