If the definition field of function f (x) is r, f (- x) = f (x), and f (2 + x) = f (2-x), try to prove that f (x) is a periodic function, and find out its period

If the definition field of function f (x) is r, f (- x) = f (x), and f (2 + x) = f (2-x), try to prove that f (x) is a periodic function, and find out its period

f(2+x)=f(2-x)
Let - a = 2 + X
x=-a-2
So 2-x = 4 + a
That is, f (- a) = f (a + 4)
So f (- x) = f (x + 4)
f(-x)=f(x)
So f (x) = f (x + 4)
So it's a periodic function, t = 4
Let a > 0, a ≠ 1, FX = loga (x ^ 2-2x + 3) have the minimum value, find the solution set of the inequality loga (x-1)
The function f (x) = ㏒ (a) [x & sup2; - 2x + 3] = ㏒ (a) [(x-1) & sup2; + 2]. ∵ (x-1) & sup2; + 2 ≥ 2, if 0 ＜ a ＜ 1, then f (x) ≤㏒ (a) 2. In this case, the function has no minimum value
The definition field of function f (x) is R. if f (x + 1) and f (x-1) are both odd functions, find the period of F (x)
It's a process
If y (x) = f (x + 1) is an odd function, then y (x) + y (- x) = 0, that is, f (x + 1) + F (- x + 1) = 0, that is, f (x + 1) = - f (- x + 1) let x = y + 1, f (y + 1) + F [- (y + 1) + 1], that is, f (y + 2) = - f (- y) t (x) = f (x-1) be an odd function, then t (x) + T (- x) = 0, that is, f (x-1) + F (- x-1) = 0
Both f (x + 1) and f (x-1) are odd functions
Then f (x + 1) = - f (- x + 1)
f（x-1）=-f（-x-1）
We get f (x + 2) = - f (- x)
f（x-2）=-f（-x）
So f (x + 2) = f (X-2)
That is, f (x) = f (x + 4)
So the period of F (x) is 4
The function definition field is r,
And f (x + 1) and f (x-1) are odd functions,
∴f(-x+1)=-f(x+1)……………… (1)
f(-x-1)=-f(x-1)………………… II.
Let - x + 1 = t be: F (T) = - f (2-T) 3.
From ② let - X-1 = t get: F (T) = - f (- 2-T) 4.
Let f (2-T) = f (- 2-T) be obtained from (3) and (4), and then let - 2-T = m be f (m) = f (4 + m)... Expansion
The function definition field is r,
And f (x + 1) and f (x-1) are odd functions,
∴f(-x+1)=-f(x+1)……………… (1)
f(-x-1)=-f(x-1)………………… II.
Let - x + 1 = t be: F (T) = - f (2-T) 3.
From ② let - X-1 = t get: F (T) = - f (- 2-T) 4.
F (2-T) = f (- 2-T) is obtained from (3) and (4), so that - 2-T = m is f (m) = f (4 + m)
So the period of the function f (x) is 4
Let a ＞ 0A ≠ 1, f (x) = loga * (x ^ 2-2x = 3) have the minimum value, then the solution set of inequality loga * (x-1) ＞ 0
That = 3 is + 3
X ^ 2-2x + 3 = (x-1) + 2, which is greater than or equal to 2
So a > 1
So the solution set is X-1 > 1
So x > 2
If an odd function defined on R has the smallest positive period a, and X belongs to (0,1), f (x) = x times of 2 / {(x times of 4) + 1}
1: Try to find the analytic solution of F (x) on [- 1,1]
2: Why is the equation f (x) = k solvable when k is a real number?
(1) If x belongs to (0,1), f (x) = 2 ^ X / (4 ^ x + 1), and F is an odd function, we get that when x belongs to (- 1,0), - x belongs to (0,1), f (x) = - f (- x) = - 2 ^ (- x) / [4 ^ (- x) + 1] = - 2 ^ X / (4 ^ x + 1), so f (x) = 2 ^ X / (4 ^ x + 1), X belongs to (0,1); f (x) = - 2 ^ X / (4 ^ x + 1), X belongs to (- 1,0) (...)
If the solution set of X ≤ - 2,2x-b ≤ 3 is - 2 ≤ x ≤ 0, then the value of A-B is
The solution set of X ≤ - 2,2x-b ≤ 3 is - 2 ≤ x ≤ 0,
And by 2x-b
F (x) is a quadratic function, and f (x-1) = the power of X-X
(1) Finding the value of F (0) and the analytic expression of F (x)
(2) If x belongs to [- 1,1], find the range of F (x)
(1) Let x = 1, then f (1-1) = 0, that is, f (0) = 0
Let X-1 = t, then x = t + 1
f(t)=(t+1)^2-(t+1)=t^2+t
That is, f (x) = x ^ 2 + X (x belongs to R)
（2） f(x)=x^2+x=(x+1/2)^2-1/4
∵ x belongs to [- 1,1]
The X + 1 / 2 belongs to [- 1 / 2,3 / 2]
(x + 1 / 2) ^ 2 belongs to [0,9 / 4]
(x + 1 / 2) ^ 2-1 / 4 belongs to [- 1 / 4,2]
That is, f (x) is [- 1 / 4,2]
If the solution set of the system of inequalities {x ≤ - 2,2x-b ≤ 3} is 0 ≤ x ≤ 1, try to find the value of a + B
Solution 2x-b ≤ 3
We get x ≤ (B + 3) / 2
Solution A-X / 2 ≤ - 2
X ≥ 2 (a + 2)
The solution of∵ inequality system is 0 ≤ x ≤ 1
∴(b+3)/2=1、2(a+2)=0
b+3=2、a+2=0
b=-1、a=-2
∴a+b=-1-2=-3
Given that the maximum value of the quadratic function f (x) = (㏒ 10a) x + 2x + 4 ㏒ 10A is 3, find the value of A
Let's first consider LGA as t, f (x) = TX & sup2; + 2x + 4T, because there is a maximum value of 3, so t < 0 (the maximum value is only when the opening is downward), 2 & sup2; - 4 * t * (4T) = 3 = = > t = - 1 / 4, so LGA = - 1 / 4 = = > the negative quarter power of a = 10
Xiao Ming mistakenly regards a number multiplied by 3 / 8 as divided by 3 / 8, and the result is 24. What is the correct answer?
The correct answer should be
24x3/8x3/8
=8x3/8
=27/8
If you don't understand this question, you can ask,
In addition, please send and click my avatar to ask me for help,
It should be 27 out of 8. It should go backwards. Multiply 24 by 3 out of 8 to get this number, and then multiply it by the correct number
24*3/8
=24*3/8
=9
9*3/8
=9*3/8
=27/8
24 times 3 / 8 divided by 3 / 8
=9 divided by 3 / 8
=24