Monotone interval of function y = log2 (x ^ 2-5x + 6)

Monotone interval of function y = log2 (x ^ 2-5x + 6)

Let x ^ 2-5x + 6 = U
Then y = log2 (U)
The base is 2, so when u increases, y increases; when u decreases, y decreases
U = x ^ 2-5x + 6 is a parabola with the opening upward and the axis of symmetry is a straight line x = 5 / 2. When x = 5 / 2, u increases
Because u > 0 (to be an integer), that is, x ^ 2-5x + 6 > 0, the solution is X3
So the increasing interval of the function u is X3
So the monotone increasing interval of function y = log2 (x ^ 2-5x + 6) is (- ∞, 2); the monotone decreasing interval is (3, ∞)
Find the value of X: X & # 178; - 121 / 49 = 0
x²-121/49=0
(x-11/7)(x+11/7)=0
X = 11 / 7 or - 11 / 7
Plus or minus 11 / 7
If there is a real number x in the set a = {1,3, X & # 178;} B = {x + 2,1} such that B & # 8838; a, if there is, find out the set a, B, if not, explain the reason
A={1,3,x²}
B={x+2,1}
Hypothesis exists
①x+2=3
So x = 1
In this case, a = {1,3,1}, which does not conform to the requirement
②x+2=x²
Then x & # 178; - X-2 = 0
So (x + 1) (X-2) = 0
So x = - 1 or x = 2
When x = - 1, a = {1,3,1}, not in accordance with
When x = 2, a = {1,3,4}, B = {1,4}, in accordance with
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When x = (), the value of 2x + 1 minus 2x-1 is 2 times larger than that of X-1
（2x+1）/3-（2x-1）/3-（x-1/6）=2
Both sides * 6
2（2x+1）-2（2x-1）-（6x-1）=12
4x+2-4x+2-6x+1=12
4x-4x-6x=12-1-2-2
-6x=7
x=-7/6
Given that f (x) is an increasing function defined on (0, + ∞), then the monotone interval of F (- X & # 178; + 5x + 6) is____
When x > 0, f (x) increases monotonically
Let - X & # 178; + 5x + 6 > 0
Namely (X-6) (x + 1)
Find the value of X in the following formulas. (1) x & # 178; = 17; (2) x & # 178; - 49 / 121 = 0
x²=17
x=±√17
x1=√17,x2=-√17
X & # 178; - 121 / 49 = 0
X & # 178; = 121 / 49
X = ± 11 / 7
X1 = 11 / 7, X2 = - 11 / 7
√ denotes the root sign
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Given the set a = {1,3, - A & # 179;}, B = {1, a + 2}, whether there is a real number a, such that a ∩ B = B? If there is, find the set a and B
Is - A & # 179;
There are two cases in this problem: the first is a + 2 = 3, that is, a = 1; the second is a + 2 = - A & # 179;, that is, (a + 1) (a ^ 2-A + 2) = 0. When a = - 1
Given x = 3Y, z = 7x (x is not equal to 0), find the value of X + y + Z of the algebraic formula 2x + 3y-z
x=3y z=7x=21y
(x+y+z)/(2x+3y-z)=(3y+y+21y)/(6y+3y-21y)=25/(-12)=-25/12
Let = 1, carry in the calculation
-25/12
Replace y and Z with X respectively, y = x / 3, z = 7x, the value of the algebraic formula is equal to - 12 / 25.
Very simple, take y = 1, so x = 3, z = 21, (x + y + Z) / (2x + 3y-z) = (3 + 1 + 21) / (6 + 3-21) = - 25 / 12
=25/12
The function f (x) defined on the interval (- 1,1) is a monotone decreasing function and satisfies f (x) + F (- x) = 0,
If f (1-A) + F (1-A & sup2;)
F (x) + F (- x) = 0 --- f (- x) = - f (x) odd function
f(1-a)a^2-1
Again - 1
From the meaning of the title, - 1
Find X in (1) x & # 178; = 16; (2) x & # 178; - 121 / 49 = 0
:(1)x²=16;
X = 4 or x = - 4
(2)x²-121/49=0.
x²=121/49
X = 11 / 7 or x = - 11 / 7