Given the function f (x) = | x + 2 | - | X-2 |, try to judge the parity of F (x)

Given the function f (x) = | x + 2 | - | X-2 |, try to judge the parity of F (x)

The definition field of F (x) is r, symmetric about the origin, and f (- x) = | - x + 2 | - | - X-2 | = | - X-2 | - | - x + 2 | = - f (x), f (x) = | - x + 2 | - | - X-2 |, are odd functions
LIM (TaNx + SiNx) / X Λ 2 x tends to zero
lim(x→∞) sinx / x^2=0
consider
|sinx/x^2-0|
≤|1/x^2|
First define the range of X: | x | 1, then there is | x | x, there is | SiNx / x ^ 2-0|
Let f (x) = x ^ 2 + ax-3a ^ 2lnx, and find the monotone interval of F (x)
And there's no domain
Solution:
f(x)=x^2+ax-3a^2lnx
Domain x > 0
be
f'(x)=2x+a - 3a^2/x
=(2x^2+ax-3a^2)/x
=(2x+3a)(x-a)/x
Let f '(x) > 0
be
(2x+3a)(x-a)/x>0
① When - 3A / 20
The increasing interval of F (x) is ((a, positive infinity)
The minus interval is (0, a)
② When - 3A / 2 > A, it means a
Let s be a set of real numbers satisfying the following conditions
Conditions: (1) 1 ∈ s; (2) if a ∈ s, then 1 / 1-A ∈ s
1. If 2 belongs to s, then there must be two other numbers in S. find out this number;
2. Proof: if a ∈ s and a ≠ 0, then 1-1 / a ∈ s;
3. Can set s contain only one element? If so, find this element; if not, explain the reason
Well, I admit it's my fault that I still have more than ten days to start reading
1 / 1-A belongs to rule a
Then 1 / [1-1 / (1-A)] also belongs to a (that is, a in 1 / 1-A is brought in with 1 / 1-A)
Does it mean that a = 1 / 1-A? In the third question, it is said that there can not be only one element in the set, that is, a ≠ 1 / 1-A. if not, who can tell me why the second question should bring a in 1 / 1-A with 1 / 1-A
If 1 / 1-A belongs to a, then 1 / [1-1 / (1-A)] also belongs to a
No. I say that ha a belongs to a, then 1 / (1-A) belongs to A. let B = 1 / (1-A) then B belongs to a, so 1 / (1-B) belongs to a so
1 / [1-1 / (1-A)] belongs to A. here a = 1 / (1-A) does not mean that the two are equal, but a meaning of "assignment". A on the left is just a sign
You're very careful. I haven't read a book all August
Finding monotone interval of function f (x) = 2lnx ax (a ∈ R)
Then make the value of the derivative greater than 0 and less than 0, and get the monotone increasing interval and decreasing interval
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Let s be a set of real numbers satisfying the following conditions: (1) 1 is not included in S (2) if a is included in s, then 1 / (1-A) is included in S
Verification: if a is included in s, then 1-1 / A is included in S
Evidence: if a is included in s, then 1 / (1-A) is included in S
If 1 / (1-A) is included in s, then 1 / (1-1 / (1-A)) is included in S
And 1 / (1-1 / (1-A)) = (A-1) / a = 1-1 / A
What is the maximum value of the function f (x) = 2x ^ 3-3x ^ 2 in the interval [- 1,4]?
Have you ever done math?
The derivative of F (x) is obtained
f（X）=6x^2-6x
When the derivative is 0, take the extremum. When f (x) = 0, x = 1 or x = 0
Then, when x = 0, f (x) takes the maximum value of 0
When x = 4, f (x) = 80
So the maximum value in the interval is 80
Zero
Is there a real number Z such that f {x} = the fourth power of X + {2-
z} The square + 2-z of X is a decreasing function on {negative infinity, 2} and a decreasing function on {negative one, 0}. There is a range for finding Z
1.1÷7×（x+14）=1÷4×（x+20） 2.1÷5×（x+15)=1÷2-1÷3×（x-7）
3.x-2÷5-x+3÷10=2x-5÷3-3 4.x-x-1÷2=2-x+2÷3
5. Given equation 1 △ 3-2 (x-1 △ 2005) = - 5 △ 3, then the value of algebraic formula 15 + 3 (x-1 △ 2005) is?
Find the function f (x) = 2x (5-3x), X belongs to the maximum value of 0,5 / 3 open interval?
f(x)=-6x²+10x
=-6(x²-5x/3)
=-6(x-5/6)(x-5/6)+25/6
The maximum value is 25 / 6 when x = 5 / 6
f(x)= －6x²　＋　10x；
F ′ (x) = - 12x + 10, Let f ′ (x) = 0, then x = 5 / 6;
On (0,5 / 6), f ′ (x) > 0, f (x) increases;
On (5 / 6, 5 / 3), f ′ (x) < 0, f (x) decreases;
Obviously, when x = 5 / 6, f (x) reaches the maximum
Use the vertex form of quadratic function, or use the mean inequality
Let f (x) = x power of 4 divided by x power of 4 + 2
1. Explore the relationship between F (x) and f (1-x)
2. Find f (1 divided by 101) + F (2 divided by 101) + F (3 divided by 101) + +F (99 divided by 101) + F (100 divided by 101)
1. F (1-x) = 4 ^ 1-x / (4 ^ 1-x + 2) = 4 / (4 + 2 * 4 ^ x) = 2 / (4 ^ x + 2), so f (x) + F (1-x) = 1
2.f(1/101)+f(2/101)+f(3/101)+------+f(99/101)+f(100/101)=50[f(1/101)+f(1/100)]=50