If the function f (x) = x2 + (a2-4a + 1) x + 2 is a decreasing function in the interval (- ∞, 1], then the value range of a is______ .

If the function f (x) = x2 + (a2-4a + 1) x + 2 is a decreasing function in the interval (- ∞, 1], then the value range of a is______ .

Since the equation of the axis of symmetry of the function f (x) = x2 + (a2-4a + 1) x + 2 is x = - A2 − 4A + 12, and the function is a decreasing function in the interval (- ∞, 1], there is - A2 − 4A + 12 ≥ 1, and 1 ≤ a ≤ 3 is obtained, so the answer is: [1, 3],
If the domain of the function f (x) = x ^ 2 + X + 1 / 2 is [n, N + 1] (n is a natural number), then there are a total of integers in the domain of the function
Let me talk about ideas,
1. First, change y = x ^ 2 + X + 1 / 2 into the formula of Y expressed by X
That is, x = how much y, y as an independent variable
2. The domain of definition is [n, N + 1], that is, the range of x = how much y
3, and then calculate the relationship between Y and N, discuss, n = 1, which integer can y take, n = 2,3 Generally, it is easy to judge
Summary: to find the range of a function, it's better to change its domain into another function's domain, and then find another function's domain, that is, the function's domain
It is known that the function f (x) = (m-1) x * x + 2mx + 3 is an odd function__ f(a*a-a+1)(a
A is r
It's even. It can't be odd
If it is an even function, then 2m = 0, M = 0
f(x)=-x^2+3
a^2-a+1=(a-1/2)^2+3/4>=3/4
So f (a ^ 2-A + 1) = f (a ^ 2-A + 1)
The problem is wrong, odd function to go through the origin, f (x) obviously but the origin
In this problem, it is impossible that the image of its odd function must pass through the origin
It's an even function. It can't be an odd function.
If it is an even function, then 2m = 0, M = 0
f(x)=-x^2+3
a^2-a+1=(a-1/2)^2+3/4>=3/4
So f (a ^ 2-A + 1) = f (a ^ 2-A + 1)
Let f (x) = x2 + X + 12 be defined as {n, N + 1} (n is a natural number), then there are two functions in the range of F (x)______ This is an integer
When n ≥ 1, f (x) is monotonically increasing on [n, N + 1], f (n + 1) - f (n) = (n + 1) 2 + (n + 1) + 12-n2-n-12 = 2n + 2, so the number of integers in the range of F (x) is 2n + 2, when n = 0, the range of values is [f (0), f (1)] = [12, 52], and there are two integers 1 and 2
Find the maximum 3Q of quadratic function f (x) = x ^ 2 + 2mx-3 (M belongs to R) in the interval [- 1.3]
F (x) = x ^ 2 + 2mx-3 (M belongs to R), the equation of axis of symmetry x = - m,
The central axis position of X in the interval [- 1,3] is x = (3-1) / 2 = 1
(1) When - M ≤ - 1, m ≥ 1, when x = 3, f (x) max = 6 + 6m,
(2) When - 1
The symmetry axis of quadratic function is x = - M
1. When - M ≤ - 1, the maximum value is f (3)
2. When - M ≥ 3, the maximum value is f (- 1)
Let f (x) be a function over the domain n *, f (1) = 1. For any natural number a and B, f (a) + F (b) = f (a + b) - AB, find f (x)
Let a = 1, B = x, f (1) + F (x) = f (x + 1) - x, and f (1) = 1, so f (x) = f (x + 1) - X-1, so f (x + 1) - f (x) = x + 1, that is: F (2) - f (1) = 1 + 1, f (3) - f (2) = 2 + 1 , f (x) - f (x-1) = X-1 + 1, left cumulative sum: so f (x) - f (1) = (1 + x-1) (x-1) / 2 + X-1, so f (x) = x (x-1) / 2 + X-1
B=1
f(a)+1=f(a+1)-a
f(a+1)-f(a)=a+1
therefore
f(a)-f(a-1)=a
f(a-1)-f(a-2)=a-1
……
f(2)-f(1)=2
Add
f(a)-f(1)=2+3+…… +a=(a+2)(a-1)/2
f(a)=(a²+a)/2
f(x)=(x²+x)/2
If the function f (x) = (m-1) x ^ 2 + 2mx + 3 is even, then the size relation of F (- & frac34;), f (A & sup2; - A + 1) (a ∈ R) is
If the function f (x) = (m-1) x ^ 2 + 2mx + 3 is even, then the size relation of F (- 3 / 4), f (a ^ 2-A + 1) (a ∈ R) is?
Because f (x) = f (- x)
So (m-1) x ^ 2 + 2mx + 3 =) = (m-1) x ^ 2-2mx + 3
So m = 0, f (x) = - x ^ 2 + 3
*Then we know the increasing and decreasing intervals of this function
Calculate the range of a ^ 2-A + 1 formula
And f (- 3 / 4) = f (3 / 4)
Compare the range of a ^ 2-A + 1 with the size of 3 / 4
According to * step, we can get the size relation of F (- 3 / 4), f (a ^ 2-A + 1) (a ∈ R)
I should know, Bo!
Given the function f (x) = log12 [(12) x − 1], (1) find the domain of definition of F (x); & nbsp; & nbsp; (2) discuss the increase and decrease of function f (x)
(1) From (12) X-1 ＞ 0, the definition field of {x | x ＜ 0}. (2) let x1 ＜ x2 ＜ 0, ∵ y = (12) X-1 be a decreasing function, ∵ f (x) = log12x be a decreasing function, ∵ log12 [(12) x1-1] ＜ log12 [(12) x2-1], ∵ f (x) = log12 [(12) x − 1] be an increasing function on (- ∞, 0)
F (x) = (M2 - 1) x2 + (m-1) x + N + 2 when m, n why is f (x) an odd function
M ^ 2-1 = 0 and M-1 is not equal to 0
So m = - 1
Given the function f (x) = 1 / x-log1 + X / 1-x, find the domain of definition and parity
If the domain of definition is x ≠ 0 and (1 + x) / (1-x) > 0, the solution is x ∈ (- 1,0) ∪ (0,1);
Judging parity: F (- x) = - 1 / X - log (1-x) / (1 + x) = - 1 / x + log (1 + x) / (1-x) = - f (x), so it is an odd function