When k is the value, the domain of the function y = LG (KX ^ 2 + 4kx + 3) is r? When k is the value, the domain is r?

When k is the value, the domain of the function y = LG (KX ^ 2 + 4kx + 3) is r? When k is the value, the domain is r?

1. The domain is r
If the logarithm is significant and the true number is more than 0, and if the definition field of the function is r, then for any real number k, KX & # 178; + 4kx + 3 is constant > 0
When k = 0, 3 > 0
When k ≠ 0, for the quadratic function f (x) = KX & # 178; + 4x + 3, the coefficient of quadratic term k > 0, for the quadratic equation KX & # 178; + 4kx + 3 = 0, the discriminant △

Function f (x) = LG (x ^ 2-2x) / (# 10004); (9-x ^ 2) domain,

Because x ^ 2-2x > 0, 9-x ^ 2 > 0,
So X2, and - 3

Let f (x) be a function defined on R with period 2. In the interval [- 1,1], f (x) = {ax + 1 (1), - 1

Yes, f (1) = f (1-2) = f (- 1)

Let f (x) be an even function defined on [a + 1,2], then f (x) = ax * x + bx-2 is an increasing function or a decreasing function on the interval [0,2]?

Let f (x) = ax * x + bx-2 be an even function defined on [a + 1,2], then is f (x) an increasing or decreasing function on the interval [0,2]?
Let f (x) be an even function and the domain of definition be symmetric about the origin, so a + 1 = - 2, a = - 3, so f (x) = - 3x * x + bx-2. Since f (x) is an even function, f (- x) = - F (x) holds for any x ∈ [- 2,2], so B = 0,
So f (x) = - 3x * x - 2, decreasing on [0,2]

Given that the tangent of the image of the function f (x) = X3 + AX2 + B at point P (1, f (1)) is 3x + Y-3 = 0. (1) find the function f (x) and monotone interval; (2) find the maximum value of the function in the interval [0, t] (T > 0)

(1) From the point P on the tangent, we get f (1) = 0, that is, point P (1, 0) on y = f (x), we get a + B = - 1 and f '(1) = - 3 {2A = - 6, so f (x) = x3-3x2 + 2F' (x) = 3x2-6x, Let f '(x) > 0, we get x ＞ 2 or X ＜ 0, the increasing interval of  f (x) is (- ∞, 0), (2, + ∞), and the decreasing interval is (?)

If f (a ^ 2-A + 1) + F (4a-5) > 0, find the value range of real number a
I begged several times, but it was not right

By using f (x) as an odd function, we get: F (a ^ 2-A + 1) > F (5-4a)
Because f (x) is decreasing, a ^ 2-A + 1 = - 1
(2)a^2-a+1

If f (a * a-a-1) + F (4a-5) > 0, find a
If f (a * a-a-1) + F (4a-5) > 0, find the range of A

F (x) is defined in the closed interval - 1,1 → - 1 ≤ a * a-a-1 ≤ 1-1 ≤ 4a-5 ≤ 1 → 1 ≤ a ≤ 1.5. ① f (x) is an odd function → - f (4a-5) = f (- (4a-5)) = f (5-4a) f (a * a-a-1) + F (4a-5) > 0 → f (a * a-a-1) > - f (4a-5) = f (5-4a) and f (x) is a decreasing function → a * a-a-1 ＜ 5-4a → (- 3 - √ 33) / 2 ＜ a

If the function f (x) is a decreasing function on (0, + ∞), then the relationship between F (a ^ 2-A + 1) and f (3 / 4) is positive____ .
If the function f (x) is a decreasing function on (0, + ∞), then the relationship between F (a ^ 2-A + 1) and f (3 / 4) is positive____ .A.f(a^2-a+1)≤f(3/4) B.f(a^2-a+1)≥f(3/4) C.f(a^2-a+1)＜f(3/4) D.f(a^2-a+1)＝f(3/4)

A ^ 2-A + 1 = (A-1 / 2) ^ 2 + 3 / 4 > = 3 / 4 > 0, so the answer is a.f (a ^ 2-A + 1) ≤ f (3 / 4)

It is known that the function f (x) is a decreasing function on the interval (0, +), and the size relationship between F (A2 (the second power of a) - A + 1) and f (3 / 4) is obtained

If A2 (the quadratic power of a) - A + 1) is greater than (3 / 4), then A2 (the quadratic power of a) - A + 1) is less than f (3 / 4) (because the function f (x) is a decreasing function on the interval (0, +), otherwise
We can use A2 (the second power of a) - A + 1 - (3 / 4) to get (A-1 / 2) 2 (the square of A-1 / 2). (A-1 / 2) 2 (the square of A-1 / 2) is an equation greater than or equal to 0. So we can get that f (A2 (the second power of a) - A + 1) is less than or equal to f (3 / 4)
Because n has not done for a long time, so only ideas. No standard process. Sorry

If the function y = f (x) is a decreasing function on R, compare the magnitude of F (a ^ 2 = a = 1) and f (3 / 4)

A ^ 2 + A + 1 > = 3 / 4, I don't need to say that
Then, because the function y = f (x) is a decreasing function on R
So the bigger x is, the smaller f (x) is
So f (a ^ 2 + A + 1)