If the function f (x) = x & # 178; + (4a + 1) x + 2 is a decreasing function in the interval 0 (- ∞, 1), then the value range of a is

If the function f (x) = x & # 178; + (4a + 1) x + 2 is a decreasing function in the interval 0 (- ∞, 1), then the value range of a is

Symmetric x = - (4a + 1) / 2 less than or equal to 1
(4a + 1) &# - 4 * 1 * 2 is less than or equal to 0
Get: 3 / 4 less than or equal to a less than or equal to (2 * radical 2-1) / 4
When x approaches 0, LIM (SiNx + TaNx) / x =?
lim(sinx+tanx)/x (x→0)
=lim(sinx)/x+lim(tanx)/x
Using equivalent infinitesimal = 2
Or using the law of lobida = limcosx + Lim1 / cos & # 178; X = 2
=lim(sinx/x+tanx/x)
If x tends to 0, then SiNx and TaNx are equivalent infinitesimals of X
So the original formula = 1 + 1 = 2
lim(sinx+tanx)/x=lim(sinx)/x+lim(tanx)/x=2lim(sinx)/x=2
Given the function f (x) = (m-1) x ^ 2-2mx + 2 (m ^ 2-1), m belongs to R. try to compare the size of F (c + 1) and f (c) (C belongs to R)
When m = 1, f (x) = - 2x, monotonically decreasing, so when f (c + 1) 1, the axis of symmetry x = m / (m-1),
(1) When m / (m-1) > = C, f (c + 1) > F (c);
(2) When m / (m-1)
Given the function f (x) = 3 ^ x, and f (a + 2) = 18, G (x) = 3 ^ ax-4 ^ x, the domain is the interval [- 1,1] to find the range of G (x)
a=log3(2)
g(x)=3^ax-4^x=2^x-4^x
Let 2 ^ X be t, GX = t-t2
The domain of T is greater than or equal to 0.5 and less than or equal to 2
GX is [- 2,0.25]
If f (x) = (M2-1) x2 + (m-1) x + (n + 2) is an odd function, then the value of M, n is ()
A. m=1，n=2B. m=-1，n=2C. m=±1，n=-2D. m=±1，n∈R
From the definition of odd function, we know that: F (- x) = (M2-1) X2 - (m-1) x + (n + 2) = - f (x) = - (M2-1) X2 - (m-1) x - (n + 2); ■ (M2-1) = - (M2-1), N + 2 = - (n + 2), ｜ M = ± 1, n = - 2; so we choose C
Given the function f (x) = ax-24-ax-1 (a > 0 and a ≠ 1); (1) find the domain of definition and value of function f (x); (2) whether there is a real number a, so that function f (x) satisfies: for any x ∈ [- 1, + ∞), there is f (x) ≤ 0? If it exists, find out the value range of a; if not, explain the reason
(1) When a ＞ 1, X ≤ loga4; when 0 ＜ a ＜ 1, X ≥ loga4. That is to say, when a ＞ 1, the domain of F (x) is (- ∞, loga4]; when 0 ＜ a ＜ 1, the domain of F (x) is [loga4, + ∞). Let t = 4-ax, then 0 ≤ t ＜ 2, ax = 4-t2, ｜ f (x) = g (T) = 4-t2-2t-1 = - (T + 1) 2 + 4. When t ≥ 0, G (x) is a monotone decreasing function of T, ｜ G（ 2) If there is a real number a such that f (x) ≤ 0 for any x ∈ [- 1, + ∞), then the interval [- 1, + ∞) is a subset of the domain. It is known from (1) that a ＞ 1 does not satisfy the condition; so 0 ＜ a ＜ 1, and loga4 ＜ - 1, that is, 14 ≤ a ＜ 1. Let t = 4-ax, it is known from (1), f (x) ＜ 1 ）=4-t2-2t-1 = - (T + 1) 2 + 4, from F (x) ≤ 0, the solution is t ≤ - 3 (rounding) or t ≥ 1, that is, there is 4-ax ≥ 1, and the solution is ax ≤ 3. From the meaning of the problem, for any x ∈ [- 1, + ∞), there is ax ≤ 3 constant, because 0 < a < 1, so for any x ∈ [- 1, + ∞), there is ax ≤ A-1. So there is A-1 ≤ 3, and the solution is a ≥ 13, that is 13 ≤ a < 1 There are f (x) ≤ 0
Given that the function y = f (x) is a decreasing function on (0, + ∞), and f (m2-2m) > F (m), the value range of M is obtained
∵ function y = f (x) is a decreasing function on (0, + ∞) ∵ f (m2-2m) > F (m) is equivalent to M2 − 2m ＞ 0m ＞ M2 − 2m ＜ M & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; the solution is 2 ＜ M ＜ 3. The value range of 〈 m is (2,3)
If the functions with the same range of values and the same domain of definitions are twin functions, how many twin functions are there with the range of y = 2x square + 1 greater than 5 and less than 19
I'm the same on the first floor, but there are only 7, 8, 9 and 10 answers. What's the matter
3
y=2x^2+1,√2
Only one (two) pair:
y=2x²+1 x∈（√2，3）
y=2x²+1 x∈（-3，-√2）
This question mainly examines the solution inequality
There are infinitely many,
According to the image, it is easy to see that the range of value on [radical 2,3] is (5,9). In (- 3, - radical 2), take any sub interval and [radical 2,3] as the domain of definition, and its range of value is (5,9), so there are infinitely many
Or in any subinterval of (radical 2,3) and
The union of (- 3, - radical 2) is the domain of definition, and there are infinitely many values in the same domain of (5,9).
To combine number with shape
There are infinitely many,
According to the image, it is easy to see that the range of value on [radical 2,3] is (5,9). In (- 3, - radical 2), take any sub interval and [radical 2,3] as the domain of definition, and its range of value is (5,9), so there are infinitely many
Or in any subinterval of (radical 2,3) and
The union of (- 3, - radical 2) is the domain of definition, and there are infinitely many values in the same domain of (5,9).
It's a combination of numbers and shapes
Get 5 from the range
If F X = 4 ^ x M2 ^ x-6m has exactly one zero point, the real number is in the range of M
f x=4^x+m2^x-6m
=(2^x+m/2)²-m²/4-6m
F X = 4 ^ x + M2 ^ x-6m has exactly one zero point
(1)2^x+m/2=0,-m²/4-6m=0
M = - 24
(2) m/2>0,-m²/4-6m0
The range of real number m is {- 24} ∪ {m | m > 0}
The definition field of function y = 2 ^ (x-1) is?, and the value field is
The definition field is R. the value field is {X / x > 0}
The image of this function is shifted to the right by y = 2 ^ x, without changing the value of the function
So apply the definition domain and value domain of exponential function directly