If the function y = ax ex has an extreme point less than zero, then the value range of real number a is () A. （0，+∞）B. （0，1）C. （-∞，1）D. （-1，1）

If the function y = ax ex has an extreme point less than zero, then the value range of real number a is () A. （0，+∞）B. （0，1）C. （-∞，1）D. （-1，1）

∵ y = ax ex, ∵ y '= a-ex. from the meaning of the title, we know that a-ex = 0 has a real root less than 0, let Y1 = ex, y2 = a, then the intersection of the two curves is in the second quadrant, combined with the image, it is easy to get 0 < a < 1, so the value range of real number a is (0, 1), so choose B
Find the sine value is 0.2873 degrees of angle ~ work related, urgent
The sine value is 4400 / 15315~
The inverse function can be calculated with the scientific calculator of windows
16 ° 41'46.97 "I calculated with a counter
16 degrees 6964
16.7 degrees
arcsin(0.2873) = 0.29140680085375
This function is used to find the inverse sine f (x) = arcsinx
The top one is the radian, the angle is 16.6964 degrees
The conversion form is 16 ° 41 ′ 46.97 ″
It is known that the sum of the first n terms of the sequence {an} is the nth power of Sn = 3. The sequence {BN} satisfies B1 = - 1, BN + 1 = BN + (2n-1) (n belongs to positive infinity)
The correct answer is as follows: because Sn = 3 to the nth power, so sn-1 = 3 to the nth power minus 1, (n greater than or equal to 2), so an = sn-sn-1 = 2 ^ 3 to the nth power minus 1 (n greater than or equal to 2), when n = 1, the 1-1 power of 2 ^ 3 = 2 does not equal S1 = A1 = 3, So an = 3 (n = 1) or the nth power of 2 ^ 3 minus 1 (n is greater than or equal to 2). How can the nth power of an = sn-sn-1 = 2 ^ 3 minus 1 (n is greater than or equal to 2) be simplified from the original nth power of an = sn-sn-1 = 3 minus 1 to get the nth power of 2 ^ 3 minus 1~
One
an=Sn-Sn-1=3^n-3^(n-1)=2 * 3^(n-1)
Two
bn+1=bn+(2n-1)
bn=bn-1+(2n-3)
..
b2=b1+1
b1=-1
Sbn=Sbn-1 -1 +[1+(2n-3)](n-1)/2
Sbn-Sbn-1=(n-1)^2-1
bn=(n-1)^2-1
If an is equal to 2, the nth power minus 1 can be obtained by using the formula of equal ratio sequence
I hope my answer can help you. Thank you! Ha, you seem to have a lot of work to do. When n ≥ 2, an = SN-S (n-1) = 3 ^ n-3 ^ (n-1) = 3 × 3 ^ (n-1) - 3 ^ (n-1) = 3 ^ (n-1) [3-1] = 2 × 3 ^ (n-1). I hope my answer can help you
If an is equal to 2, the nth power minus 1 can be obtained by using the formula of equal ratio sequence
I hope my answer can help you! Ha, you seem to have a lot of work to do
Let A1, A2 and A3 be 3-dimensional column vectors, and note that a = (A1, A2, A3) B = (a1 + A2 + a3, a1 + 2A2 + 2A3, a1 + 3a2 + 4A3). If | a | is 1, then | B|=
If the function y = e (A-1) x + 4x (x ∈ R) has an extreme point greater than zero, then the range of real number a is ()
A. a＞-3B. a＜-3C. a＞−13D. a＜−13
Because the function y = e (A-1) x + 4x, so y '= (A-1) e (A-1) x + 4 (a < 1), so the zero point of the function is x0 = 1a − 1ln41 − a, because the function y = e (A-1) x + 4x (x ∈ R) has an extreme point greater than zero, so x0 = 1a − 1ln41 − a > 0, that is, ln41 − a < 0, the solution is a < - 3
How many discontinuities are there in the function f (x) = x / ln / 1-x /?
How many discontinuities of function f (x) = x / ln (absolute value of 1-x)? How to calculate them?
The discontinuous point of elementary function is the meaningless point of function
The discontinuities of the function f (x) = x / ln | 1-x | are x = 0, x = 1, x = 2
1-x = 0, logarithmically meaningless,
The functions ln | 1-x | = 0, | X-1 | = 0, x = 0, x = 2 are meaningless
It is known that the sum of the first n terms of the sequence an is Sn, which satisfies the n-th power of an + Sn = 3-8 / 2. Let BN = 2 multiply an
The sequence BN is an arithmetic sequence
By the question: SN = 3 - 8 / 2 ^ n - ansn-1 = 3 - 8 / 2 ^ (n-1) - an-1an = Sn - sn-1 = [3 - 8 / 2 ^ n - an] - [3 - 8 / 2 ^ (n-1) - an-1] = 8 / 2 ^ (n-1) - 8 / 2 ^ n - an + an-1 both sides at the same time + an: 2An = 8 / 2 ^ (n-1) - 8 / 2 ^ n + an-1 both sides multiply by 2 ^ (n -
Let 3-order matrix A = (A1, A2, A3), where a1, A2, A3 are all 3-dimensional column vectors, and | B | = 2, matrix B = (a1 + A2 + a3, a1 + 2A2, a1 + 3a2 + a3). Then | a | =?
B=(a1+a2+a3,a1+2a2,a1+3a2+a3) =(a1,a2,a3)K = AK
K =
1 1 1
1 2 3
1 0 1
So | B | = | a | K|
That is, 2 = 2|a|
So | a | = 1
If the function FX = ax ^ 3-bx + 4, when x = 2, the function FX has the extremum - 4 / 3 1. Find the analytic expression of the function
If the function FX = ax ^ 3-bx + 4, when x = 2, the function FX has the extremum - 4 / 3 1. Find the analytic expression of the function
If there are different real roots of the equation K (k) = 3
The value range is obtained
If f (2) = 8a-2b + 4 = - 4 / 3, then f '(2) = 12a-b = 0, so a = 1 / 3, B = 4f (x) = 1 / 3x ^ 3-4x + 4, then if f' (x) = x ^ 2-4 = 0, then x = 2 or x = - 2, f (- 2) and f (x) are two extremums, and f (- 2) = 8 + 4-8 / 3 = 26 / 3, so K has three different roots between intervals (- 4 / 3, 26 / 3)
The infinite breakpoint of function f (x) = (X-2) / (LN x-1) is? Write down the specific process
First, find out the discontinuity point, that is, the point where the function has no meaning, that is, x = 2 or 0 or 1
Then find the limits f (2) and f (0)
F (2) is infinity, f (0) is also infinity (f (0 +) is positive infinity, f (0 -) is negative infinity, which is also called infinite discontinuity)
f(1)=0
So the infinite breakpoint is x = 2 or 0
I don't know if the answer is right. If not, please correct me
Removable Discontinuity
ln(1-x) = - x + 1/2 * x^2 - 1/3 * x^3 ....