Let f (x) = LG1 + 2x + 4xa3, if f (x) is meaningful when x ∈ (- ∞, 1], find the value range of real number a

Let f (x) = LG1 + 2x + 4xa3, if f (x) is meaningful when x ∈ (- ∞, 1], find the value range of real number a

When x ∈ (- ∞, 1], the function problem of F (x) = LG1 + 2x + 4xa3 is transformed into an inequality problem in which 1 + 2x + 4xa > 0 holds constant on X ∈ (- ∞, 1]. The inequality 1 + 2x + 4xa > 0 holds constant on X ∈ (- ∞, 1], that is, a ＞ - [(12) 2x + (12) x] holds constant on X ∈ (- ∞, 1]. Let t = (12) x, then t ≥ 12, and let g (T) = T2 + T, whose axis of symmetry is t = - 12  g (T) = T2 +T is an increasing function on [12, + ∞). When t = 12, G (T) has the minimum value g (12) = (12) 2 + 12 = 34, so the value range of a is a ＞ - 34
Who has a scientific calculator that can calculate degrees, minutes and seconds
59 degrees 56, 59 degrees 56 and 59 degrees 57
What is the average of these three numbers? The result should be expressed in degrees, minutes and seconds
If you add 60 degrees to each question, you'll know that
If both {an} and {BN} are arithmetic sequences, and A1 = 25, B1 = 75, A100 + B100 = 100, then the value of item 37 of {an + BN} is?
Thanks for the process
It should be 100, assuming that the deviation of each term of {an} is: X; assuming that the deviation of each term of {BN} is: y; A100 = a1 + 99 * x; B100 = B1 + 99 * y; = = > a1 + 99 * x + B1 + 99 * y = 100 = = > 25 + 99 * x + 75 + 99 * y = 100 = = > 99 * x + 99 * y = 0; = = = > x = -
Let the tolerance of an be D1 and that of BN be D2
a100+b100=25+99d1+75+99d2=100
So D1 + D2 = 0
So A37 + b37 = 25 + 36d1 + 75 + 36d2 = 100
Given that vector group A1, A2, A3, A4 are linearly independent, then vector group () a a1 + A2, A2 + a3, A3 + A4, A4 + A1 are linearly independent
B a1-a2, A2-A3, a3-a4, a4-a1 are linearly independent
C a1 + A2, A2 + a3, A3 + A4, a4-a1 are linearly independent
DA1 + A2, A2 + a3, a3-a4, a4-a1 are linearly independent
C
The function f (x) = LG [(1 + 2 ^ x + 4 ^ XA) / 3] is meaningful in X ∈ (- ∞, 1]. The value range of real number a is obtained
If a = 0, then the true number is always greater than 0
A is not equal to 0
X
Scientific calculators calculate degrees, minutes and seconds
My calculator is Kenko brand. I want to use it to calculate degrees, minutes and seconds. How to use it? For example: 55 ° 40'47 "+ 55 ° 40'40" =? How to operate?
Kenko brand, whatever you want to use
Senior high school mathematics problem (about sequence) BN = - n * 2n (the nth power of two) Sn =?
Brother, let me help you
Offset subtraction
bn=-n•2^n
b1=-1•2^1,b2=-2•2^2,b3=-3•2^3,...,bn-1=-(n-1)•2^(n-1),bn=-n•2^n
Sn=b1+b2+b3+...+bn-1+bn
Sn=-1•2^1-2•2^2-3•2^3-...-(n-1)•2^(n-1)-n•2^n.①
2Sn=-1•2^2-2•2^3-3•2^4-...-(n-1)•2^n-n•2^(n+1).②
② - 1
Sn=1•2^1+1•2^2+1•2^3+1•2^n-n•2(n+1)
Sn=2^1+2^2+2^3+...+2^n-n•2^(n+1)
Sn=2•(1-2^n)/(1-2)-n•2^(n+1)
Sn=(1-n)•2^(n+1)-2
Offset subtraction
If the sequence {an} is an arithmetic sequence and the sequence {BN} is an equal ratio sequence, the new sequence {anbn} is composed of the product of the corresponding terms of the two sequences. When calculating the sum of the first n terms of the sequence, we often use the method of multiplying each term of {anbn} by the common ratio Q of {BN}, and subtracting the wrong term from the same term of the original {anbn}
Example: find the sum of the first n terms of sequence 1,2x, 3x ^ 2,4x ^ 3,..., NX ^ (n-1)
Sn=1+2x+3x^2+4x^3+...+nx^(n-1)...①
xSn=x+2x^2+3x^3+4x^4+...+nx^n...②
① - 2
(1-x)Sn=1+x+x^2+x^3+...+x^(n-1)-nx^n
When x ≠ 1
(1-x)Sn=(1-x^n)/(1-x)-nx^n
Sn=(1-x^n)/(1-x)^2-nx^n/(1-x)=[1-(1+n)x^n+nx^n]/(1-x)^2
When x = 1
Where Sn = 1 + 2 + 3 +... + n = n (1 + n) / 2
Example: find the sum of the first n terms of sequence 1 / 2,3 / 4,5 / 8,..., (2n-1) / 2 ^ n
Sn=1/2+3/4+5/8+7/16+...+(2n-3)/2^(n-1)+(2n-1)/2^n...①
Sn/2=1/4+3/8+5/16+...+(2n-3)/2^n+(2n-1)/2^(n+1)...②
① - 2
Sn-Sn/2=1/2+(3/4-1/4)+(5/8-3/8)+(7/16-5/16)+...+[(2n-1)/2^n-(2n-3)/2^n]-(2n-1)/2^(n+1)
Sn/2=1/2+(2/4+2/8+2/16+...+2/2^n)-(2n-1)/2^(n+1)
Sn/2=1/2+[1/2+1/4+1/8+...+1/2^(n-1)]-(2n-1)/2^(n+1)
Sn/2=1/2+{(1/2)·[1-1/2^(n-1)]/(1-1/2)}-(2n-1)/2^(n+1)
Sn/2=1/2+1-1/2^(n-1)-(2n-1)/2^(n+1)
Sn/2=3/2-[4/2^(n+1)+(2n-1)/2^(n+1)]
Sn/2=3/2-(2n+3)/2^(n+1)
Sn=3-(2n+3)/2^n
Let the vector group A1, A2, A3 be linearly related and A2, A3, A4 be linearly independent. It is proved that the vector group A1 must be a linear combination of A2, A3, A4
prove:
∵ A1, A2, A3 linear correlation
The existence of numbers B1, B2, B3 which are not all zero makes
b1a1+b2a2+b3a3=0
And A2, A3, A4 are linearly independent
A 2, a 3 are linearly independent
If B1 = 0, then b2a2 + b3a3 = 0
∴b2=b3=0
It contradicts that B1, B2 and B3 are not all zero
∴b1≠0
∴a1+(b2/b1)a2+(b3/b1)a3=0
That is, A1 = - (B2 / B1) a2 - (B3 / B1) A3
A 1 can be expressed as a linear combination of a 2, a 3 and a 4
It's over
When x = 1, there is a maximum of 3, and the value of a and B is the minimum of function y
When x = 1, there is a maximum of 3, the value of a and B can be obtained, and the minimum diagram of function y can be obtained,
First question
y=ax^3+bx^2
y'=3ax^2+2bx
When 3ax ^ 2 + 2bx = 0, y has extremum
Because when x = 1, there is an extreme value of 3
So 3A + 2B = 0
a+b=3
By solving the above equations, we can get a = -- 6, B = 9
So the analytic expression of the function is y = - 6x ^ 3 + 9x ^ 2
Second question
We can see from the first
y'=-18x^2+18x
When y '= 0, x = 0 or x = 1
X x
y'=3ax^2+2bx
When x = 1, there is a maximum of 3
3a+2b=0
A + B = 3
a=-6 b=9
y=-6x^3+9x
y'=-18x^2+18x
Let y '= 0, x = 1 or x = - 1
x x
How to make Casio 9860 calculator angle output format is degree / minute / second
First press shift, then set up. The third option is angle, the fourth option is radian, and the fifth option is gradient
Press ". Key