Given that f (x) = X3 + | x | SiNx + 2 | x | + 2 (x ∈ R), the maximum value is m, and the minimum value is m, then M + M=______ .

Given that f (x) = X3 + | x | SiNx + 2 | x | + 2 (x ∈ R), the maximum value is m, and the minimum value is m, then M + M=______ .

Let f (x) = X3 + | x | SiNx + 2 | x | + 2 = x3 − SiNx | + 2 + 1. Let f (x) = g (x) + 1, so g (x) = x3 − SiNx | + 2, because g (- x) = - G (x), so g (x) is an odd function
F (x) = [(x + 1) ^ 2 + SiNx] / (x ^ 2 + 1) the maximum value of this function is m, the minimum value is m, find the value of M + M
f(x)=[(x+1)^2+sinx]/(x^2+1)
=（x^2+1+2x+sinx)/(x^2+1)
=1+(2x+sinx)/(x^2+1)
Let g (x) = f (x) - 1 = (2x + SiNx) / (x ^ 2 + 1)
Then G (- x) = - G (x) ‖ g (x) is an odd function
The maximum value of the ∵ function is m
∴g(x)max=M-1
∵ g (x) is an odd function
∴g(x)min=1-M
∵ f(x)=g(x)+1,
∴m=f(x)min=g(x)min+1=2-M
∴M+m=2
[(x+1)^2+sinx]/(x^2+1)＝1+（2x+sinx)/(x^2+1) .
After the edge is an odd function, after the maximum, minimum is the opposite number.
So the maximum of the whole is m + M = 2
Make the following function image y = x square - 2, absolute value x - 1
The even function f (- x) = f (x) of this function is symmetric about the y-axis. First, draw the image of x > 0, and then fold it about the y-axis
What is n-dimensional column vector, n-dimensional row vector
Please give me some examples
3D vector row, 3D vector column
be deeply grateful
First of all, column vector and row vector are the knowledge points of linear algebra. Row vector is called row vector because the components are horizontally arranged, and column vector is called column vector because the components are vertically arranged. There is no essential difference between them. N dimension is because vector has n components, (1,2,4) is three-dimensional row vector. If 1,2,4 is written vertically in parentheses, it is called three-dimensional column vector
In high school mathematics, the coordinates of the three vertices of △ ABC are a (4,1); B (2,3); C (6, - 3), and the median line Mn parallel to AB, then the equation of the straight line Mn is?
First, the coordinates of M and N are calculated as ((4 + 6) / 2, (1-3) / 2) and ((2 + 6) / 2, (3-3) / 2)
They are (5, - 1) and (4,0)
Then according to the point oblique formula y = (5-4) / (- 1) (x-4)
So the equation of line Mn is x + y-4 = 0
I don't know. Ask the teacher
LIM (2An + 4bn) = 8, LIM (6An BN) = 1, the value n of LIM (3an + BN) tends to infinity
Let x (2An + 4bn) + y (6An BN) = 3an + BN, then 2x + 6y = 3, 4x-y = 1, x = 9 / 26, y = 5 / 13,
So LIM (3an + BN) = 72 / 26 + 5 / 13 = 41 / 13
Two thirds of the root - 4 × 216 - 42 × one sixth of the root
√(2/3)-4×√216-42×√(1/6)
=√2/√3-4×6√6-42×√6/6
=√6/3-24√6-7√6
=-92√3/3
The results are as follows
Molecule: - 92 √ 3
Denominator: 3
If you have any questions, please ask. If you are satisfied, please choose it as the satisfactory answer!
√(2/3)-4×√216-42×√(1/6)
=√2/√3-4×6√6-42×√6/6
=√6/3-24√6-7√6
=-92√6/3
N-dimensional vector computation
It is known that A1, A2, B1, B2 and y are all three-dimensional column vectors, and the determinant is the same
|a1,b2,y|=|a1,b2,y|=|a2,b1,y|=|a2,b2,y|=3
So | - 2Y, a1 + A2, B1 + 2B2 | =?
|-2Y, a1 + A2, B1 + 2B2 | = - 2 | y, a1 + A2, B1 + 2B2 | (PS is proposed from the first column, 2) = - 2 | y, A1, B1 + 2B2 | - 2 |, A2, B1 + 2B2 | (Ps. the second column can be divided)
=-2 | y, A1, B1 | - 4 | y, A1, B2 | - 2 |, A2, B1 | - 4 |, A2, B2 | (PS assign the third column)
|y. A1, B1 | = | A1, B1, y |
So = - 2 * 3-4 * 3-2 * 3-4 * 3 = - 36
Given that the three vertices of △ ABC are a (1,2), B (3,0), C (7,4), then the equation of the straight line of the median line parallel to AB in △ ABC is ()
A. x+y-7=0B. x+y+3=0C. x+y-5=0D. x+y-2=0
Let the median line of AB side be de, d be the midpoint of AC, and E be the midpoint of BC, then the slope of the line where the median line De is located is k = the slope of the line where AB is located is 2 − 01 − 3 = - 1. According to the midpoint coordinate formula, D (1 + 72, 2 + 42), i.e. (4, 3), so the general equation of the line is Y-3 = - (x-4) and simplified to x + Y-7 = 0
LIM (3an + 4bn) = 8 LIM (6An BN) = 1 to find LIM (3an + BN), let 3an + 4bn = m 6An BN = t
In the second problem, if an = (5-3x) ^ n 1) an has a limit, find the range of X. 2) an limit is zero, find the range of X
Let the limit of an be m and the limit of BN be t
lim(3an+4bn)=8 3m+4t=8
lim(6an-bn)=1 6m-t=1
m=4/9 t=5/3
lim(3an+bn) =3m+t=3
If an = (5-3x) ^ n 1) an has a limit, - 1