Any real number x 1, x 2 min {x 1, x 2} denotes the smaller one If f (n) = 2-N & # 178; G (n) = n, what is the maximum value of Min {f (n), G (n)} I don't quite understand

Any real number x 1, x 2 min {x 1, x 2} denotes the smaller one If f (n) = 2-N & # 178; G (n) = n, what is the maximum value of Min {f (n), G (n)} I don't quite understand

This problem is more intuitive by graphic method
Firstly, the functions f (n) = 2-N & # 178; and G (n) = n are given on the same coordinate axis;
Then find the coordinates of the intersection point
The coordinates at the intersection point are solved by solving the quadratic equation of 2-N & # = n, and the coordinates are (- 2, - 2), (1,1);
It can be seen from the image that:
(1) When N1, f (n) = 2-N & # 178; the image is always below the image g (n) = n, so f (n)

How to calculate x1. ^ x2 in MATLAB? X1 and X2 are two matrices

Find the power of x2 matrix elements at the same position of X1 matrix elements;
a=[2 2];
b=[3 3];
a.^b
ans =
8 8

Calculate the value of: (2 + 1) (the 2nd power of 2 + 1 (the 4th power of 2 + 1) ······ (the 2nth power of 2 + 1)

Because 2-1 = 1, the value of the original formula multiplied by (2-1) remains unchanged
The original formula = (2-1) (2 + 1) (2 & # 178; + 1) (2 ^ 4 + 1) (2 ^ 2n + 1) repeated use of square difference formula
=(2²-1)(2²+1)(2^4+1)…… (2^2n+1)
=(2^4-1)(2^4+1)…… (2^2n+1)
=(2^8-1)…… (2^2n+1)
=(2^4n-1)

On the hyperbola 4x2-y2 + 64 = 0, the distance from a point P to one of its focal points is equal to 1, then the distance from point P to another focal point is equal to ()
A. 17B. 16C. 15D. 13

∵ hyperbola 4x2-y2 + 64 = 0, ∵ the standard equation of hyperbola is y264 − x216 = 1, ∵ a = 8, C = 43, the distance from a point P to one of its focal points is equal to 1, let the distance from point P to another focal point be x, then we can know from the definition of hyperbola: | X-1 | = 16, the solution is x = 17, or x = - 15 (rounding)

Several factorization problems in grade two of junior high school
1. Let y = (x-1) (x-3) (x-4) (X-6) + 10 prove that no matter x is any real number, the value of Y is always greater than 0
2. Factorization: x ^ 2 + 4xy + 4Y ^ 2-4x-8y + 3
3. If a ^ 2 + Ba + 12 can be decomposed into the product of two first-order factors and B is an integer, then B =?
② If a + 12a + B can be decomposed into the product of two linear factors and B is a positive integer, then B =?
4. Prove that the square difference of two adjacent odd numbers is a multiple of 8

1.y=(x-1)(x-3)(x-4)(x-6)+10=(x-1)(x-6)(x-3)(x-4)+10=(x2-7x+6)(x2-7x+12)+10=(x2-7x)^2+18(x2-7x)+82=(x2-7x)^2+18(x2-7x)+81+1=(x2-7x+9)^2+1>02.x^2+4xy+4y^2-4x-8y+3=(x+y)^2-4(x-y)+3=(x+y-1)(x+y-3)3.①12=3...

[x + 1] / 3-x / 2 ≥ [x + 1] / 4 17-3x / 4 ≤ 5 [x + 11] / 8-4
1.【x+1】/3-x/2≥【x+1】/4
2.17-3X/4≤5【x+11】/8-4
It's a system of inequalities

1. Direct general division: (x + 1) / 4 + X / 2 - (x + 1) / 3 ≤ 0
(5x-1)/12≤0
x≤1/5
2.5(x+11)/8+3x/4-4-17≥0
x≥113/11

It is known that the parabola y = ax & # 178; + BX + C intersects the y-axis at a (0,3), and intersects the x-axis at B (1,0) and C (5,0), respectively
(1) Find the analytical formula of this parabola
(2) If point D is a trisection point of line OA, the analytic expression of straight line DC is obtained
(3) If a moving point P starts from the midpoint of OA, first reaches a point on the x-axis (set as point E), then reaches a point on the symmetry axis of the parabola (set as point F), and finally moves to point A. find the coordinates of points E and f that make the total path of point P shortest, and find the length of the shortest path

Answer: 1) the parabola y = ax & # 178; + BX + C intersects the y-axis at point a (0,3), and intersects the x-axis at B (1,0) and C (5,0) respectively. Suppose the parabola y = a (x-1) (X-5), point a is substituted into 5A = 3, and the solution is a = 3 / 5. Therefore, if the parabola is y = (3 / 5) (x + 1) (X-5) 2) point D is the trisection of OA = 3, then point D is (0,1) or (0,1), 2) According to the intercept formula, we can know that the DC straight line is: X / 5 + y = 1 or X / 5 + Y / 2 = 1, so: y = - X / 5 + 1 or y = - 2x / 5 + 23) as shown in the figure, make the symmetry point P1 (0, - 3 / 2) of point P about X axis, make the symmetry point A1 (6,3) of point a about x = 3 of parabola symmetry axis, connect a1p1, intersect X axis at point E, intersect symmetry axis X = 3 at point F, and the straight line a1p1 is: Y-3 = [(- 3 / 2-3) / (0-6)] * (X-6) y = 3 (X-2) / 4, so: point E is (2,0), point F is (3, 3 / 4) the length of the shortest path = a1p1 = √ [(- 3 / 2-3) ^ 2 + (0-6) ^ 2] = 15 / 2, so: the length of the shortest path is 15 / 2

The coordinates of the intersection point of parabolic y = 2x & # 178; - 4x and X axis

(2.0). (0.0) give me a good comment

To solve an applied problem
An engineering team originally planned to excavate at least 600 cubic meters of earth in 10 days. After completing 120 cubic meters in the first two days, it required to excavate two days in advance. How much earth would it excavate in the next few days?
Let's just list out the equation. It should be an equation with unequal sign. My result is that at least 60 cubic meters of earth will be excavated every day on average. Suppose: in the next few days, at least x cubic meters of earth will be excavated every day on average
The equation is: 120 × 2 + 6x > 600
Did I do it right

(10-2-2)x≥600-120
x≥80

The point to ground trajectory equation with the square difference of the distance between O (0,0), a (C, 0) and the ground constant C is obtained

Let P (x, y) be such a point
Then Po ^ 2 = (x-0) ^ 2 + (y-0) ^ 2 = x ^ 2 + y ^ 2
PA^2=(x-c)^2+(y-0)^2=x^2-2cx+c^2+y^2
So | Po ^ 2-pa ^ 2 | = C
|2cx-c^2|=c
Square on both sides
4c^2x^2-4c^3x+c^4=c^2
If C = 0, then OA coincides, which is obviously not satisfactory, because then p is the origin
So C is not equal to 0
therefore
4x^2-4cx+c^2-1=0
[2x-(c+1)][2x-(c-1)]=0
x=(c+1)/2,x=(c-1)/2
So these are two straight lines
X = (c + 1) / 2 and x = (C-1) / 2