Given the function f (x) = x2 + 2x-3, the set M = {(x, y) | f (x) + F (y) ≤ 0}, the set n = {(x, y) | f (x) - f (y) ≥ 0} Finding the intersection of M and n How to make f (x) - f (y) = 0?

Given the function f (x) = x2 + 2x-3, the set M = {(x, y) | f (x) + F (y) ≤ 0}, the set n = {(x, y) | f (x) - f (y) ≥ 0} Finding the intersection of M and n How to make f (x) - f (y) = 0?

For the set M = {(x, y) | f (x) + F (y) ≤ 0}, f (x) ≤ - f (y)
Since f (x) = x2 + 2x-3 is a parabola with the opening upward, f (x) and - f (y) are symmetrical about the X axis (or Y axis, so that X axis and Y axis coincide)
So when f (x) ≤ - f (y), f (x) = x2 + 2x-3 ≤ 0, the solution is - 3 ≤ x ≤ 1, that is - 3 ≤ M = {x, y} ≤ 1
For the set n = {(x, y) | f (x) - f (y) ≥ 0}, f (x) ≥ f (y)
Because f (x) = x2 + 2x-3 = (x + 1) ^ 2-4, the symmetry axis of the curve is x = - 1
When x, y ≤ - 1, the function f (x) is a monotone decreasing function. If f (x) ≥ f (y), X ≤ y is necessary
When x, y ≥ - 1, the function f (x) is a monotone increasing function. If f (x) ≥ f (y), X ≥ y is necessary
That is, the solution of set n is x ≤ y ≤ - 1 or X ≥ y ≥ - 1
In conclusion, the intersection of M and N is - 3 ≤ x ≤ y ≤ - 1 or - 1 ≤ y ≤ x ≤ 1

Given the function f (x = | x-a | - 9 / x + A, X belongs to the closed interval of 1,6, when a belongs to the open interval of 1,6, the expression m (a) for finding the minimum value of function f (x)

f(x)={x-9/x;x≥a
2a-(x+9/x);x≤a
In [a, 6], the function f (x) increases monotonically, where f (x) min = A-9 / A
In [1, a], the maximum value of G (x) = x + 9 / X is g (1) = 10, where f (x) min = 2a-10
On (1,6), 2a-10 < A-9 / A
∴f(x)min=M(a)=2a-10

We know that the function f (x) = X3 + MX2 + NX + 1 has extremum at x = - 2 / 3 and x = 1 (1) find the value of real number m and n (2) find the monotone decreasing interval of function f (x)

（1）f（x）=x3+mx2+nx+1
f'(x)=3x^2+2mx+n
f'(x)=0,3x^2+2mx+n=0 (#)
The extremum is obtained at x = - 2 and x = 1
X = - 2, x = 1 is the root of (?)
According to Weida's theorem:
-2m/3=-2/3+1=-1/3,m=1/2
n/3=-2/3,n=-2
Then f '(x) = 3 (x + 2 / 3) (x-1)
X = - 2 and x = 1 are extreme points
∴m=1/2 ,n=-2
(2) From F '(x) = 3 (x + 2 / 3) (x-1)

If f (x) = x ^ 3 + 3mx ^ 2 + NX + m ^ 2 has an extreme value of 0 when x = - 1, then M =? N =?

m=2,n=9

The function y = 2-m / 3x + M-4 is an inverse proportional function of X

If the function y = 2-m / 3x + M-4 is an inverse proportion function of X, M-4 = 0, 2-m ≠ 0, and M = - 2, then when m is - 2, the function y = 2-m / 3x + M-4 is an inverse proportion function of X
Adopt

X (y + 2) = 1, y = 1 / x + 1, y = 1 / x square, y = negative 2x, y = negative X / 2, y = 1 / 3x, where y is the inverse proportion function of X?

Y is the inverse scale function of X: y = negative 2x

Is y the inverse proportion function of X in y = 1 / (3x)

Yes
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Monotone increasing interval of function y = 2tan (3x + π / 3)

The period of TaNx is π
The period of 2tan (3x + π / 3) is π / 3
The monotone increasing interval of the function 2tan (3x + π / 3) is: (K π - π / 6, K π - π / 6) (k integer)

Monotone increasing interval of y = 2tan (3x + π / 4)

Monotone increasing interval of y = 2tan (3x + π / 4)
-π/2+kπ

F (x) = ax ˇ 2 - (3a-1) x + a ˇ 2 is an increasing function on [1, + ∞]. The range of a is obtained

F (x) = ax ˇ 2 - (3a-1) x + a ˇ 2 is an increasing function on [1, + ∞],
A0 axis of symmetry x = (3a-1) / 2A