It is proved that the function y = x2 + 1 is a decreasing function in the interval (0, + ∞), Y = - x2 + 1 is a decreasing function in the interval (0, + ∞). Sorry, wrong number

It is proved that the function y = x2 + 1 is a decreasing function in the interval (0, + ∞), Y = - x2 + 1 is a decreasing function in the interval (0, + ∞). Sorry, wrong number

Is there any mistake?
Y = x ^ 2 + 1 is an increasing function on (0, + ∞)!
Note (1): if y = - x2 + 1 in the interval (0, + ∞)
Let: x + △ x > x > 0
f(x+△x)-f(x)=[-(x+△x)^2+1]-[-x^2+1]
=[-(x+△x)^2+1]-[-x^2+1]
=-(x+△x)^2+x^2
=-[2 △ x * x + (△ x) ^ 2] < 0, it is proved
Note 2: y = 1 / x ^ 2 + 1 is a decreasing function
Let: x + △ x > x > 0
f(x+△x)=1/(x+△x)^2+1
f(x+△x)-f(x)
=[1/(x+△x)^2+1]-[1/x^2+1]
=1/(x+△x)^2-1/x^2
=[x^2-(x+△x)^2]/x^2(x+△x)^2
=-[2 △ x * x + (△ x) ^ 2] / x ^ 2 (x + △ x) ^ 2 < 0

How to prove that y = x / 1 + x 2 is a bounded function

Divide X and discuss the denominator range, but note the explanation of x = 0

Customize an isosceles trapezoid with a wooden strip, pull it to make it into a triangle, which are 6cm, 8cm and 10cm respectively, and find the top and bottom of the trapezoid

4 cm

It is known that points a (m, 2) and B (2, n) are on the image of inverse scale function y = m + 3x. (1) find the values of M and N; (2) if the line y = mx-n intersects with the X axis at point C, find the coordinates of the symmetric point C 'of point c about the Y axis

(1) Substituting the coordinate of a into y = m + 3x, we can get 2 = m + 3M, the solution is m = 3, y = 6x, substituting the coordinate of B into y = 6x, we can get n = 62 = 3, i.e. M = 3, n = 3. (2) substituting m = 3, n = 3 into the straight line, y = mx-n, we can get y = 3x-3, substituting y = 0 into y = 3x-3, we can get 0 = 3x-3, x = 1, i.e. the coordinate of C is (1, 0), the coordinate of C about y

Given the quadratic function f [x] = AX2 + BX [A is not equal to 0], satisfy f [X-1] = f [3-x] and equation f [x] = 2x, have equal roots, find f [x]

From ①, a (x-1) ^ 2 + B (x-1) = a (3-x) ^ 2 + B (3-x),
∴（x-2)(2a+b)=0,b=-2a.
From the two equal points of 2, ax ^ 2 + (b-2) x = 0, a = - 1, B = 2
(1)f(x)=-x^2+2x.

As shown in the figure, the spiral in the figure is composed of a series of right triangles, and the area of the nth right triangle is:______ ．

According to the Pythagorean theorem: in the first triangle: oa12 = 1 + 1, S1 = 1 × 1 ／ 2; in the second triangle: oa22 = oa12 + 1 = 1 + 1 + 1, S2 = OA1 × 1 ／ 2 = 1 + 1 × 1 ／ 2; in the third triangle: oa32 = oa22 + 1 = 1 + 1 + 1 + 1 + 1, S3 = oa2 × 1 ／ 2 = 1 + 1 + 1 × 1 ／ 2 In the nth triangle: SN = n × 1 ^

If the coordinates of the intersection point of the images of the linear functions y = 7-x and y = 1-x are? Then the solution of the equations {4x + y = 7, x + y = 1 is?

The linear functions y = 7 - X and y = 1 - X are two parallel lines without intersection
The point of intersection (2, - 1) of linear function y = 7-4x and y = 1-x, equations 4x + y = 7, x + y = 1, x = 2, y = - 1

U = ln (1 / (x + √ (y ^ 2 + Z ^ 2))), find the partial derivatives of u to x, u to y, u to Z,
Is u = ln (x + √ (y ^ 2 + Z ^ 2)). There is no one in two.

u=ln(1/(x+√（y^2+z^2)))du/dx=1/ 1/(x+√（y^2+z^2)) * -1/(x+√（y^2+z^2))^2 =-(x+√（y^2+z^2))dx/(x+√（y^2+z^2))^2=-1/(x+√（y^2+z^2))du/dy=1/ 1/(x+√（y^2+z^2)) * -1/(x+√（y^2+z^2))^2 *1/2√(y^2+z^...

Fill in the words as you like
Black paste black () () black () () white () () white () ()
One heart and one mind one () one () one () one () one () one () one () one ()

It's dark, it's white, it's white
One day, one night, one surprise, one life, one plant and one tree

In the same rectangular coordinate system, draw a function y = - 2x-1, y = - 2x

In the same rectangular coordinate system
The graph of linear function y = - 2x-1 and y = - 2x is shown in the following figure