In this paper, the function y = (2x-1) / (x + 1) is transformed properly, and its image is made by using the translation of the image, and the range of the function is written out

In this paper, the function y = (2x-1) / (x + 1) is transformed properly, and its image is made by using the translation of the image, and the range of the function is written out

The range y ≠ 2
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Find the definition and range of function y = x + 1 / x, and study its image

The definition field is {all real numbers whose x | x is not equal to 0} and the value field is (negative infinity, - 2] and [2, positive infinity]; when x > 0, x + 1 / x > is equal to (x · 1 / x) = 2 under double root sign, (the equal sign holds if and only if x = 1) X

Given the function y = (1 / 2) ^ (| X-1 |), find its domain of definition, range of value, and make its image
No more images.

The domain of exponential function is r, so x ∈ R, so the domain is r;
|x-1|≥0,(1/2)^|x-1|≤(1/2)^0=1
So the domain is (- ∞, 1]

When K (k belongs to R), the function y = (x ^ 2 + 2x + m) / x-kx has a zero point, and the zero point is obtained

Is m a known number?
When there is zero point, that is, (x ^ 2 + 2x + m) / x-kx = 0 (x is not equal to 0)
The transformation can get x ^ 2 + 2x + M = KX ^ 2 (1-k) x ^ 2 + 2x + M = 0
If △ = 4-4 * (1-k) * m is greater than or equal to 0, the value of K can be obtained

It is known that the edge BC of rectangle ABCD is on the x-axis, e is the midpoint of diagonal BD, and the image of function y = 2 / X (x > 0) passes through points a and E,
The ordinate of point E is m
1. Point a coordinate (expressed in M)
2. Whether there is a real number m, it is a quadrilateral, ABCD is a square, if it exists, find m, it does not exist, explain the reason

1. The abscissa of point E is 2 / m, so the coordinate of point E is (2 / m, m)
The ordinate of point a is 2m, the abscissa is 2 / 2m = 1 / m, so the coordinate of point a is (1 / m, 2m)
2. When the difference between abscissa and ordinate is equal, ABCD is a square
2/m-1/m=2m-m
m=1

It is known that point (1,3) is on the image of function y = KX (x > 0), the edge BC of rectangle ABCD is on the X axis, e is the midpoint of diagonal BD, the image of function y = KX (x > 0) passes through two points a and E, and the abscissa of point E is m. The following problems are solved: (1) find the value of K; (2) find the abscissa of point C; (3) find the value of m when ∠ abd = 45 °

(1) If the function y = KX passes through point (1,3), then the coordinates of point (1,3) can be substituted into y = KX to get k = 3; (2) if AC is connected, then AC passes through e, eg ⊥ BC is made through e, and BC intersects with point G ⊥ E. the abscissa of point E is m, e is on hyperbola y = 3x, the ordinate of point E is y = 3M, e is the midpoint of BD

As shown in the figure, the edge BC of rectangle ABCD is on the x-axis, e is the midpoint of diagonal BD, and the image of function y = 3 / X passes through two points AE
(1) Find the coordinate of point C (expressed by M)
(2) When ∠ abd = 45 °, calculate the coordinate of point C
(point B does not coincide with the origin)
The abscissa of point E is m

(1) Because the abscissa of E is m, the ordinate is 3 / m, then the ordinate of a is 2 times of 3 / m, that is 6 / m, so the abscissa of point a is m / 2, that is, the abscissa of point B is m / 2. So the abscissa of point C is 2m-m / 2 = 3m / 2, so the coordinate of point C is (3m / 2,0). (2) if ∠ abd = 45 °, then the rectangle is square

As shown in the figure, if the diagonal BD of rectangular ABCD passes through o point, BC ‖ X axis, and a (2, - 1), then the analytic expression of the inverse scale function passing through C point is______ ．

∵ the coordinates of point a are (2, - 1), the edges of rectangular ABCD are parallel to the coordinate axis, the abscissa of point B is 2, and the ordinate of point D is - 1. Let the coordinates of point d be (a, - 1), and the coordinates of point B be (2, b), then the coordinates of point C are (a, b), ∵ the diagonal BD of rectangular ABCD passes through the coordinate origin o, and the resolution of straight line BD

Given the inverse scale function y = 3 / X (x is greater than 0), in rectangular ABCD, BC is on the x-axis, e is the midpoint of diagonal BD, and the transverse mark is m. m is used to find the transverse mark of C
When ∠ abd = 45 °, the value of M

Since the abscissa of point E is m, substituting y = K / x, the ordinate of point E is K / m
So e (m, K / M)
E is also the midpoint of BD, so the ordinate of a is twice that of E, which is 2K / m, thus a (M / 2,2k / M) is obtained
We also know that half of the sum of abscissa of C and abscissa of a is equal to m (because e is in the middle of B and C, and abscissa of B is abscissa of a)
So the abscissa of C is 3m / 2
When ∠ abd = 45 degrees
ABCD is a square
Then AB = BC = 2ob
6/m=(m/2)*2=m
m^2=6
M = radical 6 or M = - radical 6

F (x) = x ^ 2 + 4x + 3, TR, the function g (T) represents the minimum value of function f (x) in the interval [T, t + 1]. Find the expression of G (T)

f(x)=x^2+4x+3
The axis of symmetry is x = - 2
The function g (T) represents the minimum value of the function f (x) in the interval [T, t + 1]
The following is a classified discussion:
(1) If t + 1 ＜ - 2, that is t ＜ - 3
Then G (T) = f (T + 1) = (T + 1) ^ 2 + 4 (T + 1) + 3 = T ^ 2 + 6T + 8
(2) If t ≤ - 2 ≤ T + 1, that is - 3 ≤ t ≤ - 2
Then G (T) = f (- 2) = (- 2) ^ 2 + 4 * (- 2) + 3 = - 1
(3) If t ＞ - 2
Then G (T) = f (T) = T ^ 2 + 4T + 3
So g (T) = T ^ 2 + 6T + 8 (T < - 3)
=-1 (-3≤t≤-2)
=t^2+4t+3 (t＞-2)
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