It is known that the abscissa of an intersection of the inverse scale function y = K / X and the positive scale function y = 2x is 2 The coordinates of an intersection point

It is known that the abscissa of an intersection of the inverse scale function y = K / X and the positive scale function y = 2x is 2 The coordinates of an intersection point

Substituting x = 2, K / x = 2x = K / 2 = 4, we get k = 8
The original formula is 8 / x = 2x, and the other solution of X is - 2,
The other intersection coordinates are (- 2, - 4)
y=8/x
(-2,-4)
On the inequality of X (- 1 / 2) x & # 178; + 2x ＞ MX, the solution set is 0 ＜ x ＜ 2, find the value of M
Move the MX term to the left and use the quadratic function to be less than 0
If the images of linear functions y = 2 / 3x + P and y = - 1 / 2x + Q pass through point a (- 2,0), and intersect with y axis at two points B and C respectively, the area of triangle ABC can be calculated
Substituting the coordinates of point a into the linear functions y = 2 / 3x + P and y = - 1 / 2x + Q, P = 4 / 3 and q = - 1 are obtained. Then we know that the coordinates of point B and point C are B (0,4 / 3) and C (0, - 1), respectively. Thus, the area of triangle ABC is 2 * (4 / 3 + 1) / 2 = 7 / 3
S=4*2*1/2=4
It is known that the maximum value of the function f (x) = - 2x2 + 3Tx + T (x, t ∈ R) is u (T). When u (T) has a minimum value, the value of T is ()
A. 94B. −94C. 49D. −49
F (x) = - 2x2 + 3Tx + T = − 2 (x − 3t4) 2 + 98t2 + T, the opening of parabola is downward, when x = 3t4, f (x) gets the maximum value 98t2 + T, that is, u (T) = 98t2 + T, ∵ U (t) = 98t2 + T = 98 (T2 + 89t) = 98 (T + 49) 2 − 29, ∵ t ∈ R, ∵ when t = − 49, u (T) has the minimum value - 29
Given a positive scale function and a linear function, their images pass through the point P (- 2,1), and the image of the linear function intersects the Y axis at the point Q (0,3)
Analytic expressions of two functions
Let: the analytic formula of positive proportion function be y = KX; the analytic formula of primary function be y = k'x + B
Because the positive proportional function passes through P
So 1 = - 2K
k=-1/2
That is y = - 1 / 2x;
And because the first-order function passes through P and Q points
So 1 = - 2K '+ B
3=0+b
B=3
So K '= 1
That is y = x + 3
in summary
y=-1/2x
y=x+3
It is known that the maximum value of the function f (x) = - 2x2 + 3Tx + T (x, t ∈ R) is u (T). When u (T) reaches the minimum value, the value of T is______ .
∵ the maximum value of quadratic function f (x) = - 2x2 + 3Tx + T (x, t ∈ R) is u (T), ∵ U (T) = 4 × (− 2) × t − 9t24 × (− 2) = 9t2 + 8t8 = 98t2 + T, when t = - 49, u (T) gets the minimum value; ∵ t takes - 49, so the answer is: - 49
Images and properties of positive scale function
1. Domain: R (real number set)
2. Range: R (real number set)
3. Parity: odd function
4. Monotonicity: when k > 0, the image is in the first and third quadrants, and Y increases with the increase of X (monotonic increase); when k > 0, the image is in the first and third quadrants
The image passes through the origin and two quadrants
The minimum value of quadratic function f (x) = 2x ^ 2 + 3Tx + 3T is g (T), then when G (T) takes the maximum value, the value of T is g (T)
First calculate the expression of the minimum value:
f（X）=2X^2+3TX+3T = 2(X^2+3TX/2+3T/2) =2(X+3T/4)^2 - 9T^2/8+3T > = - 9T^2/8+3T
g(T)= - 9T^2/8+3T
Take the maximum at the axis of symmetry
T=-3/(2*(-9/8))=4/3
Images and properties of positive scale function
Properties of positive proportion function
1. Domain: R (real number set)
2. Range: R (real number set)
3. Parity: odd function
4. Monotonicity: when k > 0, the image is in the first and third quadrants, and Y increases with the increase of X (monotonic increase); when k > 0, the image is in the first and third quadrants
Properties of positive proportion function
1. Domain: R (real number set)
2. Range: R (real number set)
3. Parity: odd function
4. Monotonicity: when k > 0, the image is in the first and third quadrants, and Y increases with the increase of X (monotonic increase); when K0, the image is in the first and third quadrants, and Y increases with the increase of X (monotonic increase); when K0, y increases with the increase of X
(2) When K0, y increases with the increase of X
(2) When K0, the image is located in the first and third quadrants, y increases with the increase of X (monotonic increase); when K0, the image is located in the first and third quadrants, y increases with the increase of X (monotonic increase); when K0, y increases with the increase of X
(2) When K0, y increases with the increase of X
(2) When K0, y increases with the increase of X
(2) When k
Let the minimum value of the function f (x) = 2x ^ 2 + 3Tx + 2T be g (T), find the analytic expression of G (T), and find the maximum value of G (T) when t is the value
Hope as soon as possible!
f(x)=2(x+3t/4)^2+2t-9t^2/8
So when x = - 3t / 4, f (x) has the minimum value, and the minimum value is 2t-9t ^ 2 / 8 = g (T)
G (T) = - 8 / 9 (T-8 / 9) ^ 2-8 / 9
So when t = 8 / 9, G (T) reaches the maximum value of - 8 / 9
When the opening of quadratic function is upward and x = - B / 2a, take the minimum direct, that is, x = - 3t / 4, G (T) = 2t-9t / 8, G (T) is also the opening of quadratic function is downward and T = - B / 2a, that is, take the maximum direct when t = 8 / 9