Find the value range of the independent variables of the following functions (to make the function meaningful) （1）y=3\2x-1 （2）y=2x+1\3x-1

Find the value range of the independent variables of the following functions (to make the function meaningful) （1）y=3\2x-1 （2）y=2x+1\3x-1

X is not equal to 1 / 2, X is not equal to 1 / 3

Given that the value range of the independent variable x in a function y = 1-3x is x greater than or equal to - 1 and less than or equal to 2, then the value range of the corresponding function is---------------

The function decreases with the increase of X, only when x = - 1, y = 4, x = 2, y = - 5
Then the range of function value is y greater than or equal to - 5 and less than or equal to 4

It is known that the value range of the independent variable x in the function y = KX + B is - 3x less than or equal to x less than or equal to 6
The value range of corresponding function value is - 5 less than or equal to y less than or equal to - 2

-3 less than or equal to x less than or equal to 6
The value range of corresponding function value is - 5 less than or equal to y less than or equal to - 2
x=-3 y=-5 -5=-3k+b
x=6 y=-2 -2=6k+b
k=1/3
b=-4
Analytical formula y = x / 3-4

In the linear function y = 3x-9, when the function value y is greater than or equal to - 3, what is the value range of the independent variable x?

3x-9≥-3
x≥2

Find the linear equation tangent to the curve y = x ^ 3 + 3x ^ 2-1 and perpendicular to the line x-3y + 2 = 0
Need process, thank you!

X-3y + 2 = 0y = x / 3 + 2 / 3, slope is 1 / 3, tangent is perpendicular to it, so tangent slope = - 3Y = x ^ 3 + 3x ^ 2-1y '= 3x ^ 2 + 6x derivative is tangent slope, tangent slope = - 3, so 3x ^ 2 + 6x = - 3x ^ 2 + 2x + 1 = 0x = - 1y = x ^ 3 + 3x ^ 2-1 = - 1 + 3-1 = 1, so tangent point is (- 1,1), so straight line Y-1 = - 3 (x + 1) 3x + y + 2 = 0

Quadratic function y = x ^ 2-2x + C (0

The axis of symmetry x = 1 is at 0

The minimum value of quadratic function y = - X & sup2; - 2x-m is 5. Then M=
Don't give me the answer directly. You need to solve the problem. There is no answer 6 or tell me the minimum formula of quadratic function

596481482a：
The minimum is the lowest point of quadratic function
Vertex coordinates (- B / 2a, (4ac-b & sup2;) / 4A]
That is (4ac-b & sup2;) / 4A = 5
[4×(-1)×(-m)-(-2)²]/[4×(-1)]=5
(4m-4)/(-4)=5
4m-4=5×(-4)
4m-4=-20
4m=-16
m=-4
The value of M is - 4

Solution equation: 2x * x + 3x-3 = 0

2X * x + 3x-3 = 02x * x + 3x = 3x * x + 3x / 2 = 3 / 2 (x + 3 / 4) ^ 2 = 3 / 2 + 9 / 16 (x + 3 / 4) ^ 2 = 33 / 16x + 3 / 4 = ± √ 33 / 4x = (- 3 ± √ 33) / 4 if you don't understand, please continue to ask

Solution equation: 2x + 4 = - 3x + 15

2x+4=-3x+15
2x+3x=15-4
5x=11
x=11÷5
x=2.2

(2x-15) / (3x-15) = 25% to solve the equation

(2x—15)/(3x—15)=25%
2x—15=25%(3x—15)
2x-15=0.75x-15/4
1.25x=45/4
x=9