Find the range of function: y = 3x-1 / X-2

Find the range of function: y = 3x-1 / X-2

y=(3x-1)/(x-2)
3x-1=yx-2y
(y-3)x=(2y-1)
x=(2y-1)/(y-3)
Because of the function
X = (2y-1) / (Y-3) makes sense, so,
y-3≠0
y≠3
The range of the original function is: (- ∞, 3) ∪ (3, + ∞)
What is the principle of discriminant method to find the range ∵ x ∈ R, ∵ Δ≥ 0?
This easy to do, in fact, is the transformation thought of senior one. We know that for any real number y, its sufficient and necessary condition in the range of function f (x) is that the equation y = xxxxx about X has a real number solution (the function is transformed into the equation)
Good question! Students should ask a clear!
∵ x ∈ R, according to the function definition, for any real number x, there is always a unique real number y corresponding to it, and all the values of Y constitute a range; in other words, if and only if y takes the value in the range, the equation about X must have a solution, while the quadratic equation with one variable has a solution
“∴Δ≥0 ”
Greater than or equal to 0 under root sign
How to use discriminant method to calculate range
Also, please don't make a long speech``````
That's to transform the equation into an equation about X, and then treat y as an unknown constant, such as ax squared + BX + C = 0
In this paper, we discuss the case of a = 0 and a is not equal to 0. When a is not equal to 0, we can get the range of Y by using B square-4ac greater than 0
such as
y=6/(x*2-3x+2)
The discriminant method can be used
y(x^2-3x+2)=6
yx^2-3xy+2y-6=0
Y is not equal to 0
There is a solution,
So the discriminant > = 0
therefore
9y^2-4y(2y-6)>=0
9y^2-8y^2+24y>=0
y^2+24y>=0
y(y+24)>=0
Y=0
Y0
So the range is: Y0
Another example
y=3x/x*2+4
(x^2+4)y=3x
yx^2+4y+3x=0
y=0,x=0
establish
Y is not equal to 0
There is a discriminant
9-16y^2>=0
16y^2
For example, y = x ^ 2 + 3x + 6
Find its range
It can be reduced to x ^ 2 + 3x + 6-y = 0
Because there is a solution, the discriminant is > = 0
That is 9-4 (6-y) > = 0
The solution is Y > = 15 / 4
That's how it works
As for the specific topic, we need to analyze it in detail!!
Careful!!
How to find the range of y = 3x? What is the range?
The range is R (all real numbers)
Range is the range of function value (function value changes with the change of independent variable)
For example, if y = 1 / x, then the range is y ≠ 0, because x cannot take 0
Find the range of y = 3x ^ 2-x + 2 (x ∈ [- 1,3])
Min 23 △ 12 'max 26
What is the range of y = 3x-1 / x + 1
Please write clearly, with brackets, what are your numerator and denominator?
(-∞，+∞)
It's not equal to three
Find the range of y = (3x + 3x + 1) / (x + x-1)?
I don't quite understand how to ask,
We can use discriminant method to multiply the denominator, and then treat it as a quadratic equation of one variable with respect to X. then we can find out the range of y value with discriminant greater than or equal to zero
What is the range of y = x + 1 / 3x + 2?
y=(x+1)/(3x+2)
3xy+2y=x+1
(3y-1)x=1-2y
x=(1-2y)/(3y-1)
∵3y-1≠0,
∴y≠1/3
So the range of the original function is {y | y ≠ 1 / 3}
Two cases of X ＆ gt; 0 and X ＆ lt; 0 are discussed by using the basic inequality
F (x) = x square - 3x + 4
f(x)=x^2-3x+4=(x-3/2)^2+7/4
Range [7 / 4, positive infinity)
F (x) = the square of 3x-5x + 2, where is the range of X ∈ [0,2]?
There should be a way to solve the problem
f(x)=3x^2-5x+2
=3(x^2-5x/3)+2
=3(x^2-5x/3+25/36-25/36)+2
=3(x-5/6)^2-3*25/36+2
=3(x-5/6)^2-1/12
Zero
Find the vertex coordinates first
x=-b/2a=5/6
On [0,2]
So the minimum value is x = 5 / 6
y=-7/3
The maximum value only needs to be compared
The size of F (0) and f (2)
f(0)=2
f(2)=4
therefore
The range is [- 7 / 3, 4]