Find the analytic expression of Y with respect to X by knowing x = 2y-2 divided by 1 + y When x = 0, - 1,3,..., the value of function y The value range of independent variable x Wrong, wrong... It's x = 2y-1 divided by 1 + y

Find the analytic expression of Y with respect to X by knowing x = 2y-2 divided by 1 + y When x = 0, - 1,3,..., the value of function y The value range of independent variable x Wrong, wrong... It's x = 2y-1 divided by 1 + y

Expand: x + X * y = 2 * Y-2
Transfer: y * (2-x) = 2 + X
So y = (2 + x) / (2-x)
When x = 0, - 1,3, y is 1,1 / 3, - 5 respectively
The value range of X is that x is not equal to 2
Dizzy, that's not the same way
Expansion: x + X * y = 2y-1
Y*(2-X)=X+1
So y = (x + 1) / (2-x)
When x = 0, - 1,3,..., the values of Y are 0.5,0, - 4 respectively
The value range of X is not equal to 2

Is y = 2x + 1 the same function as x = 2Y + 1? Why?

In the same function, the first x y has nothing to do with the second x y, they are just unknowns

Let x > - 1, find the maximum value of the function y = (x ^ 2 + 7x + 10) / (x + 1), and let x

Another a = x + 1, because x > - 1, so a > 0
Then y = ((A-1) ^ 2 + 7 (A-1) + 10) / a = (a ^ 2 + 5A + 4) / a = a + 4 / A + 5 > = 5 + 2 √ (a * 4 / a) = 9
If and only if a = 2, take the equal sign, then x = 1. Then 9 is his minimum. Similarly, we can find X

C programming language, calculate the following piecewise function y value: y = 2x + 1 (x > 0); y = 10 (x = 0); y = - 7x & # 179; = 3 (x > 0)

You don't divide the interval of X correctly, X & lt; 0 doesn't
Is the last expression - 3 or = 3

The range of the function y = x-x on [1,2] is

Y = X-1 / X on [1,2]
Ymin=Y（1）=0
Ymax=Y（2）=3/2
The range is [0,3 / 2]

What is the range of the function y = x + X-1

Basic inequality y = (x-1) + 2 / (x-1) + 1
X1, y ≥ 2 roots 2 + 1, if and only if x = roots 2 + 1, take the equal sign
The range is (- infinity, - 2 roots 2 + 1] ∪ [2 roots 2 + 1, + infinity)

The range of the function y = x + 2 / 1 is

Because x ^ 2 > = 0
Function y = x ^ 2 + 2 / 1 > = 1 / 2
The range of function y = x + 2 / 1 is [1 / 2, + OO]

Find the range of function (expressed by interval) 1. Y = 8 / 2 of x ^ 2 (1 ≤ x ≤ 2) 2. Y = - x under the root sign, X belongs to [0, + infinity)

In the first function, the denominator increases as x increases
When x = 1, y = 8. When x = 2, y = 2
So the first function is [2,8]
The second function is y = √ X image about X axis symmetry image
The maximum value of y = - √ x is y = 0 when it is taken as 0
X increases and √ x decreases because x ∈ [0, + ∞)
-√x∈（-∞,0]
The range is (- ∞, 0]

If the range of function y = 2 / (X-2) (x ∈ a) is {y | y

Y = 2 / (X-2): X-2 = 2 / y, x = 2 / y + 2
∵y

Finding the range of the function y = x + (x + 1)

y=x+1/(x+1)=(x+1)+1/(x+1)-1
When x + 1 > 0, (x + 1) + 1 / (x + 1) > = 2
∴y>=1
When x + 1