Y X X 1 X 2X 3 evaluation range. Why is the answer x 3 and x 1 not defining the domain?

The denominator of y=x+x+1 x-2x-3 is not 0, i.e., x-2x-3 is not 0, x-2x-3=(x+1)(x-3), i.e., x≠3 and x≠-1 is the defining domain.

When x≠3 and x≠-1, x +x+1=(x+1 2) 2+3 4, the numerator must be positive; x -2x-3 = (x-1) 2-4, when x 3 or x -1, the range of y is positive, and when -1 x 3, the range of y is negative.

When it comes to defining a domain, it must refer to the x in the function parentheses, whether it is an x or not, or other things such as x+1 or x 2+1, all refer to x

Then the range of the whole in parentheses is as found in the title.

1 x 2+1 10), in which case this range refers to the domain of x in simple f(x).

If you don't understand, please continue to ask.

y=x²+x+1/x²-2x-3

y=x²+x+x-x+1/x²-2x+1-4y=(x²+2x+1)-1/(x²-2x+1)-4y=(x+1)²-1/(x-1)²-4

Division.

The dividend cannot be = 0

Or it doesn't make sense.

That is, when (x-1) -4≠0x=3.

x=-1.

So x≠3 and x≠-1

Which is the definition domain.

y(x²-2x-3)=x²+x+1

y-1)x-(2y+1)x-(3y+1)=0x is a real number, then the equation has a solution.

So the discriminant is greater than or equal to 0

2y+1)²+4(y-1)(3y+1)>=04y²+3y-1>=0

4y-1)(y+1)>=0

y<=-1,y>=1/4

y=x²-2x-3=(x-1)²-4。

The domain is defined by x r, which belongs to the set of all real numbers.

Because (x-1) 0, y=(x-1) -4 -4 and the range is y [-4,+

Introduction to Function Definition Domain: A mathematical noun, which is one of the three elements of a function (definition domain, value range, and corresponding law), and the object of the corresponding law. Refers to the range of values of the independent variables of the function, that is, for two non-empty sets d and m with function correspondence, any number in the set d has a definite number corresponding to it in the set m, then the set d is called the function definition domain.

Parse: y=x -2x-3 definition domain: r

Range: [-4,+.]

Find the hail hidden definition domain and analyze the denominator.

The evaluation range can be evaluated using the judgment method of Li Bei.

For reference, please smile.

y = 2x state shed 5x 6) (x 2x 3), the definition domain is r

y = 2x²＋4x＋6 + x )/x²＋2x＋3) =2 + x/(x²＋2x＋3)

y'=x 2x 3) -x(2x+2)] x sail then 2x 3) 2 = x +3) (x 2x 3) 2

Gain station x = 3,y( 3) =12+5 3) (6+2 3) =12+5 3)(3- 3) 12

21+3√3)/12 = 7+√3)/4,y(-√3) =12-5√3)/(6-2√3) =12-5√3)(3+√3)/12

21-3√3)/12 = 7-√3)/4,lim(2x²＋5x＋6)/(x²＋2x＋3) =2

Range (7- 3) 4 y 7+ 3) spike 4

Requiring the function y = x -x) x - 1) to define the domain, we need to pay attention to two aspects:

The denominator cannot be zero, i.e. x ≠ 1.

The number of squares to be opened must be greater than or equal to 0, i.e. x orange collapse - x 0. Note that x(x-1) =0 corresponds to the two real number solutions x=0 and x=1, which are the zero points of the square number to be opened, so the value of x must be x 0 or x 1.

In summary, the domain of the function is defined as (-0] 1, +

x 2 defines the domain as , minus 3 and of course , so y defines the domain as

y=x^2-3

is a quadratic function whose domain is defined as all real numbers.

The definition domain is all real numbers.

Summary. Hello dear, happy to answer for you.

y=-2x 2-x+1 defines the domain as r

2（x^2-x/2+1/16)+9/8

2（x-1/4)^2+9/8

What is the domain of y=2x -x+1?

Hello dear, I'm glad to answer the wild talk for you: y=-2x 2-x+1 defines the domain as r=-2 (shoot no x 2-x 2+1 16)+9 8=-2(x-1 4) 2+9 8

Dear, I hope mine is helpful to you, and I wish you a happy life! 

According to the definition of square root, then there is:

2x - x³ ≥0

x (x² -2) ≤0

x (x + 2) (x - 2) ≤0

Because: x - 2 x x + 2

The product of the three of them is negative, so there are two cases where the above inequality holds:

x - 2 0 x Note: Only one of the 3 factors is a non-positive number.

x - 2 x x + 2 0 Note: All 3 factors are negative.

From the 1st inequality we get: 0 x 2 from the 2nd inequality we get: x 2

So, this function is defined in the following domains:

y=(x-3)(x-1).

is the whole real number r

Hello: The domain of this function is a solid number.

;y-derivative = 2x (x +x+1)-(x -1)*(2x+1) (x +x+1) 2= [2x*(x +x+1)-(x -1)*(2x+1)] x +x+1) 2=(x 2+4x+1) (x +x+1) 2, it is easy to get: increasing on (-2- 3,-), decreasing on (-2- 3,-2+ 3), and increasing on (-2+ 3,+).

x = -2 - 3, y has a maximum value of 2 3 3, x = -2 + 3, y has a minimum value of -2 3 3, and when x approaches +, y approaches 1

The value range is [-2 3 3,2 3 3].

y-lead=2+(1 2)*(4) 13-4x)=2-2 (13-4x);

When 0,8,y=1-(x+2) (x +x+1) let (x+2) (x +x 1)=k, then kx kx k x 2(x x 1≠0) has a solution, k=0, you can k≠0, apply the δ of the judgment to destroy the solution, remember to remove the solution with a denominator of zero.

The second, y=3+5 (x-1), is not followed by zero, so it is obvious that the range is y≠3, which can be expressed as **. ,1,y=(x²-1)/(x²+x+1)

Analysis: y(x 2+x+1)=x 2-1==>1-y)x 2-xy-y-1=0

y^2+4(1-y)(y+1)>=0==>y^2<=4/3

2√3/3<=y<=2√3/3

y=2x-3+ or Yu Feng (13-4x).

Analysis: The function definition domain is x<=13 4

Let y'=2-4 [2 (13-4x)]=0==>1....0, evaluation range y=(x -1) (x +x+1); y=2x-3+√(13-4x); y=(3x+2)/(x-1)