Find the range of y 2x 1 1 x

[2(x+1)-4]/(x+1)

2-4/(x+1)

y=(2x-1) (1+x): y≠2, i.e., (-2) (2, ).

2) If x belongs to [3,5] the maximum value = 4 3 and the minimum value = 1

y=(2x-1) (1+x)=2-3 (1+x) can be obtained as all real numbers.

3=1/2=<3/(1+x)=<3/4

5/4=<2-3/(1+x)=<3/2

Therefore, its maximum is 3 2 and the minimum is 5 4

(2x+2-3)/(1+x)

2-3/(1+x)

So the range is {y|y is not equal to 2}

Solution: y=2x-1 x+1

y=2(x+1)-3/x+1

y=2-(3/x+1)

Therefore, 3 x + 1 is not equal to 0, and the value of the function is not equal to 22) when x belongs to [3,5].

3 x+1 belongs to [3 4, 3 2].

3 x+1 belongs to [-3 2, -3 4].

2- (3 x+1) belongs to [1 2, 5 4].

The equation can be reduced to y=2x2+x-1

a>0 opens upwards.

When x=-1 4, y has a minimum plant-9 8

So the range is [-9 8 + infinity].

Summary. Hello dear, it is a pleasure to serve you. When the domain of this function is x> its range is, (0, root 3-1: 4).

Find the range of y=x-1 x +2(x 1).

Hello dear, it is a pleasure to serve you. When the domain of this function is x> its range is, (0, root 3-1: 4).

Because it has a maximum point, then it is a maximum point of its value, which is the maximum value at the time of the OnePlus root number 3.

Yes can be equal to the root number 3-1 4

y=(2x-1) (x+1).

First, determine the range of x, the range of x is x≠ 1, y=(2x-1) (x+1) can be converted to y=2-3 (x+1), because x+1≠0, so y≠2, then the range is.

The definition field is x not equal to 0

You can find the inverse function:

x=1/(1-y)

The domain of the original function is the domain of the definition of the inverse function.

It can be concluded that y is not equal to 1

So the range of y=x-1 x is (- 1) (1 +

=2/x-1/x^2

1-(1 x-1) 2, the range of which is (- 1).

Note: 1 x-1 can be 1, and 0 does not need to be removed

=-(1/x²)+2/x-1+1

(1/x-1)²+1<=1

And because x≠0

So (1 x-1) ≠ 1

The y≠0 range is (-0) (0,1).

y=1 (x +2x-4), then.

yx²+2yx-(4y+1)=0.

Discriminant 0, and y≠0

i.e. (2y) +4y(4y+1) 0 and y≠0 y>0 or y -1 5

Therefore, the range of the function is .

Geometric method: two absolute values represent the distance from x to -1 and to 2 respectively, take two points -1 and 2 on the number line, according to the geometric meaning of the absolute value, the minimum value of y is 3, so the value range is greater than or equal to 3.

Algebraic Methods: A Categorical Discussion.

When x 2, y = 2x-1

When -1 x 2, y=3

When x -1, y = 3-2x

Then, a piecewise function graph is drawn, and the value range is greater than or equal to 3.

Algebraic Methods: A Categorical Discussion.

When x 2, y = 2x-1

When -1 x 2, y=3

When x -1, y = 3-2x

Then, a piecewise function graph is drawn, and the value range is greater than or equal to 3.

Geometric method: two absolute values represent the distance from x to -1 and to 2 respectively, take two points -1 and 2 on the number line, according to the geometric meaning of the absolute value, then the minimum value of y is 3, so the value range.

is greater than or equal to 3.

By: y=|x+1|+|x-2|

It is learned that y is greater than or equal to 0

Deforming the above equation yields: y 2=(x+1) 2

x-2)^2

2*|x+1|*|x-2|

y=root(x+1) 2

x-2)^2

2*|x+1|*|x-2|

1) When -12, y=root number (

4x^2-4x+1

According to the quadratic parabolic image, the range of y is y>3

Combining the two cases, we get that y has a range of 3 to infinity, including 3

I'm bored, let's talk about a way.

x-a|Indicates the distance from x to x=a.

So |x+1|denotes the distance from x to -1, |x-2|Indicates that the distance from x to 2 is easy to find, and when -1 x 2, ymin=|-1-2|=3 When x<-1 or x>2, y keeps increasing.

In this way, ymin=3 and no maximum.

Intuitively, this equation can be seen as the sum of the distances from one point (x,0) to the fixed point (-1,0) and (2,0) on the x-axis.

So if x=2, y=3+|x+1|or y=3+|x-2|, at this time y is all "=3".

If -1<=x<=2, then y=3

So from the two points together, it is y>=3