1 y x x 1 2 y x 1 x to find the range of the function

Let's do the second question first, the first method: when x>0, we can get y=(sqrt(x)-sqrt(1 x)) =2;When x<0, y=-(sqrt(-x)-sqrt(-1 x)) =2

In the second method, y=x+1 x is equivalent to x 2-yx+1=0 has roots, so the discriminant is greater than or equal to zero, i.e., y 2>=4, and the solution is still the same answer. In fact, if you learn the mean inequality in the future, this kind of problem will be very simple. Considering that you are only in your first year of high school, this method will not be talked about.

In the first problem: method one, y=x (x+1)=(x+1-1) (x+1)=1-1 (x+1), obviously, 1 (x+1) can be any number except 0, i.e. y is not equal to 1

In fact, you can also do it according to the method you said, and the specific operation is method 2.

Method 2: y=x (x+1) is equivalent to yx+y=x, x=y (1-y), so that x=-1, then y=y-1, we can know that any y is not satisfied, that is, the reverse y can be any number. The combined denominator 1-y is not equal to 0, and the intersection of y is not equal to 1

If you don't understand, please ask.

You can use the "inverse function method", I don't know if you have learned it? Find x swap x, y and note that the domain of the inverse function is equal to the range of the original function (this method applies to the function once)! Your question - it only makes sense to find a value domain within a defined domain!

Hope you understand!

The first question is y=1-1 (x+1) because 1 (x+1) cannot be 0, so the range is y≠1

The second question is ab(a,b 0) under the root sign of the mean theorem a+b so x+1 x is greater than or equal to 2 (when x is greater than 0).

When x is less than 0, -x is greater than 0 -x-1 x is greater than or equal to 2 x+1 x is less than or equal to -2 This gives the final result.

x +x+1=(x+1) -x=(x+1) -x+1)+1y=(x +x+1) (x+1)=x+1+1 (x+1)-1x+1+1 (x+1) is the hook function, and the x+1+1 (x+1) is greater than or equal to 2 or less than or equal to -2 from the image

Then y=(x +x+1) (x+1)=x+1+1 (x+1)-1 is greater than or equal to 1 or less than or equal to -3

The range is (negative infinity, -3) and on [1, positive infinity).

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

For which Sakura -1 3,1). Because.

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

1-1/〔(x-1/2)^2+3/4〕

(x-1 2) 2+3 4 3 4

Then, 1 (x-1 2) 2+3 Li Cong4 4 31-1 (x-1 2) 2+3 4 -1 3 so y -1 3.

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

Obtained by long division.

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

When x-> y->1

Consider the discriminant of the denominator.

x^2-x+1)

Find the derivative. y'= 2x-1)/(x^2-x+1)^2y'=0.

x=1 2 can be used.

y'|x=1 2+ <0 and.

y'|x=1/2- >0

Which skin to get out of the difference x =1 2 is the maximum.

The largest f(x) = f(1 2) = 1 + 1 ( 1 4 -1 2 +1) =1 + 4 3 = 7 3

Range = 1, 7 3].

Answer: y=(x 2-x) (x 2-x+1) Let t=x 2-x=(x-1 2) 2-1 4>=-1 4 original function as:

y=t/(t+1),t>=-1/4

t+1-1)/(t+1)

1-1/(t+1)

Because: Wang Jingli t>=-1 4, t+1> manuscript = 3 Sleepy 4 so: 0

x +x+1=(x+1) -x=(x+1) -x+1)+1y=(x +x+1) (x+1)=x+1+1 (x+1)-1x+1+1 (x+1) is the hook function, and the x+1+1 (x+1) is greater than or equal to 2 or less than or equal to -2 from the image

Then y=(x +x+1) (x+1)=x+1+1 (x+1)-1 is greater than or equal to 1 or less than or equal to -3

The range is (negative infinity, -3) and on [1, positive infinity).

Let's start with defining domains.

It is found that it is all real numbers except -1.

Multiply the denominator and treat y as a constant.

y(x+1)=x²+x+1

Sort it out. x²+(1-y)x+1-y=0

When x=-1 doesn't fit the topic.

When x≠-1 = (1-y) 2-4(1-y) 0 solves y 1

Or. y≤-3

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 +

From the meaning of the title: f(x)=1 y=x 2+1 2x[x is not equal to 0], f(x)=0[

x is equal to 0] when the stool or x belongs to [-1,0), f(x)=1 y=x 2+1 2x=-[x 2-1 2x]<=2*root number [-x 2*(-1 2x)]=2*1 2=-1 If and only if -x 2=-1 2x, i.e., x=-1, and the function f(x)=1 y=x 2+1 2x is a monotonically decreasing function on x belonging to [-1,0], so the range of the function at this time is [-1,0), so the function y=2x x +1 is at [-1,0] The range on is Jujube Collapse (-00, -1), and when x belongs to (0, 2), f(x)=1 y=x 2+1 2x>=2*root number [x 2*(1 2x)]=2*1 2=1 is true if and only if x 2=1 2x, i.e., x=1.

The function f(x)=1 y=x 2+1 2x is a monotonically decreasing function on (0,1], and the minimum value of the monotonically increasing function is 1 and the maximum value is rHence 0

Define the domain x r

x^2+x+1

x^2+x+1/4+3/4

x+1/2)^2+3/4.=3/4

The domain of the function y= (x +x+1) [ 3 2, + infinity).

First of all, you should keep in mind that the value range under the root number and the value range under the non-root number are the same, and the slightest difference will not affect the problem. Just remember that the difference with a root number is that there is a limit to the definition domain, and y must be greater than 0, and the process will not be described in detail.

y=(x-1)/(x²-x+1)

Method 1: Discriminant method.

The highest number of numerator denominators is 2 times, and the discriminant method can usually be used.

Use 0 to find the range of y.

Method 2: Construct a basic inequality method.

y=(x-1)/(x²-x+1)

x-1) [(x-1) +x-1)+1] when x=1, y=0 (1) When x≠1, the numerator and denominator are divided by x-1

We get y=1 [(x-1)+1 (x-1)+1]When x>1, (x-1)+1 (x-1) 2 So 0 When x<1, (x-1)+1 (x-1) -2 So -1 y<0 (3).

The above (1), (2), and (3) are combined to obtain the value range of [ 1, 1 3].

The answer is wrong, it should be (minus infinity, 1), if nothing else.

Hello, you put some places in parentheses in the formula, okay?