Ask a question about defining a domain in a higher level

f(x+1)=x 2-2x+1, define the domain [-2,6], find f(x-1) to define the domain and increase or decrease interval.

f(x+1)=x 2-2x+1=(x-1) 2f(x)=(x-2) 2 defines the domain [-1,7]( 2<=x-1<=6).

f(x-1)=(x-3) 2 defines the domain [0,8] (1<=x-1<=7).

When x is at [0,3) is minus [3,8] is increased.

Let x+1=t, i.e., x=t-1

So f(t)=(t-1) 2-2(t-1)+1=t 2-4t+4, (t belongs to [-2,6]).

and t=x-1

f(x-1)=(x-1) 2-4(x-1)+4 (x-1) belongs to [-2,6].

Later, I will formulate it myself and find the interval

f(x+1)=x 2-2x+1 = (x-1) 2 = (x+1-2) 2 defines the domain[-2,6].

The original function f(x)= (x-2) 2 defines the domain [-1,7]f(x-1) = ((x-1)-2) 2 =(x-3) 2 and defines the domain [0,8].

Increasing intervals [3,8].

f（x+1）=x^2-2x+1=(x-1)^2=(x+1-2)^2f(x)=(x-2)^2

f(x-1)=(x-3)^2

The f(x+1) domain is [-2,6].

That is, f(x-1) defines the domain as [0,8].

Ignore the case of defining the domain.

f(x)=(x-2) 2 is a subtractive function over the interval (- 2).

On the interval [2,+ is the increment function.

f(x-1)=(x-3) 2 is a subtraction function over the interval (- 3).

On the interval [3,+ is the increment function.

f(x-1) defines the domain as [0,8].

f(x-1)=(x-3) 2 is a subtractive function over the interval [0,3].

On the interval [3,8] is the increment function.

x+1 0, the solution is x -1, so the domain is defined as (-1, +

Solution: For the above equation to be meaningful, it is necessary to satisfy: 1+x>0;

then x -1.

Therefore, the definition field of x is: x -1 or x (-1, if you are satisfied with the above reply, you want to be .

The expression for the true position of the logarithmic function should be greater than 0.

So 1+x>0, x>-1

So the domain of this function is defined as (-1,+).

This question is not for you to evaluate, what is the use of you to wait?

The original function f(x) defines the domain as (2+x) (2-x)>0 (true numbers must be greater than 0), and the solution gives -2=-2=<2 x<2 and -2=, and the domain is [-4,-1] (1,4).

Instead of directly substituting it, first determine the large range, that is, (2-x)*(2+x)>0, and then substituting it.

Solve the definition domain of f(x) from 2-2 x+2>=0, and then let 2 x and x 2 be in this range at the same time, and then solve the range to which x belongs, which is the definition domain of the latter.

The domain of f(x -1) is [- 3, 3], i.e. x [-3, 3], x -1) 1,2].

The question asks about the domain of f(x).

Then you have to pay attention to what f(x) means here. The x in f(x -1) is not the same as the x in f(x -1) above, where the x in f(x) represents the x -1 in f(x -1), and the domain of f(x) is [-1,2].

I don't know how to ask again.

It is known that the domain of f(x -1) is [- 3, 3], that is, x belongs to [- 3, 3], so x -1 belongs to [-1,2], and replacing x -1 with x requires x to belong to [-1,2], so the domain of f(x) is [-1,2].

1 < x < 1

0 x < 1

2 -2x 0

1 "Nian Rotten -2x +1 1

So the domain of f(x) is (-1 , 1].

From g(x)=f(ax)+f(x a):

x < 3 (2a), x < 3a 2 when a>1, the defined field is < x < 3 (2a) when 0

At 1, -a 2 < 1 2a , 3 2a<3a 2 g(x) = f(ax) + f(x a)(a>0): (-1 2a, 3 2a).

When a 1, -a 2 1 2a , 3 2a 3a 2

g(x)=f(ax)+f(x a)(a>0): (-a2,3a2).

Solution: (1) y = (4x+b)+1 x

4x+b≥0，x≥-b/4

The domain of the x≠0 primitive function is x -b 4 and x≠0(2)y=arccos(x-1) 2+1

x-1) 2 1, then x 3

x-1) 2 -1, then x -1

The domain of the original function is -1 x 3

3) f(x) = x 2-1) + lg(3-x) x 2-1 0, then x 1 or x -1

3-x>0, then x<3

The original function defines the domain as x -1 or 1 x<3

4）f（x）=e^x，x≤1；f(x)=x+2, and the domain of the original function of x>1 is x r

The domain of f(x) is [-2,1], that is, the value of x in f(x) is [-2,1].

f(2x-1): then it must be -2 2x-1 1, and the solution is -1 2 x 1

f(-1 2x+3) is defined as [1,3], which also refers to -1 2x

In 3, the value of x is [1,3].

So, 3 2 -1 2x

3 5 2, if f(-1 2x+3)=f(t), then the range of t is [3 2,5 2].

For a function to find f(x), there is no essential difference between using x or t to represent the hole, so f(x) defines the domain.

i.e. [3 2, 5 2].

f(-1 2x+3) is [1,3], then "2" has been obtained, and f(x) is [3 2,5 2].

f(2x-1) Nabi sedan defines the domain: that is, 3 2 2x-1 5 2, and the solution is 5 4 x 7 4

The answer upstairs, except for the one on the second floor, is completely wrong, and I don't even know what a function is and what is the definition domain of a function!!

Now I will use Question 3 to revisit the problem of functions and their defined domains!

Let's say that the function y=f(x), y or f(x) is a function of x, and x is the domain of the function!

Defining a domain always refers to the range of values of the independent variables in the function expression.

Therefore, for question 3, f(-1 2x+3) is defined in the domain [1,3], which means -1 2x

In 3, the value of x is [1,3].

So, 3 2 -1 2x

3 5 2, if f(-1 2x+3)=f(t), then the range of t is [3 2,5 2].

The reason why we say this is that we can see from the relationship between the function and its defined domain that f(x)--x, that is, the definition domain of f(x) refers to the range of values of x;

f(-1/2x+3)

1 2x+3, then f(-1 2x+3)=f(t), t=-1 2x+3, isn't the range of t?

So, no, f(t).

t, obviously, t is the independent variable of the function f(t), and the value of t is the domain of f(t).

x, just what quantity is used to represent the self-variation of the function is the difference in question, so it can be seen that the domain of f(x) is t=-1 2x+3, that is, [3 2,5 2].

Therefore, in Question 3, we first find the domain of f(x) as follows: [3 2,5 2].

Then find the domain of f(2x-1) and in turn, let t = 2x-1 and f(2x-1) = f(t).

Since the value range of t is: [3 2,5 2], isn't it true that the value range of 2x-1 is [3 2,5 2]?

Therefore, according to the solution of the inequality, it can be obtained that in 2x-1, the range of the value of x is the domain of f(2x-1).

Having said all that, I can only go so far, I hope it can help you!

With regard to a function of x, the argument is x, not any algebraic expression about x, which is invariant.

Therefore 2x+1 is not an independent variable of f(2x+1), the independent variable is x, and the defined domain is the range of values of x, not any other algebraic range of values about x.

Tell the definition range of f(x), which is the range of values of x, and then ask the definition range of f(2x+1), which also requires the range of values of x. Since 2x+1 is on the definition domain, the range of x is different, and the definition domain changes.

You can think of it this way: if the definition domain is the range of values of the algebraic formula, then no matter how the algebraic formula changes, the definition domain will not change, which is obviously wrong.

The independent variable of f(2x+1) is x, not 2x+1. The independent variable is always an unknown algebra.

But if we say y=2x+1, then the independent variable in f(y) is y, and you ask the other way around, then it is actually the domain of the definition of f(x), and the domain of f(y) is knownSince y=2x+1, then x=(y-1) 2, so the problem is actually the domain of f((y-1) 2) is known, and the domain of f(y) is known.

I think the answer to the second question should be (1,3).