Questions about the domain of high school math function definitions

In high school functions, if there is no special requirement in the problem, the domain range is defined in a meaningful way.

Generally, there is no need for a special process to solve the defined domain, but to mark the defined domain after the function. Or a summary will do. You don't have to write a few steps to define a domain.

Generally, no points are deducted, because even if the definition field is written incorrectly, it does not matter whether the result is correct in the end. It's important not to make mistakes with the function.

Of course, the definition domain is generally the entire set of real numbers, etc., if the definition domain is not, at least 1 point will be deducted if the error is made.

Think of it this way!

f(x), 2 "(x)"3

Remove the brackets:

f( ), 2" ( 3

Fill in the parentheses with x+1

Yes. f(x+1), 2 "(x+1)"3

So, 1" x "2

Therefore, f(x+1) is defined in the domain [1,2].

In the same way, you put (x-1) or something in parentheses, and then you can solve the inequality ......

In fact, the above method of understanding uses the idea ...... of substitution

Formal solution process:

Solution: (x+1) in 2 "x+1 "3 solution 1 "x" 2. Therefore, define the domain as [1,2].

f(x-1)2"x-1"3 solves 3"x"4 Therefore, the domain is defined as [3,4] I don't think these two questions will be confused!

It's easy to get the mistake here:

If the domain of f(x+1) is [2,3], find the domain of f(x-1)!

If you understand this, then this kind of question is basically no problem!

The way to understand it is similar! Try the ...... for yourself）

Hope it solves your problem.

Look at the question, if the question gives some range of numbers, you need to pay attention to the definition field.

If you use discriminants, you can also calculate the definition domain ......

You misunderstood, f(x) is defined in the domain (1,5).

Y=f(2x-1) replaces x with 2x-1, and obviously defines the domain in the same way.

Conversely, if the domain of y=f(2x-1) is (1,5), then the domain of f(x) is not, but as you say: 1<=2x-1<=5.

The definition domain refers to the algebra in parentheses and not just x, and here 2x-1 is to be treated as a whole, f(x)=f(2x-1) so.

The definition domain does not change.

The crosses are multiplied wrong, right?! If you break it down like this, the coefficient of the term is 4. It should be 3 2x x = (1 x) (3+x).

Or extract the minus sign first:

x +2x 3) 0, then x +2x 3 0 multiply by the cross: (x 1)(x+3) 0

The definition domain is [-3,1].

Solution: It can be seen from the problem setting, 0 x 1===>-1≤2x-1＜1.===>-1≤1-3x＜1.===>0＜x≤2/3.The domain of the composite function f(1-3x) is defined as (0,2 3).

then 0 x 1

Get -1 2x-1 1

f(x) is defined in the domain [-1,0).

1≤1-3x＜1

0≤x＜2/3

f(1-3x) [0,2,3).

In fact, many times you can set x to confuse people, you can set f(t), because this is a composite function, that is to say, the inner functions are bound by the outer functions. You must have done such a question with a sign (x+2).Seeking satisfaction with x, the same truth.

For example, f(t) = with the sign x, then no matter what is in it, it must meet the condition greater than zero. Just like f (3-x) and f (2n+3) they both share f(t), i.e. t and 3-x and 2n+3 all have the same range. But the actual definition domain is the range of t x n, so it can be found that way.

The method of finding the domain of the definition of f(x+1) is 2"x+1"3 to get 1"x"2

The method of finding the domain of the definition of f(x-1) is as follows: 2 "x-1" 3 solution 3 "x" 4

Rule: The range of the whole in parentheses is the same, and the definition field must be the range of the value of x itself.

The function f defines the domain as [2,3], f(x+t), and x+t is in the range of 2 to 3, then the range of x is [2-t,3-t],

It is equivalent to the two things in parentheses are regarded as the same, but they are seen from different angles, but they are actually the same, so as a whole, it is the same to define the domain.

That is, the things in parentheses are equivalent, and the first x defines the domain [2,3]. Then, the second is that the range of x+1 is [2,3], the domain of x is [1,2], then the third is that the range of x-1 is [2,3], then the domain of f(x-1) is [3,4]. Note:

The definition domain mentioned in the question is the range of x.

The bracketed range is the same, x-1 is also [2,3], and the x of the solution is [3,4].

y=f(x) is the original function.

y=f(2x) is a composite function, and u=2x is an intermediate function.

When we define y=f(u), we mean that for the law of functions f, the independent variable is u

However, when u is another function, the law of the function changes, not the original f

For example, y=f(x)=2 x is an exponential function, which is an elementary function.

However, y=2 (x+2) is not an exponential function, just an exponential function, it is a composite function.

It can also be understood as, y=f(x+2), and the intermediate function is u=x+2

Obviously, y=2 x and y=2 (x+2) are not the same as functions, because the laws of functions are different.

Can you accept this explanation?

In addition, the three elements of a function are: a defined domain, a value range, and a law of functions.

Therefore, the sufficient and necessary conditions for the two functions to be the same are: the definition domain, the value range, and the function rules are all the same, and none of them are indispensable.

Your question: Why is the value range of x equal to the value range of 2x?

The function expression y=f(x), the argument is x, and the choice of the argument letter has nothing to do with the function relationship, so even if.

y=f(t), which is the same range as x.

However, for the function rule f of y=f(x), y=f(u), etc., the independent variable is something in parentheses, even if it is complex, for example, in f(ax+b), for f, its independent variable is a whole large block (ax+b) inside the parentheses

Therefore, f(x) and f(2x) and x and 2x can be in the same range.

This is the difference between the original function and the composite function.

The x of f(x) is the same as the 2x of f(2x)!

The function is the same as the requirements 1, the definition field is the same, and the range of x is the same.

2. The operation relationship is the same and f is the same.

When doing this kind of problem, remember that the set of ranges equal to x in parentheses is the definition field.

1.The definition domain of the function refers to the input value, that is, the range of x in 2x-1, so it is 0 x 1, so the parentheses in -1 2x-1 1 refer to the closed interval, that is, there is equal to and the small parenthesis on the right is the open interval, so it is less than The range of objects applied by the same correspondence rule is the same, so the range of 2x-1 and 1-3x is the same, so -1 1-3x 1 solves the value of x to be 0 x 2 3

2.Think of it this way: the status of x in parentheses is the same as the status of 2x+1, and the required definition domain is the range of x in 2x+1.

3.This problem is the same as the first one, [-2,3] is the range of x, not the range of x+1, so we need to find the range of x+1 first, because x+1 and 2x-1 are the same correspondence rule, so they are the same range, so -1 2x-1 4 can then find the definition domain.

All the same, as long as you know that the domain of f(2x-1) is [0,1), which refers to the range of x in 2x-1, x first becomes t=2x-1, and then substituted into f(t), f(2x-1) defines the domain of [0,1) so 0 x<1 so -1 t=2x-1<1, so f(t) defines the domain.

1, because f(2x-1) defines the domain as [0,1) so 0 x<1 so 0<=2x<2 -1 2x-1<1

2, The domain of the function f(x) is [-1,4], i.e. -1 x 4 so -1 2x+1 4 so -1 x 3 2

3, the domain of the function f(x+1) is [-2,3], i.e. -2 x 3, so -1 x+1 4, so -1, 2x-1, 4, so 0 x

P.S. The defined domain in the known condition is the restriction of x, and the range of the whole parenthesis is found according to the known condition, and then the range of x in the parentheses is the defined domain of the requirement.

x refers to the domain in which f(x) is generated, and the x in "2x-1" f(x).

Solution: Because :f (2x-1) defines the kernel domain of the rotten ridge digging [1,3], 2x-1 belongs to [1,5].

So (2x-x squared) belongs to [1,5].

So f(2x-xsquared) defines the domain.

The value range of 2x-1 is [-1,5], so in f(2x-x), the vertical Qi domain of 2x-x is also [-1,5], so it is easy to get that the definition domain of (2x-x) is [1-root number 2,1 + root number 2].

The domain [1,3] is defined by the f(2x-1) of the pin, and x is [1,2].

by (2x-x squared) = x (deficit 2x-1).

Bringing x in, the (2x-x squared) Dingwu Li Yiyu is [0,1].

Solution: Because Dan filial piety is: f (modulo liter 2x-1) defines the domain [1,3], so 2x-1 belongs to [1,5].

So (2x-x squared) belongs to [1,5].

So f(2x-xsquared) defines the domain.

Author: zuoweiqin - Level 5 2010-7-25 22:43This is right and cautious, I promise.