# Functions are consistent in the range of their domains

### If two functions have the same domain of definition and the same correspondence, are the two functions the same?

Right. The domain of definition and the law of correspondence are the two elements of a function, and two functions that have the same two elements are the same function.

### We say that two functions are equal if their domains are the same and their correspondences are identical. What is the meaning of this statement?

First of all, we need to understand that the three elements of a function are the domain of definition, the law of correspondence and the domain of value. If the three elements are the same we say that the two functions are equal. Definitional domain is like the original product, the correspondence is the processing plant, the original product is the same, the processing is the same, the processing out of things, that is, the value domain is also the same.

### Under the same correspondence law f, the range of both x in f(x) and g(x) in f[g(x)] should be the same?

From my years of teaching experience, the domain of definition of an abstract function is a difficult problem to understand.

On the question of the domain of definition of an abstract function, let’s stick to one point: [the range of the parentheses should be the same]

[Principle]: for the same law of correspondence, the equations put in parentheses are the values that make sense for f (i.e., the ranges are the same).

For example: the definition domain of f(x) is [0,2], that is, the range of the parentheses is [0,2], then whatever is replaced with in the parentheses must belong to [0,2], so after replacing x+1, x+1∈[0,2], so x∈[-1,1], so the definition domain of f(x+1) is [-1,1]

Another example: the definition domain of f(x+1) is [0,2], i.e. the range of the bracket is [1,3],then whatever is replaced with in the bracket must belong to [1,3], so after replacing it with x, x ∈ [1,3], so the domain of definition of f(x) is [1,3].

Please experience these two examples carefully.

Understanding the above, let’s discuss your question next:

First of all: the domain of definition of a function always refers to the range of values of x.

(1) The range of x in f(x) is the domain of definition of f(x).

(2) The range of x in f[g(x)] is the domain of definition of f[g(x)].

I’m sure you can understand that both of the above definitional domains are ranges of x.

So now the question is what is the relationship between the domains of definition of the two functions f(x) and f[g(x)].

We might as well assume that the domain of definition of f(x) is the set A, so this suggests that whatever the parentheses are replaced with must belong to the set A.

Now f[g(x)] is replacing f(x) parentheses by g(x), so g(x) must belong to the set A. But g(x) it’s a function, and a function belongs to a set, which means that all of the values that the function can take on All the values of g(x) belong to set A, i.e., the domain of the function g(x) is set A. Then solve for the range of values of x in g(x), and this range of x is the domain of definition of f[g(x)].

In summary, it can be seen that the domain of definition of f(x) is not the domain of definition of f[g(x)], but the range of values of g(x) in f[g(x)].

### Why is the domain of definition the same in f()? The domain of definition of f(x+1) refers to x’s, but why isn’t it the domain of x’s when it comes to solving for f(2x-1)

Remember two things about these kinds of questions: the domain of definition always refers to the range of values of the independent variable (that is, x); and with f(), the overall range is the same in the parentheses.

The domain of definition is always the range of x.

Post a question I answered before for you

If the domain of definition of the function y=f(3x-1) is [1, 3], then the domain of definition of y=f(x) is? Views: 104Reward Points: 10|Solution Time: 2011-9-1713:44|Questioner: lcbboy1528|Prosecution

∵1≤X≤3

∴2≤3X-1≤8

∴2≤x≤8

Answer is [2, 8] I know.

But:1≤X≤3 why is this.

Isn’t the domain of definition of f(3x-1) supposed to be 1≤3x-1≤3?

Why is it 2≤3X-1≤8 when it’s clearly 2≤3X-1≤8

2≤3X-1≤8 I get, but why is x also greater than or equal to 2 less than or equal to 8

It feels like it’s all backwards?

The best answer is to remember that the domain of definition is always the range of values of the independent variable, that is to say, it is generally the range of values of x;

So the condition of the question says that “the domain of definition of the function y=f(3x-1) is [1, 3]”, which refers to the range of values of x. Therefore, 2 ≤ 3x-1 ≤ 8;

So 2 ≤ 3x-1 ≤ 8, according to f (), the parentheses within the overall range of the same principle, so the definition of the domain of f (x) is [2, 8]

“The definition of the domain is the range of values of x”, keep in mind this sentence

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### Is the same function with the same domain of definition and value

No. Look at the graph, the boxed variable is the range of the domain of value and the domain of definition, and the red line is the curve of the function

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