If the domain of the function y f 3x 1 is 1,3, then the domain of y f x is ?

Remember: the definition domain is always the range of values of the independent variable, that is, it is generally the range of values of x;

Therefore, the condition of the question says that "the domain of the function y=f(3x-1) is [1,3]", which refers to the range of values of x, so 2 3x-1 8;

Then, according to the principle that the overall scope in parentheses is the same as f( ), the domain of f(x) is defined as [2,8].

The domain of a function is the range of values of the independent variable x that makes the function relation meaningful.

Then: the domain of the function y=f(3x+1) is [1,3], which means that f(3x+1) is meaningful when x [1,3], that is, the independent variable of the function y=f(3x+1) is x, not 3x+1, y=f(3x+1) as"Composite functions", 3x+1 is just an intermediate variable.

Let u=3x+1, then y=f(3x+1), x [1,3] can be expressed as: y=f(u), u=3x+1, x [1,3].

Let's solve this problem: When 1

The domain of the function y=f(3x-1) is [1,3], and if u= 3x-1, then when 1 x 3, there is: 2 u 8

So the domain of y=f(u) is [2,8].

That is, the domain of y=f(x) is [2,8].

In the function y=f(3x-1), the variable is x, and the definition field is said to the variable.

After y=f(x), you can think of it as another function y=f(u), but the function rules are the same, the variable is u, and the definition range of u is the value range of u=3x-1.

Solution: Let t=3x-1, let x be at [1,3], so t is at [2,8], according to the function, it is independent of the independent variable, that is, there is: y=f(t) definition domain [2,8], that is, y=f(x) definition domain [2,8].

1≤x≤3 2≤3x-1≤8

2 x 8 The answer is [2,8] I know.

According to this solution, don't think much about the rest, relax and relax, don't be too stressed!

The domain equivalent to f(3x-1)=5x+x 2 is [1,3], so it is 1 x 3

Don't you notice the two x cases above.

It's hard to tell.

Let t=3x-1, then x=(t+1) 3

Since y=f(3x-1) defines the domain of Tongyuan ascending worm as [-1,3], -1 x=(t+1) 3 3

3≤t+1≤9

4 t 8 is the definitive laugh semantic domain of the function y=f(x) is [-4,8].

Solution: f(x+1) is defined in the domain [-2,3].

2 x 3 (the definition field refers to the range of values of x, not the range of values back to x+1).

1≤x+1≤4

2x-1 is in the definition field, and the answer -1 2x-1 4 (2x-1 is considered as a whole, and this whole can only be taken in [-1,4]).

0 x 5 2 (find the range of x, because the definition field is the range of x, not 2x-1).

y=f(2x-1) is defined in the domain of [0,5 2] (the range of values of x is the definition domain).

Remember: the definition domain is always the range of values of the independent variable, that is, it is generally the range of values of x;

Therefore, the condition of the question says that "the domain of the function y=f(3x-1) is [1,3]", which refers to the range of values of x, so 2 3x-1 8;

Then, according to the principle that the overall scope in parentheses is the same as f( ), the domain of f(x) is defined as [2,8].

The domain of the function y=f(3x-1) is .

1,3], let u=

3x-1, then 1 x 3

, there are: 2

u 8 thus.

y=f(u)

The definition domain is.

i.e. y=f(x).

The definition domain is.

Don't you notice the two x cases above.

It's hard to tell.

-1=1/3=3^(-1)<=3^x<=3^1=3

So the domain of the function y=f(x) is [1, 3,3].

Let t=3x-1, then x=(t+1) 3

Since y=f(3x-1) is defined in the domain of [-1,3], -1 x=(t+1) 3 3

3≤t+1≤9

4 t 8 means that the domain of the function y=f(x) is [-4,8].

Have fun.

y=f(3x-1) is defined in the domain of -1,3

then -1 x 3

3≤3x≤9

4≤3x-1≤8

So y=f(x) is defined in the domain of -4,8

When x [1,c], f(x) is a monotonic increasing function f(1)f(3) with a maximum value of f(c).

So f(x) takes the maximum value at x=c.

Defines the domain as the set of all the values of the function arguments.

Therefore, the value range of x in y=f(3x-1) is [-1,2], so the value range of 3x-1 is [-4,5], that is, the set of values in y=f(t) is [-4,5], so the definition domain is [-4,5].

Hello: We know that the domain of the function y=f(3x-1) is [-1,2], find the domain of y=f(x).

Let t=3x-1, then x=(t+1) 3

y=f（3x-1）=f(t)

is defined as [-1,2].

1≤x≤24≤x≤5

That is, the domain of f(t) is [-4,5] and the domain of f(x) is [-4,5].