The definition domain of f x is 0,3, and the definition domain of f x 1 is found

Solution: is defined in the domain of [0,3], and the domain of f(x-1) is defined in [0-1,3-1], i.e., [-1,2].

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

3.If there is f(x)-2f(-x)=9x+2 for any x r, then f(x)-2f(-x)=3kx-b

3k=9, k=3, b=-2, so f(x)=3x-2

Given f(1+2x)=x -4x-1, let f(x)=ax +bx+c, f(1+2x)=a(1+2x) +b(1+2x)+c=4ax +(4a+2b)x+a+b+c

From 4a=1 we get a=1 4, from 4a+2b=-4 we get b=-5 2, from a+b+c=-1 we get c=5 4, so f(x)=x 4-5x 2+5 4

So f(3-4x) = (3-4x) 4-5(3-4x) 2+5 4=4x +4x+11 4

4.Knowing that f(x) is a quadratic function, and f(2x)+f(3x+1)=13x +6x-1, let f(x)=ax +bx+c

f(2x)+f(3x+1)=a(2x)²+b(2x)+c+a(3x+1)²+b(3x+1)+c=13ax²+(5b+6a)x+a+b+2c

From 13a=13 we get a=1, from 5b+6a=6 we get b=0, from a+b+2c=-1 we get c=-1, so f(x)=x -1

Less than or equal to x-1 less than or equal to 3, so 0+1 less than or equal to x less than or equal to 3+1, that is, 1 less than or equal to x less than or equal to 4

2.Ibid. [1-3, 1+3].

3.(1) If it is an even function, f(x)=-9x-2: if it is an odd function, f(x)=3x+2 3

2) Let 2x+1=t, x=t 2-1 2, f(t)=(t 2) 4-5 2t+1 2, f(3-4x)=4x 2-6x+19 4

4.Let f(x)=ax 2+bx+c, and substitute f(2x)+f(3x+1) to get a=1, b=0, c=-1, f(x)=x 2-1

<=x-1=<3 [1,4]

2.[1-root-3, 1+root-3].

3.(1) If it is an even function, f(x)=-9x-2: if it is an odd function, f(x)=3x+2 3

2) Let 2x+1=t, x=t 2-1 2, f(t)=(t 2) 4-5 2t+1 2, f(3-4x)=4x 2-6x+19 4

4.Let f(x)=ax 2+bx+c, and substitute f(2x)+f(3x+1) to get a=1, b=0, c=-1, f(x)=x 2-1

The domain of f(x) is the domain of f[ (x+1)] in the interval [0,3].

f[ (x+1)] is defined by [0,3].

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

This is a composite function to find a domain of the question.

x+3∈（0,1）

x∈（-3,-2）

Therefore, f(x+3) defines the domain hall trace as (-3, -2) He Da.

The function f(1-3x) can be split into :

y=f(t)

t=1-3x, because.

0 x 1 so.

0≤3x≤3

3≤-3x≤0

2≤1-3x≤1

Namely. 2≤t≤1

So the domain of the function f(t) is defined as -2,1

Because the function f(t) and the function f(x) are the same function, so.

f(x) is defined in the domain of -2,1

The definition domain of f(x-1) is [0,3], and the definition domain of f(x-1) is [0-1,3-1], i.e., [-1,2].

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

3.If there is f(x)-2f(-x)=9x+2 for any x r, then f(x)-2f(-x)=3kx-b

3k=9, k=3, b=-2, so f(x)=3x-2

Given f(1+2x)=x -4x-1, let f(x)=ax +bx+c, f(1+2x)=a(1+2x) +b(1+2x)+c=4ax +(4a+2b)x+a+b+c

From 4a=1 we get a=1 4, from 4a+2b=-4 we get b=-5 2, from a+b+c=-1 we get c=5 4, so f(x)=x 4-5x 2+5 4

So f(3-4x) = (3-4x) 4-5(3-4x) 2+5 4=4x +4x+11 4

4.Knowing that f(x) is a quadratic function, and f(2x)+f(3x+1)=13x +6x-1, let f(x)=ax +bx+c

f(2x)+f(3x+1)=a(2x)²+b(2x)+c+a(3x+1)²+b(3x+1)+c=13ax²+(5b+6a)x+a+b+2c

From 13a=13 we get a=1, from 5b+6a=6 we get b=0, from a+b+2c=-1 we get c=-1, so f(x)=x -1

Can it solve your problem?

f(x-1) is defined in the domain [0,3].

x∈[0,3]

x-1∈[1,4]

f(x) is defined in [1,4].

The domain of f(x-1) is [0,3], i.e., 0"x"3, so -1"x-1"2, which is the domain of f(x) [-1,2].

The definition domain of f(x-1) is [0,3], that is, x [0,3], and the x in f(x) is the substitution.

x-1), equivalent substitution, so the range of the value of x in f(x) is the range of the value of (x-1) in f(x-1), and because of x [0,3].

Therefore, the value of x in f(x) ranges from -1 x 2

No, the range of values of the arguments in a function is called the domain of the function.

The key to solving this kind of problem is to understand that the defined domain is the scope of independent variables, and to distinguish between the domain of abstract function definition and the scope of function function.

For the defined domain of f(1+x) is [-2,3], find the defined domain of f[x].

1) The definition domain of f(1+x) is [-2,3], which refers to the range of x, because the independent variable is the range of the whole f() bracket (1+x), so the domain of f(x) is t=1+x.

2) Solution: f(x) definition domain = f(t) definition domain, that is, the value range of t place. t=1+x。The solution is [-1,4].

For the function f(2x-1) where the domain is [0,1], find the domain of the function f(1-3x).

1) The domain of the function f(2x-1) is defined as [0,1) which refers to the range of x, because the independent variable is x. Same as above, first find the domain defined by f(x) as [-1,1].

2) The domain defined by f(x) can be known as f(1-3x), and the (1-3x) range is [-1,1]. The reason for this is that the value of the scope of the abstract function is the same. The definition domain of the function f(1-3x) is (0,2 3】

Defining a domain is the scope of x.

So the first one.

2<=x<=3

then -1<=1+x<=4

So the f(x domain is [-1,4].

The second 0<=x<1

So -1<=2x-1<1

f(x) defines the domain [-1,1).

So -1<=1-3x<1

2<=-3x<0

0 so f(1-3x) defines the domain (0,2 3].

y=f(2^x-3)

0<=2^x-3<=1

log(2)(3)<=x<=2

f(2 x-3) is defined as [log(2)(3),2].

log(2)(3) is the logarithm of 2 with 2 as the base 3.

To solve similar problems, grasp a principle:

That is, for the same function f(x), its value range and definition domain are fixed!

That is, no matter what is in (), in short, the value range of () is certain, that is, the definition domain!

Knowing that the domain of y=f(x+1) is [-2,3], when you find the domain of f(x), (x 1) is a whole, which is equivalent to the (x) in f(x) that you require

So the range of () is the range of (x 1)!

The x in y=f(x+1) belongs to [-2,3], and obviously the (x) in f(x) is the range of x+1, which is [-1,4].

Knowing the domain of f(x) [-1,4], when finding the domain of f(2x+1), (2x+1) is a whole, which is equivalent to (x) in f(x).

The value range of (x) in f(x) is [-1,4], so the value range of (2x+1) in f(2x+1) is [-1,4], and the value range of x is the value range of x in f(2x+1), that is, the definition range of f(2x+1) is [-1,3 2].

First of all, it is clear that the domain refers to the range of independent variables, for f(x), the independent variables are x, and the independent variables of f(2x-1) are also x, not 2x-1

The domain of f(x) is [1,3], i.e. x [1,3], so the range of values in parentheses that makes the function f() meaningful is the [1,3] keypoint.

So for f(2x-1), there is 2x-1 [1,3] to get x [1,2], which defines the domain.

Note: The key to such a problem is to clearly define it, and everything else is easy to do.