High school math to find problems that define the domain, high school math to define the field of pr

The function f(x)= (x -9) and log (x-1) are defined in the domain of

Solution: The way the question is written is not very clear, and there can be two understandings:

The log (x-1) on the denominator is not in the root number, i.e. f(x)=[ (x -9)] log (x-1), and the definition domain is determined by the following group of inequalities:

x²-9=(x+3)(x-3)≧0...1), x-1>0 and x≠2....2)

x -3 or x 3 from (1).from (2) yields x>1 and x≠2; Therefore, the domain is defined as (1) (2) x 3;

The log (x-1) on the denominator is inside the root number, i.e. f(x) = [(x -9) log (x-1)], and the definition domain is determined by the following inequality:

x -9) log (x-1)] 0, which is divided into two cases:

a) x -9 0....1)，log₂(x-1)>0...2)

(x+3)(x-3) 0 from (1), i.e., x -3 or x 3;from (2) to obtain x-1>0, and x-1≠1, i.e., x>1, and x≠2;

Take their intersection to get x 3.

b) x -9 0....3)，log₂(x-1)<0...4)

(x+3)(x-3) 0 from (3), i.e., -3 x 3;From (4) to get 0, the domain is defined as (1) (2) =

The formula under the root number should be "=0", that is, (x -9) log2(x-1) >=0and (x-1) in the logarithm > 0Get 1 = 3The denominator cannot be 0So x is not equal to 2So the final result is 1=3

First of all, x>1 and the denominator is unequal to 0, then x-1 is unequal to 1, so x is unequal to 2, because the root number is greater than or equal to 0, when x>=3, the numerator and denominator are greater than zero, and when the numerator and denominator are less than 0, it can be 1=3

I can't see clearly, man, can you be more detailed?

The domain problem starts at the outermost level: first, the number in the root number should be greater than or equal to 0, log(>=0, log(is the decreasing function, (x>0). When x=1, log(

So 1>=(3x-2)>0

That's right.

3x-2﹚≥0

3x-2＞0

Find the intersection of this group of inequalities.

Solution, 2 3 x 1

According to the title, it can be known.

1=>3x-2>0 is fine.

1.f(x) is defined in the domain [-2,1], so the range of 2x-1 is [-2,1], so the range of x is [-1 2,1].

The domain is [1,3], i.e., the range of x is [1,3], so the range of -1 2x+3 is [3 2 ,5 2], i.e. f(x) defines the domain as [3 2 ,5 2].

The domain is defined as [1,3], i.e. the range of x is [1,3], so the range of -1 2x+3 is [3 2 ,5 2], so the range of 2x-1 is [3 2 ,5 2], so the range of x is [4 5 ,4 7].

1.Let t=2x-1, then 02Let t=-1 2x+3, and the solution is x=1 2(t-3), which is given by 13From the result of 2, -7 2<2x-1<-19 6, and the solution of inequality yields -5 4

Let u=x+1 get -1 u 4 from -2 x 3, let 2x-1=v, -1 v 4, i.e. -1 2x-1 4, get 0 x 5 2

Note that defining a domain is a function that converts it to the same type and calculates it.

Satisfied. In fact, the subject regards f(2x) as the component (y), and y belongs to r.

Let 2x=y then x=1 2y

f(y)=f(2x)=4x+1=4*1 2y+1=2y+1 when y=1

f(1)=3

Hence the choice D

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!

Let t=x 2-3, so f(t)=lg( (t-3) t+3)).

Since the domain of the function lg(n) is defined as n>0; So (t-3) t+3)>0 gives t>3 or t<-3

So f(x) is defined in the domain of negative infinity to -3 or 3 to positive infinity.

Find the domain of the function f(x 2-3) = x 2 (x 2-6).

Analysis: Let t= x 2-3==> x 2=t+3f(t)=(t+3) (t-3)==t≠3, that is, x 2-3≠3==> x 2≠6==>x≠- 6 or x≠ 6 function f(x 2-3)=x 2 (x 2-6) is defined as x≠-6 or x≠ 6