# Homework Help: Determine inverse

1. Dec 15, 2004

### aisha

1/(-3y+9) x cannot = 3 is the inverse y=(x+1/9)/(1/-3)? The 1 in the numerator is confusing me also how will I know if the inverse is a function?

2. Dec 15, 2004

### learningphysics

That doesn't look right. When you find the inverse, you can check that you did it right by evaluating f-1(f(x)). It should come out to x. If it doesn't, that means your inverse is wrong.

3. Dec 16, 2004

### delton

There are two easy ways to check.
Are the graphs of those functions symetrical about y=x? If so they are inverses.
AND EVEN EASIER
An inverse should have the x and y values of the original function switched.
Thats how one finds inverses: by switching x with y and solving for y.

4. Dec 16, 2004

### HallsofIvy

I'm not sure what you mean by y=(x+1/9)/(1/-3). Is that "x plus 1/9 divided by -1/3? If so, then it is easier to write it as -3(x+ 1/9)= -3x- 1/3.
If so, then, for example, if x= 1, then y= -3(1)- 1/3)= -3- 1/3= -10/3. And then
1/(-3((-10/3)+9) = 1/(10+9)= 1/19, not 1. So the two functions are certainly NOT inverse to one another.

What you do to find the inverse of a function like this is to "swap" x and y.
Your original function is y= -3x- 1/3. To find the inverse, swap x and y:
x= -3y- 1/3. Now solve for y: x+ 1/3= -3y so y= (-1/3)x- 1/9. Notice the negative signs!!!

5. Dec 17, 2004

### aisha

I know how to find the inverse but the 1 in the numerator is confusing me, I dont know how to solve for y after switching x and y. f(x)=1/(-3y+9) how do i get rid of the numerator? Also since the denominator can be factored should I, or do i not have to?

6. Dec 18, 2004

### t_unit92003

first of all, it should be x=1/(-3y+9). Then you solve for y by multiplying both sides by (-3y+9). then divide by x. then subtract 9 from both sides and divide both sides by -3. you should get y=(-3/x)-9.

7. Dec 20, 2004

### aisha

I did all of that and I understand it well, but my answer is y=(x-9)/-3 or -3(x-9) how is the answer (-3/x)-9? :uhh: :uhh:

8. Dec 20, 2004

### HallsofIvy

How can you do "all of that" and "understand it well" and not know whether the answer is (x-9)/(-3) or -3(x-9)???

Your original function was y= 1/(-3x+ 9). Swapping x and y gives x= 1/(-3y+ 9)

That is the same as x(-3y+9)= 1 or, as t unit92003 said, -3y+ 9= 1/x. Subtracting 9 from both sides gives -3y= 1/x- 9 so y= (1/x- 9)/(-3)= 3- 1/(3x), not the (-3/x)- 9 that t unit92003 then gave.

9. Dec 22, 2004

### aisha

I GOT UP TO THE LAST STEP, y=(1/x-9)/-3 Can someone please tell me how this became 3-1/(3x)? This is the only part im stuck on now plz help me understand this please....

10. Dec 22, 2004

### HallsofIvy

You did the algebra correctly but couldn't do the arithmetic??

(1/x- 9)/(-3)= (1/x)/(-3)- 9/(-3) by the "distributive property".

(1/x)/(-3)= -1/(3x) and -9/(-3)= 3.

(1/x- 9)/(-3)= (1/x)/(-3)- 9/(-3)= -1/(3x)+ 3= 3- 1/(3x).

11. Dec 25, 2004

### aisha

THANKS FINALLY AFTER A LONG TIME OF TRYING I GOT IT!!! I had another question like that and was able to solve it thanks again everyone, esp Mentor.
The content of the passage is a series of forum posts discussing the process of finding the inverse of a function, specifically \( y = \frac{1}{-3x + 9} \). The posts include a student's initial confusion, explanations from other users, and a detailed step-by-step solution from a mentor.
multiple_questions_with_answers

1. the initial post asks if the inverse of \( y = \frac{1}{-3x + 9} \) is \( y = \frac{x + \frac{1}{9}}{\frac{1}{-3}} \) and how to determine if the inverse is a function.
2. subsequent posts provide explanations and methods to check if the functions are inverses.
3. the final posts clarify the algebraic steps to find the correct inverse and address the student's confusion about the final form of the answer.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# Homework Help: Determine inverse

1. Dec 15, 2004

### aisha

1/(-3y+9) x cannot = 3 is the inverse y=(x+1/9)/(1/-3)? The 1 in the numerator is confusing me also how will I know if the inverse is a function?

2. Dec 15, 2004

### learningphysics

That doesn't look right. When you find the inverse, you can check that you did it right by evaluating f-1(f(x)). It should come out to x. If it doesn't, that means your inverse is wrong.

3. Dec 16, 2004

### delton

There are two easy ways to check.
Are the graphs of those functions symetrical about y=x? If so they are inverses.
AND EVEN EASIER
An inverse should have the x and y values of the original function switched.
Thats how one finds inverses: by switching x with y and solving for y.

4. Dec 16, 2004

### HallsofIvy

I'm not sure what you mean by y=(x+1/9)/(1/-3). Is that "x plus 1/9 divided by -1/3? If so, then it is easier to write it as -3(x+ 1/9)= -3x- 1/3.
If so, then, for example, if x= 1, then y= -3(1)- 1/3)= -3- 1/3= -10/3. And then
1/(-3((-10/3)+9) = 1/(10+9)= 1/19, not 1. So the two functions are certainly NOT inverse to one another.

What you do to find the inverse of a function like this is to "swap" x and y.
Your original function is y= -3x- 1/3. To find the inverse, swap x and y:
x= -3y- 1/3. Now solve for y: x+ 1/3= -3y so y= (-1/3)x- 1/9. Notice the negative signs!!!

5. Dec 17, 2004

### aisha

I know how to find the inverse but the 1 in the numerator is confusing me, I dont know how to solve for y after switching x and y. f(x)=1/(-3y+9) how do i get rid of the numerator? Also since the denominator can be factored should I, or do i not have to?

6. Dec 18, 2004

### t_unit92003

first of all, it should be x=1/(-3y+9). Then you solve for y by multiplying both sides by (-3y+9). then divide by x. then subtract 9 from both sides and divide both sides by -3. you should get y=(-3/x)-9.

7. Dec 20, 2004

### aisha

I did all of that and I understand it well, but my answer is y=(x-9)/-3 or -3(x-9) how is the answer (-3/x)-9? :uhh: :uhh:

8. Dec 20, 2004

### HallsofIvy

How can you do "all of that" and "understand it well" and not know whether the answer is (x-9)/(-3) or -3(x-9)???

Your original function was y= 1/(-3x+ 9). Swapping x and y gives x= 1/(-3y+ 9)

That is the same as x(-3y+9)= 1 or, as t unit92003 said, -3y+ 9= 1/x. Subtracting 9 from both sides gives -3y= 1/x- 9 so y= (1/x- 9)/(-3)= 3- 1/(3x), not the (-3/x)- 9 that t unit92003 then gave.

9. Dec 22, 2004

### aisha

I GOT UP TO THE LAST STEP, y=(1/x-9)/-3 Can someone please tell me how this became 3-1/(3x)? This is the only part im stuck on now plz help me understand this please....

10. Dec 22, 2004

### HallsofIvy

You did the algebra correctly but couldn't do the arithmetic??

(1/x- 9)/(-3)= (1/x)/(-3)- 9/(-3) by the "distributive property".

(1/x)/(-3)= -1/(3x) and -9/(-3)= 3.

(1/x- 9)/(-3)= (1/x)/(-3)- 9/(-3)= -1/(3x)+ 3= 3- 1/(3x).

11. Dec 25, 2004

### aisha

THANKS FINALLY AFTER A LONG TIME OF TRYING I GOT IT!!! I had another question like that and was able to solve it thanks again everyone, esp Mentor.
user
The content of the passage is a series of forum posts discussing the process of finding the inverse of a function, specifically \( y = \frac{1}{-3x + 9} \). The posts include a student's initial confusion, explanations from other users, and a detailed step-by-step solution from a mentor.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS

1. The initial post asks if the inverse of \( y = \frac{1}{-3x + 9} \) is \( y = \frac{x + \frac{1}{9}}{\frac{1}{-3}} \) and how to determine if the inverse is a function.
2. Subsequent posts provide explanations and methods to check if the functions are inverses.
3. The final posts clarify the algebraic steps to find the correct inverse and address the student's confusion about the final form of the answer.