# Homework Help: Function in terms of x with 2 y variables

1. Feb 17, 2010

### Asphyxiated

1. The problem statement, all variables and given/known data

$$\ x = y^4 + y^2 + 1, y \geq 0$$

Find the inverse function of y in terms of x and its derivative dy/dx

2. Relevant equations

3. The attempt at a solution
I don't really understand how to find y isolated when there is two y variables, this is the best that I have =

$$\ x = y^4 + y^2 + 1$$
$$\ -y^4 + x = y^2 + 1$$
$$\ -y^4 = y^2 - x + 1$$
$$\ -y = ( y^2 - x + 1)^{1/4}$$
$$\ y = (-y^2 + x - 1)^{1/4}$$

If I could find the correct formula I could find the derivative and the inverse but I really have no idea what to do.

2. Feb 17, 2010

### Dick

Try a warmup exercise. Suppose I give you the relation x=y^3 and you want to find f(x)=y in terms of x. That's f(x)=x^(1/3), right? That makes the inverse function to the cube root the cube or, f^(-1)(x)=x^3, right? Everything ok so far? Now notice you could have gotten the inverse function just by interchanging x and y in the original relation, x=y^3, making it y=x^3. That's a general thing. You can find the inverse relation just by interchanging x and y in the original relation. Now take another look at your problem.

3. Feb 17, 2010

### Asphyxiated

So I take it that you are saying that:

$$\ x = y^4 + y^2 + 1$$

is just:

$$\ y = x^4 + x^2 +1$$

in terms of x right? This would also be the inverse?

4. Feb 17, 2010

### Dick

Yes, the inverse of the relation x=y^4+y^2+1 is y=x^4+x^2+1. In a more usual problem they will tell you to find the inverse function by interchanging x and y and then solving for y. In this case, if you interchange x and y to get the inverse, it's already solved for y. Makes life easy, yes?

5. Feb 17, 2010

### Asphyxiated

Indeed thats great, probably just another thing I forgot from high school algebra or something! I have been hitting those blocks lately as I try to learn calculus on my own, tell me what you think of this though. I posted this same problem on the Facebook Physics forum and this is the answer someone else gave me.

$$\ x = y^4 + y^2 + 1$$

$$\ 4x = 4y^4 + 4y^2 + 4$$

$$\ 4x = (2y^{2})^{2} + 2(2y^{2}) + 4$$

$$\ 4x - 4 = (2y^{2})^{2} + 2(2y^{2})$$

$$\ 4x - 4 + 1 = (2y^{2})^{2} + 2(2y^{2}) + 1$$

$$\ 4x -3 = (2y^{2} + 1)^{2}$$

$$\ \sqrt{4x-3} = 2y^{2} + 1$$

$$\ \sqrt{4x - 3} - 1 = 2y^{2}$$

$$\ \frac {\sqrt{ 4x - 3} -1} {2} = y^{2}$$

$$\ \frac {\sqrt{(\sqrt{4x-3}-1})} {\sqrt{2}} = y$$

He said that to do the problem to had to complete the squares, any information as to why he is wrong and you are right or you are both right for different reasons or whats up?

Last edited: Feb 18, 2010
6. Feb 17, 2010

### Dick

I'm impressed with the Facebook Physics forum. That's not bad. And it's correct. Now you have y(x) assuming all the root signs are correct. The problem is that the question isn't about y(x), it's about the INVERSE of y(x). Now you have to interchange x and y to get the inverse function and solve for y. Now you have to reverse all of those clever steps and get back to y=x^4+x^2+1. I think the original question is more about realizing, hey, I didn't have to do any of that. Assuming I've read it correctly.

7. Feb 17, 2010

### Asphyxiated

OH I get it now, I didn't say anything about having to find the inverse of the original equation just that I needed to solved for y, which is probably why he gave me that answer instead. Anyway, I know this is going above and beyond what I originally asked but could you possibly explain what he means by "completing the squares". I vaguely remember the term from school but nothing about what was suppose to be done to use it.

thanks either way!

8. Feb 17, 2010

### Dick

9. Feb 17, 2010

### Asphyxiated

Took a quick look at the wiki, I think I got it, thanks again for all your help

10. Feb 18, 2010

### Asphyxiated

Hey dick, or anyone else, can I solve this problem in the way that I did to complete the square?

Question: $$\ x = y^{2} + 3y -1$$

Now I understand that I should just swap the x and y to get the inverse but I wanted to get the original equation in terms of x instead of y so this is what I tried:

$$\ x = y^{2} + 3y -1$$

$$\ x = (y + \frac {3} {2})^{2} - \frac {13} {4}$$

$$\ x + \frac{13} {4} = (y + \frac {3} {2})^{2}$$

$$\ \pm \sqrt{x + \frac {13} {4}} = y + \frac {3} {2}$$

$$\ y = \pm \sqrt{x + \frac {13} {4}} - \frac {3} {2}$$

Last edited: Feb 19, 2010
11. Feb 18, 2010

### Dick

That's just fine. Don't forget the +/- on the square root. It can be either sign.

12. Feb 18, 2010

### Asphyxiated

Thanks, I changed it to reflect +/-, sorry for the PM i wasn't sure if it would tell you about my post since the original topic was already solved.

13. Feb 19, 2010

### Dick

That's ok. I already get an email notice when you post onto a thread I've been working on. You don't need to send the extra PM. I'll notice.