# Homework Help: Need to find the intervals where the function increase, decrease & concavity

1. Aug 9, 2010

### 619snake

the function is:
x/(x^2 - 9)
i have found the horizontal asymptote(y=0) and vertical asymptote (x=3, x=-3)
the first derivative is giving me problems =(

2. Aug 9, 2010

### ╔(σ_σ)╝

Have you tried the quotient rule ? Please post any sort of attempt at a solution.

3. Aug 9, 2010

### 619snake

I ve done the quotient rule but i cant find the points that are 0

4. Aug 9, 2010

### Staff: Mentor

Show us what you have done.

5. Aug 9, 2010

### ehild

A function can be increasing in an interval and decreasing in an other, without having extrema. Find out the sign of the first derivative and sketch the function knowing its asymptotes. .

ehild

Last edited: Aug 9, 2010
6. Aug 9, 2010

### ╔(σ_σ)╝

If you have done the first derievative you should not have problems finding out where the function is increasing or decreasing; as long as you know the definitions .

This is why I want to see your solution so that I can point out any errors.

I have no idea what you mean by " i can't find the points that are 0" .

7. Aug 9, 2010

### 619snake

this is what i get as
f'(x)=-x^(2)-9/(x^(2)-9)^2
sry to use so many parenthesis but im writing from an ipod.

8. Aug 9, 2010

### Dick

Actually you should have used an another parentheses if you mean (-x^2-9)/(x^2-9)^2. That's the right derivative. Your numerator is always negative and your denominator is always positive (unless x=3 or x=(-3)), right? Are there ANY points where the derivative is zero?

9. Aug 9, 2010

### 619snake

I think the answer is: no points that makes the derivative 0 except x=3, x=-3
but still gonna do the process to verify

10. Aug 9, 2010

### Staff: Mentor

How do x = 3 or x = -3 make the derivative 0?

11. Aug 9, 2010

### 619snake

Ok... I think I got confused... I was writing while in a class...
this is what I got:

horizontal asymptote: (y=0) *don't know if it's right*
Vertical asymptote: x=3, x=(-3)
the domain is: (-$$\infty$$,-3) U (-3,3) U (3,$$\infty$$)

derivative:
f'(x)= (-x^(2)-9)/(x^(2)-9)^2

what I need to find is the sign of the first derivative... do I need to use the points between those of the domain?... I'm not too sure about this... the professor gives like two or three examples, and it's hard follow up... goes so fast

12. Aug 9, 2010

### ╔(σ_σ)╝

What you have so far is correct but remember the definitions.

The function has a horizontal asymptote at y=0 because the limit as x --> +/- infinity =0

A function is increasing on an interval (a,b) or (a,b) if f '(x)>0 on (a,b)
A function is decreasing on an interval (a,b) or (a,b) if f '(x)< 0 on (a,b)

What can you tell me about f'(x)?

is f' (x) >0 anywhere ?
is f ' (x) < 0 anywhere ?

13. Aug 9, 2010

### Staff: Mentor

I don't understand what you're asking here.

14. Aug 9, 2010

### ╔(σ_σ)╝

I believe he wants to find where the derievative is positive and negative using the intervals of the domain.

15. Aug 9, 2010

### 619snake

ok, if I'm not mistaken, f'(x) < 0 (-3,3)

16. Aug 9, 2010

### Dick

You aren't mistaken about that. But that's not the only interval where f'(x)<0. What are the others?

17. Aug 9, 2010

### 619snake

ok the others are f'(x) < 0 (-$$\infty$$,-3), (3,$$\infty$$)
I guess it's because of the asymptotes :uhh:

18. Aug 9, 2010

### Dick

Yes, thank you. It's decreasing EVERYWHERE except at the vertical asymptotes where it makes a huge jump. Can you sketch a graph of the function? Now you have to do concavity, right?

19. Aug 10, 2010

### 619snake

Yes I need to do concavity... it is a bit weird... because the second derivative is pretty long, and have really big numbers

20. Aug 10, 2010

### Dick

It's not that bad if you keep the x^2-9 factored. As before, it would be nice if you'd show us your result.