Need to find the intervals where the function increase, decrease & concavity

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 =(

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 =(

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

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

Mark44
Mentor
Show us what you have done.

ehild
Homework Helper
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

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

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" .

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.

Dick
Homework Helper
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.

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?

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

Mark44
Mentor
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

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

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

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

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 ?

Mark44
Mentor
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 don't understand what you're asking here.

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

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

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

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

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

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

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

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?

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

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

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.

well... i got something like this... hope Im not mistaken xD

f"(x)= ((-2x(x^(2)-9)^2)+(x^(2)-9)(4x^(3)-36x))/(x^(8)-36x^(6)+486^(4)-2916x^(2)+6561)

Mark44
Mentor
When you expanded the denominator, you did a lot of extra work for no advantage. As Dick suggested, you should leave the x^2 - 9 stuff unexpanded. This will allow you to simplify the numerator and denominator somewhat, because both have a factor of x^2 - 9

I see two sign errors in the numerator.

im sorry about those signs errors
now i have something like this, after correcting the errors
f"(x)=(-2x+(x^(2)+9)(4x^(3)-36x))/(x^(2)-9)^2

im sorry about those signs errors
now i have something like this, after correcting the errors
f"(x)=(-2x+(x^(2)+9)(4x^(3)-36x))/(x^(2)-9)^2
EDIT **

Your derievative is incorrect.

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but why? i simplified already