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

[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 =(

 

[╔(σ_σ)╝]

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.

 

[619snake]

I ve done the quotient rule but i can't find the points that are 0

 

[Mark44]

Show us what you have done.

 

[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

 

[╔(σ_σ)╝]

I ve done the quotient rule but i can't 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" .

 

[619snake]

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

 

[Dick]

this is what i get as
f'(x)=-x^(2)-9/(x^(2)-9)^2
sry to use so many parenthesis but I am 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?

 

[619snake]

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

 

[Mark44]

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

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

 

[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

 

[╔(σ_σ)╝]

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]

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.

 

[╔(σ_σ)╝]

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.

 

[619snake]

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

 

[Dick]

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?

 

[619snake]

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

 

[Dick]

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?

 

[619snake]

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

 

[Dick]

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.

 

[619snake]

well... i got something like this... hope I am 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]

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.

 

[619snake]

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

 

[╔(σ_σ)╝]

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.

 

[619snake]

but why? i simplified already

 

[╔(σ_σ)╝]

but why? i simplified already

Check your work again.

if
f"(x)=(-2x+(x^(2)+9)(4x^(3)-36x))/(x^(2)-9)^2


(4x^(3)-36x) = 4x(x^2 -9)

f"(x)= 4x(-2x+(x^(2)+9)/(x^(2)-9)

Which I highly doubt considering the numerator is of a higher degree than the denominator.

You answer should look like
k(x)( ax^2 + b) / (x^2 -9)^3 for some integers k,a,b.

 

[619snake]

ok without that thing i did, the derivative is like this:
f"(x)=(-2x(x^(2)-9)^2)+(x^(2)-9)(4x^(3)-36x)/(x^(2)-9)^4

 

[Mark44]

Edit: fixed exponent.
Unsimplified, the second derivative is
$$f''(x) = \frac{(x^2 - 9)^2(-2x) + (x^2 + 9)(4x(x^2 - 9))}{(x^2 - 9)^4}$$

This can be simplified somewhat by removing common factors of the numerator and denominator.

I can see that you are being diligent about putting parentheses in, but there are some you use that are not needed. Any time an exponent is a positive integer, you don't need parentheses around it. Doing so just makes what you have written harder to read and understand.

For example, there is no reason to write x^(2). x^2 means the same thing and is clearer. On the other hand, you need parentheses for x^(-1) and e^(2x).

 

[619snake]

must ask, when the second derivative:
the second part of the term (-2x)(x^2-9) doesn't get squared like this:
"(-2x)(x^2-9)^2"?

 

[Mark44]

Right, that factor in the first term should be (x^2 - 9)^2. I had it in my notes, but neglected to write it in my post. I've gone back and edited my post.

When you simplify the numerator, resist the urge to multiply everything out. The best way to do things is to find write the numerator in factored form. That way it will be easier to determine the sign of f''(x).

 

[619snake]

so... what is left now is to try to simplify and evaluate the second derivative right?

if I'm not mistaken, after simplifying, the derivative ends like this:
f"(x)=(2x(x^2 + 27))/(x^2 - 9)^3

 

[Mark44]

Yes, that looks fine.

 

[619snake]

yay! now i can sketch the graph!