Finding Horiz. Asymptotes and Sketching Curve: 2nd Derivative Troubleshooting

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Homework Help Overview

The discussion revolves around finding horizontal asymptotes and sketching the curve of the function f(x) = x/(x^2 + 1). Participants are troubleshooting issues related to the second derivative and its implications for concavity.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to differentiate the function and are specifically focused on the second derivative. There is a discussion about the correct application of the quotient rule and chain rule. Some participants question the accuracy of their algebraic manipulations and the expansion of terms.

Discussion Status

Several participants are actively engaged in clarifying their understanding of the differentiation process. Guidance has been provided regarding the separation of terms for clearer calculations. There is an acknowledgment of algebraic errors, but no consensus has been reached on the correct form of the second derivative.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance they can receive. There is an ongoing exploration of potential mistakes in algebraic expansions and differentiation techniques.

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Homework Statement



f(x)=x/x^2+1
Find horiz. asymptotes and use them w/ concavity and intervals to sketch the curve

Homework Equations





The Attempt at a Solution

f'(x)= 1-x^2/(x^2+1)^2
But then I can't seem to work through taking the 2nd derivative, perhaps I am not using the chain rule right.
I get -4x^5-2x^3-2x/(x^2+1)^4
But that's not right... please help!
 
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Assuming you meant f(x)=x/(x^2+1) and not f(x)=(x/x^2)+1, the first derivative is correct. So again using the quotient rule, the second derivative will be
[tex]\frac{ (1 - x^2)' (x^2 + 1)^2 - (1 - x^2) ((x^2 + 1)^2 )' }{(x^2+1)^4}[/tex]
Can you work out the two derivatives that are there separately?
 
So I get that f''(x)= ((-2x)((x^2+1)^2)-1-x^2)((2)(x^2+1))(2x))/(x^2+1)^4
I think that's right but not sure about distributing through.

I get (2x^5-2x^3-4x)/(x^2+1)^4

But the posted answer is (2x^5-4x^3-6x)/(x^2+1)^4
 
The first line is correct, so you probably did something wrong in the expansion.
Try working out
((-2x)((x^2+1)^2)
and
-(1-x^2)((2)(x^2+1))(2x))
separately, and only then adding them.
Forgetting about the denominator (x^2 + 1)^4 for a while, you should get 2 x^5 - 4x^3 - 6x.

And look at the bright side: probably it's some stupid writing error, at least you know you can differentiate :smile:
 
Thanks for all the help, most of my errors are in the algebra...
on the left side I get (-2x^5-2x) and the right -((1-x^2)(4x^3+4x)) which goes to -(4x^3+4x-4x^5-4x^3)... Left - Right I get 2x^5-6x! still missing the -4x^3 I don't know where i am going wrong.
 
ande1717 said:
on the left side I get (-2x^5-2x)
You do? I don't.
What does (x^2 + 1)^2 expand to? It's not x^4 + 1!
If you want, write it out: (a + b)^2 = (a + b)(a + b) = ...
Then remember that formula (or at least, remember to remember that something is going on whenever you see it) forever :smile:
 
now I understand...I was convinced my mistake was on the right not the left. Thanks a ton.
 

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