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

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In summary, the homework statement is that there is an attempt to find horiz. asymptotes and use them to sketch the curve. The equations that are given are for a function that is f(x)=x/x^2+1. The first derivative is correct, so the second derivative is\frac{ (1-x^2)' (x^2 + 1)^2 - (1-x^2) ((x^2+1)^2 )' }{(x^2+1)^4}. The first line is correct, so you probably did something wrong in the expansion. The second line is-(1-x^2)((2)(x^2+
  • #1
ande1717
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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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  • #2
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?
 
  • #3
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
 
  • #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:
 
  • #5
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.
 
  • #6
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:
 
  • #7
now I understand...I was convinced my mistake was on the right not the left. Thanks a ton.
 

1. What is the second derivative?

The second derivative is a mathematical concept that represents the rate of change of the slope of a curve. It is the derivative of the derivative, meaning it measures how the slope of a curve is changing at a particular point.

2. What is "2nd derivative trouble"?

"2nd derivative trouble" refers to difficulties or challenges that arise when trying to calculate or interpret the second derivative of a function or data set.

3. What causes "2nd derivative trouble"?

There are several possible causes of "2nd derivative trouble," including non-smooth functions, discontinuities, and undefined or infinite values in the first derivative. These can make it difficult or impossible to calculate the second derivative accurately.

4. How can "2nd derivative trouble" be avoided?

To avoid "2nd derivative trouble," it is important to carefully select and analyze the data or function being used. Ensuring that the first derivative is well-behaved and does not contain any abrupt changes or undefined values can also help avoid issues when calculating the second derivative.

5. Why is the second derivative important?

The second derivative has several practical applications in science and engineering, including optimization, curve fitting, and determining the concavity of a function. It can also provide insights into the rate of change of a system and help make predictions about future behavior.

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