Concavity of a rational function

  • Thread starter Thread starter realism877
  • Start date Start date
  • Tags Tags
    Function Rational
Click For Summary

Homework Help Overview

The discussion revolves around the concavity and curve sketching of the rational function \( \frac{x^3}{x^2-1} \). Participants are analyzing the second derivative to determine intervals of concavity and discussing the behavior of the function as \( |x| \) approaches infinity.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the intervals of concavity derived from the second derivative and question the identification of inflection points and relative extrema. There is also an exploration of the function's behavior at infinity.

Discussion Status

Some participants have provided guidance on the intervals of concavity and suggested methods for analyzing the function's behavior at infinity. Multiple interpretations regarding inflection points and local extrema are being explored, but there is no explicit consensus on the correctness of the identified points.

Contextual Notes

Participants reference asymptotes at \( x = -1 \) and \( x = 1 \) as part of their analysis. There is a mention of specific values for local extrema, but the accuracy of these values is under discussion.

realism877
Messages
80
Reaction score
0
I have to curve sketch this function, ((x)3))/((x)2-1)

I did all of the steps, and I got this as the second derivative: ((2x(x2+3))/((x2-1)3)

I got concave up:(-1, 0)u(1, inifinity)

Concave down:(-infinity, -1)u(0,1)


Am I right?

I use -1 and 1 as the interval since they are the asymptotes.
 
Physics news on Phys.org
Hi realism877! :smile:

realism877 said:
I have to curve sketch this function, ((x)3))/((x)2-1)

I did all of the steps, and I got this as the second derivative: ((2x(x2+3))/((x2-1)3)

I got concave up:(-1, 0)u(1, inifinity)

Concave down:(-infinity, -1)u(0,1)


Am I right?

I use -1 and 1 as the interval since they are the asymptotes.

That is correct.
 
To have a more complete sketch, how does the function behave for |x| → ∞ ?

Where are inflection points and relative extrema?
 
the point of inflection is (0,0)
 
To see how the function behaves for |x| → ∞, it might help to write [itex]\displaystyle \frac{x^3}{x^2-1}\ \ \text{ as }\ \ x+\frac{x}{x^2-1}[/itex]
 
SammyS said:
To have a more complete sketch, how does the function behave for |x| → ∞ ?

Where are inflection points and relative extrema?

The local max=(-sqr(3), -2.598)

Local min=(sqr(3), 2.598)

Correct?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
8K