Concavity of a rational function

  • Thread starter realism877
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  • #1
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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.
 

Answers and Replies

  • #2
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Hi realism877! :smile:

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.
 
  • #3
SammyS
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To have a more complete sketch, how does the function behave for |x| → ∞ ?

Where are inflection points and relative extrema?
 
  • #4
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the point of inflection is (0,0)
 
  • #5
SammyS
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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]
 
  • #6
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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?
 

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