Concavity of a rational function

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The discussion focuses on the concavity of the rational function f(x) = (x^3)/(x^2 - 1). The second derivative is calculated as f''(x) = (2x(x^2 + 3))/(x^2 - 1)^3, leading to intervals of concavity: concave up on (-1, 0) ∪ (1, ∞) and concave down on (-∞, -1) ∪ (0, 1). The point of inflection is identified at (0, 0), with local extrema at (-√3, -2.598) for the maximum and (√3, 2.598) for the minimum. The asymptotes at x = -1 and x = 1 are crucial for determining these intervals.

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realism877
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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.
 
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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 \displaystyle \frac{x^3}{x^2-1}\ \ \text{ as }\ \ x+\frac{x}{x^2-1}
 
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?
 

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