- #1

- 80

- 0

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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- Thread starter realism877
- Start date

- #1

- 80

- 0

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.

- #2

- 22,129

- 3,297

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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Science Advisor

Homework Helper

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Where are inflection points and relative extrema?

- #4

- 80

- 0

the point of inflection is (0,0)

- #5

SammyS

Staff Emeritus

Science Advisor

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Gold Member

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- #6

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- 0

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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