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Calc Help +K

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TheBlueMtSkier
Posts: 909 - Karma: 42
I know this is usually a hit or miss with NSG, but hell I've been stuck on this problem for thirty minutes. And yes, I did google. Even tried tutorvista, but they literally brought me halfway through the problem and then wanted me to buy the full package.

Problem: Determine where the function is concave upward and where it is concave downward.
F(x) = (1)/(2+x^2)

Steps: 1. Find first deriv
2. Find second deriv
3. Set equal to 0 and find where it switches
4. Use test points to determine what areas are negative and positive



Answer: Concave upward on 1. (-Infinity, -sqrt(6)/3) 2. (sqrt(6)/3, Infinity)
Concave Downward on (-sqrt(6)/3, sqrt(6), 3)




Thanks!
Mar 18 2012 5:52PM
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Crossfade
Posts: 743 - Karma: 24
Reading that made my head hurt.  Good luck.
Mar 18 2012 5:54PM
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shocker611
Posts: 2599 - Karma: 423
use division rule for derivatives (lo d'high-hi d'low)/low^2. Do that twice as both times you will still have a fraction. Set = 0 like you said and use test points. Are you just having problems with the derivative or what?
Mar 18 2012 5:57PM
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TheBlueMtSkier
Posts: 909 - Karma: 42
I'm having problems with the second derivative. For some reason it's not coming out right.


I end up with

((-4-2x^2) - 16x^2 - 8x^4) / ((2+x^2)^4)
Mar 18 2012 6:00PM
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TheBlueMtSkier
Posts: 909 - Karma: 42
Found an error in that.

Now I got.

(8x^4 - 6x^2 + 8) / (2+x^2)^4
Mar 18 2012 6:04PM
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huge-go
Posts: 1451 - Karma: 33
I'd say screw the quotient rule, move the denominator up to the top and make the exponent negative. Then chain rule for first derivative and chain/product for second.
Mar 18 2012 6:23PM
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