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March 3, 2015

March 3, 2015

Posted by **Lisa** on Sunday, March 10, 2013 at 7:46pm.

I took the second derivative and solved for zeros. I got x=-1 and x=1. However when plugged into the orig. equation this results in 4/0 (undefined). Does this mean there is no concavity?

- Concavity-Calculus -
**Steve**, Sunday, March 10, 2013 at 8:46pmif f = 4/(x^2-1)

f'' = 8(3x^2+1)/(x^2-1)^3

f'' is never zero.

f'' is discontinuous at x=1 and -1, and changes sign there

so, f ix concave down for |x|<1 and concave up for |x|>1

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