posted by Lisa on .
Determine the concavity for f(x)= 4/x^2-1
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?
if 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