Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 9

f(x)=(x^2+1)/(x^2-1)
differentiating by applying quotient rule,
f'(x)=((x^2-1)(2x)-(x^2+1)(2x))/(x^2-1)^2
f'(x)=(2x^3-2x-2x^3-2x)/(x^2-1)^2
f'(x)=(-4x)/(x^2-1)^2
differentiating again,
f''(x)=((x^2-1)^2(-4)-(-4x)(2)(x^2-1)(2x))/(x^2-1)^4
f''(x)=(-4(x^2-1)(x^2-1-4x^2))/(x^2-1)^4
f''(x)=(4(3x^2+1))/(x^2-1)^3
In order to determine the concavity , determine when f''(x)=0 , however there are no points at which f''(x)=0 but at x=+- 1 the function is not continuous.
So test for concavity in the intervals (-oo ,-1) , (-1,1) , (1,oo ) by plugging in the test values in f''(x).
f''(0)=(4(3*0^2+1))/(0^2-1)^3=-4
f''(-2)=(4(3*(-2)^2+1))/((-2)^2-1)^3=52/27
f''(2)=(4(3*2^2+1))/(2^2-1)^3=52/27
Since f''(2) and f''(-2) are positive , so the function is concave upward in the interval (-oo ,-1) and (1,oo )
f''(0) is negative , so the function is concave downward in the interval (-1,1).