Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 11

g(x)=(x^2+4)/(4-x^2)
differentiating by applying quotient rule,
g'(x)=((4-x^2)(2x)-(x^2+4)(-2x))/(4-x^2)^2
g'(x)=(8x-2x^3+2x^3+8x)/(4-x^2)^2
g'(x)=(16x)/(4-x^2)^2
differentiating again by applying quotient rule,
g''(x)=16((4-x^2)^2-x(2)(4-x^2)(-2x))/(4-x^2)^4
g''(x)=(16(4-x^2)(4-x^2+4x^2))/(4-x^2)^4
g''(x)=(16(3x^2+4))/(4-x^2)^3
There are no points at g''(x)=0,but at x=2 and x=-2 the function is not continuous.
So test for concavity in the intervals (-oo ,-2) , (-2,2) and (2,oo )
g''(-3)=-496/125
g''(0)=1
g''(3)=-496/125
Since g''(-3) and g''(3) are less than 0 , so the graph is concave downward in the intervals (-oo ,-2) and(2,oo )
and g''(0) is greater than 0 , so the graph is concave upward in the interval (-2,2).