Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 8

f(x)=(2x^2)/(3x^2+1)
differentiating by applying quotient rule,
f'(x)=((3x^2+1)(4x)-(2x^2)(6x))/(3x^2+1)^2
f'(x)=(12x^3+4x-12x^3)/(3x^2+1)^2
f'(x)=(4x)/(3x^2+1)^2
differentiating again,
f''(x)=((3x^2+1)^2(4)-4x(2)(3x^2+1)(6x))/(3x^2+1)^4
f''(x)=(4(3x^2+1)(3x^2+1-12x^2))/(3x^2+1)^4
f''(x)=(4(1-9x^2))/(3x^2+1)^3
In order to determine the concavity , first determine when f''(x)=0,
4(1-9x^2)/(3x^2+1)^3=0
1-9x^2=0
x=+-1/3
Now let us consider the intervals (-oo ,-1/3) , (-1/3,1/3) and (1/3,oo ) and determine the signs of f''(x) by plugging in the test values in f''(x).
f''(0)=(4(1-9*0^2))/(3*0^2+1)^3=4
f''(-1)=(4(1-9*(-1)^2))/(3*(-1)^2+1)^3=-1/2
f''(1)=(4(1-9*(1)^2))/(3*1^2+1)^3=-1/2
So the graph is concave downward in the interval (-oo ,-1/3) and (1/3,oo )
graph is concave upward in the interval (-1/3,1/3).