In order to investigate concavity, we shall look at the sign of the second derivative.
f(x) = (5x^2)/(x^2 + 4)
f'(x) = (-40x)/(x^2+4)^2
Finding f''(x):
f''(x) = (-40(x^2+4)^2 -(-40x)2(x^2+4)(2x))/(x^2+4)^4
= ((x^2+4)[-40(x^2+4) + 160x^2])/(x^2+4)^4
= [-40(x^2+4) + 160x^2]/(x^2+4)^3
So:
f''(x) = (120x^2 - 160)/(x^2+4)^3
In general, a function can change sign by either crossing the x axis (being equal to 0), or by being discontinuous (teleporting across the x axis).
In this case f''(x) is never discontinuous for real x, and
f''(x) = 0 when 3x^2-4 = 0, so x= +-2/sqrt3
Cut the number line into intervals and test each interval:
(-oo, -2/sqrt3) f''(x) is positive, so the graph of f is concave up
(-2/sqrt3, 2/sqrt3) f''(x) is negative, so the graph of f is concave down
(2/sqrt3, oo) f''(x) is positive, so the graph of f is concave up
