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