If a function is differentiable twice, we know that it is concave up if #f''(x) >0# and concave down viceversa.

So let us calculate:

#f(x) = 4/(x^2+1) = 4(x^2+1)^(-2)#

#f'(x) = -16x(x^2+1)^(-3)#

#f''(x) = 96x^2(x^2+1)^(-4)-16(x^2+1)^(-3) = frac(96x^2-16(x^2+1)) ((x^2+1)^4)= frac(96x^2-16x^2-16) ((x^2+1)^4)=16frac (5x^2-1) ((x^2+1)^4)#

Evidently the denominator of #f''(x)# is always positive, so the sign of #f''(x)# is the sign of the numerator and we see that:

#f''(x) < 0# for #|x|< 1/sqrt(5)#

So #f(x)# is concave up in the intervals #(-oo,-1/sqrt(5))# and #(1/sqrt(5),+oo)# and concave down for #x# in the interval #(-1/sqrt(5),1/sqrt(5))#.

graph{4/(x^2+1) [-10, 10, -5, 5]}