To determine concavity of a function, you must find the second derivative #f''(x)#, determine its critical points (places where #f''(x)=0# or is not defined), and then examine the sign of #f''(x)# in all intervals defined by those critical points.
#f(x) = x^4 - x^2#
#f'(x) = 4x^3 - 2x#
#f''(x) = 12x^2- 2#
From examination #f''(x)# is a quadratic function, and therefore there are no places where it is undefined. Thus, our only critical points will come from the zeroes of #f''(x)#:
#12x^2-2 = 0#
#12x^2 = 2#
#x^2 = 1/6 :. x=+-1/sqrt(6)~~+-0.408#
Having two zeroes means we have 3 intervals to examine:
#(-oo,-1/sqrt(6)), (-1/sqrt(6),1/sqrt(6)), (1/sqrt(6),oo)#
Choose a single #x# value inside of each interval and evaluate #f''(x)# at that value. If the result is positive, the function #f(x)# is concave up in that interval; if the result is negative, the function is concave down. For simplicity, choose "easy" values of #x# to evaluate:
#f''(-1) = 12(-1)^2-2 = 12-2 = 10 > 0 :. "concave up"#
#f''(0) = 12(0)^2-2 = 0-2 = -2 < 0 :. "concave down"#
#f''(1) = 12(1)^2-2 = 12-2 = 10 > 0 :. "concave up"#
graph{x^4-x^2 [-1.923, 1.923, -0.963, 0.959]}
