This function has no discontinuities . A polynomial is always continuous.
To find intervals of concavity and determine inflection points, take the second derivative, set it equal to #0,# and solve for #x# :
#y'=4x^3-(2)(3)x=4x^3-6x#
#y''=(3)(4)x^2-6=12x^2-6#
#12x^2-6=0#
#12x^2=6#
#x^2=6/12=1/2#
#x=+-sqrt(1/2)=+-1/sqrt(2)#
Now, we want to break up the domain of #y# around these #x#-values. The domain of #y# is #(-∞,∞)#. Breaking it up yields:
#(-∞,-1/sqrt(2)),(-1/sqrt(2),1/sqrt(2)),(1/sqrt(2),∞)#
Now, we want to determine whether #y''# is positive or negative in each of these intervals. If #y''>0# in an interval, then #y# is concave up on that interval; if #y''<0# on an interval, then #y# is concave down on that interval. If #y''# changes signs at a certain value of #x#, then there is an inflection point at that #x#-value.
#(-∞,-1/sqrt(2)):#
#y''(-1)=12(-1)^2-6=12-6>0#
#y# is concave up on #(-∞,-1/sqrt(2))#
#(-1/sqrt(2),1/sqrt(2))#:
#y''(0)=12(0^2)-6=-6<0#
#y# is concave down on #(-1/sqrt(2),1/sqrt(2))# and has changed signs at #x=-1/sqrt(2)#; thus, there is an inflection point at #x=-1/sqrt(2)#.
#(1/sqrt(2),∞):#
#y''(1)=12(1^2)-6=12-6>0#
#y# is concave up on #(1/sqrt(2),∞)# and has changed signs at #x=1/sqrt(2)#; thus, there is an inflection point at #x=1/sqrt(2).#
