Take the first derivative:
#y'=18 x^2-12 x^3#
Take the second derivative:
#y''=36x-36 x^2#
Let's factor #y''# a bit to simplify it:
#y''=36x(1-x)#
Set #y''# equal to zero and solve for #x#:
#36x(1-x)=0#
#36x=0 #
So #x=0# is our first potential inflection point's #x#-value.
#1-x=0#
So #x=1# is our second potential inflection point's #x#-value.
We must test values of #y''# in the following intervals (but not at the endpoints, hence the parentheses):
#(-∞, 0)#
#(0, 1)#
#(1, ∞)#
#(-∞,0)#
#y''(-1)=-36(2) <0 #
In the interval #(-∞, 0#), #y'' < 0# and so the graph of y is concave down.
#(0,1)#
#y''(1/2)=18(1/2) >0 #
In the interval #(0,1)#, #y''>0# and so the graph of y is concave up. We've switched concavity. This means we have an inflection point at #x=0#.
Let's plug #x=0# back into our original function to get the inflection point's coordinates:
#y(0)=6(0)^3-3(0)^4=0#
#(0,0)# is an inflection point.
#(1,∞)#
#y''(2)=72(-1)<0 #
In the interval #(1,∞)#, #y''<0# and so the graph of y is concave down. Again, we've switched concavity. We also have an inflection point at #x=1#.
Plug #x=1# into the original function to get the second inflection point's coordinates:
#y(1)=6-3=3#
#(1,3)# is an inflection point.
