What are the inflection points for #y = 6x^3 - 3x^4#?

1 Answer
Feb 20, 2018

#(0,0), (1,3)#

Explanation:

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.