How do you graph #y = 2x^3 - 4x + 1# using the first and second derivative?

1 Answer
Gió
Feb 21, 2015

You can start by using the first derivative to find points of maximum and minimum:
#y'=6x^2-4#
You set the first derivative equal to zero:
#6x^2-4=0# which gives you:
#x=+-sqrt(2/3)#
enter image source here
So:
maximum: #x=-sqrt(2/3)# and #y=3.17#
minimum: #x=sqrt(2/3)# and #y=-1.17#

You use the second derivative to find inflection points:
#y''=12x#
Setting it equal to zero you get the point of inflection:
#x=0# #y=1#
enter image source here

And finally the graph:
enter image source here

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