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

Related questions
See all questions in Examples of Curve Sketching
Impact of this question
3664 views around the world
You can reuse this answer
Creative Commons License