What is the x-coordinate of the point of inflection on the graph of #y=1/10x^(5)+1/2X^(4)-3/10#?

1 Answer
Lucy
Jul 14, 2018

The x-coordinate of the inflexion point is #-3#

Explanation:

#y=1/10x^5+1/2x^4-3/10#

#(dy)/(dx)=1/2x^4+2x^3#

#(d^2y)/(dx^2)=2x^3+6x^2#

For second derivative, #(d^2y)/(dx^2)=0#

#2x^3+6x^2=0#
#2x^2(x+3)=0#
#x=0# or #x=-3#

Then you must these points for concavity
Test #(0,-3/10)#
When #x=-1#, #(d^2y)/(dx^2)=4#
When #x=0#, #(d^2y)/(dx^2)=0#
When #x=1#, #(d^2y)/(dx^2)=8#
Therefore, #(0,-3/10)# is not a point of inflexion as the concavity doesn't change

Test #(-3,15 9/10)#,
When #x=-4#, #(d^2y)/(dx^2)=-32#
When #x=-3#, #(d^2y)/(dx^2)=0#
When #x=-2#, #(d^2y)/(dx^2)=8#
Therefore, #(-3,15 9/10)# is a point of inflexion as the concavity changes

Related questions
See all questions in Determining Points of Inflection for a Function
Impact of this question
7480 views around the world
You can reuse this answer
Creative Commons License