How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for #y=-x^5+2x^3+4#?

1 Answer
RedRobin9688
Jun 24, 2018

No discontinuities.
Points of inflection at #x= 0, +-sqrt(3/5)#
CCU on (inf,#-sqrt(3/5)#) and (0,#sqrt(3/5)#)
CCD on (#-sqrt(3/5)#,0) and (#sqrt(3/5)#,inf)

Explanation:

#f(x) # is a polynomial so continuous on reals.

By power rule we can find second derivative to be

#f''(x)=-20x^3+12x#

Setting equal to zero we can find Possible Inflection Points

#-20x^3+12x=0#
#x=0, +-sqrt(3/5)#

Testing concavity between points using #x=-1,-1/2,1/2,1# gives positive concavity between -inf and #-sqrt(3/5)# negative concavity from #-sqrt(3/5)# to 0 positive concavity from 0 to #sqrt(3/5)# and negative concavity from #sqrt(3/5)# to inf.

Related questions
See all questions in Determining Points of Inflection for a Function
Impact of this question
1517 views around the world
You can reuse this answer
Creative Commons License