How do you find points of inflection and determine the intervals of concavity given #y=2x^(1/3)+3#?

Question

1361 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-23T17:00:46

Investigate the sign of #y''#

#y' = 2/3x^(-2/3)#

#y'' = -4/9x^(-5/3) = (-4)/(9x^(5/3))#

Note that the sign of #x^(5/3)# is the same as that of #x#, so

#y''# is positive left of #x=0# and the graph is concave up (convex)

and #y''# is negative right of #0# and the graph is concave down (concave).

The concavity changes at #x = 0# whic is in the domain of the function, so there is a inflection point at #x = 0# which make #y = 3#

#(0,3)# is the only inflection point.