If # f(x) = ( x^3 )/( x^2 - 25 )#, what are the points of inflection, concavity and critical points?

1 Answer
Nov 30, 2016

Critical numbers: #0# and #+-sqrt75#

Explanation:

Concave down on #(-oo,-5)# and on #(0,5)#

Concave up on #(-5,0)# and on #(5,oo)#

Inflection point #(0,0)#

#f'(x) = (x^2(x^2-75))/(x^2-25)^2#

#f'(x)# DNE at #x=+-5# but those are not in the domain, so they are not critical.

#f'(x) = 0# at #x=0# and at #x + +-sqrt75# which are in the domain, so they are all critical numbers.

#f''(x) = (50x(x^2+75))/(x^2-25)^3# could change sign at #0# and at #+-5#

On #(-oo,-5)#, #f''(x) < 0#, so #f# is concave down.
On #(-5,0)#, #f''(x) > 0#, so #f# is concave up.
On #(0,5)#, #f''(x) < 0#, so #f# is concave down.
On #(5,oo)#, #f''(x) < 0#, so #f# is concave up.

The concavity changes at #x=-5#, #0# and #5#. An inflection point is a point of the graph where concavity changes. Since #-5# and #5# are not in the domain of #f#, there are no IPs there, but #f(0) = 0#, so #(0,0)# is an Infle pt.