What are the points of inflection of #f(x)=1/(5x^2+3) #?

1 Answer
Jun 6, 2018

No inflection points

Explanation:

Calculate the zeroes of #f'(x)#. The points of inflection are these points where #f'(x)=0# and also #f''(x)=0#.

#f'(x)=-(10x)/(5x^2+3)^2#

#f'(x)=0 rArr -10x=0 rArr x=0#, a single possible point of inflection. Note that both #f# and #f'# are well-defined at this point.

Calculate #f''(x)# to determine the nature of the point:

#f''(x)=(-10(5x^2+3)^2+10x(2(5x^2+3)*10x))/(5x^2+3)^4#
#=(-10(5x^2+3)^2+200x^2(5x^2+3))/(5x^2+3)^4#
#=(-10(5x^2+3)+200x^2)/(5x^2+3)^3#

Therefore #f''(0)=(-30+0)/3^3=-30/27<0#. So this point is a local maximum of #f#, not a point of inflection.

Compare with the graph of the function for a visual sanity check:
graph{1/(5x^2+3) [-2.5, 2.5, -1.25, 1.25]}