How do you find points of inflection and determine the intervals of concavity given #y=3/(x^2+4)#?

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Answer 1 · 2018-04-03T05:03:21

Below

#y=3/(x^2+4)#
#dy/(dx)=-(6x)/(x^2+4)^2#
#(d^2y)/(dx^2)=-6times(4-3x^2)/(x^2+4)^3#

For stationary points, #dy/(dx)=-(6x)/(x^2+4)^2=0#

ie #-6x=0#
#x=0#

Test #x=0#
#(d^2y)/(dx^2)=-3/8 <0#

Therefore, it is a maximum and concave down at #x=0# #(0,3/4)#

For point of inflexion, #(d^2y)/(dx^2)=0#

#-6times(4-3x^2)/(x^2+4)^3=0#

#4-3x^2=0#
#4/3=x^2#
#x=+-2/sqrt3#

Test when #x=-2/sqrt3#

#x=-1.5# #(d^2y)/(dx^2)=1056/15625#

#x=-2/sqrt3# #(d^2y)/(dx^2)=0#

#x=-1# #(d^2y)/(dx^2)=-6/125#

Therefore, there is a change in concavity so there is a point of inflexion at #x=-2/sqrt3# #(-2/sqrt3,9/16)#

Test when #x=2/sqrt3#

#x=1# #(d^2y)/(dx^2)=-6/125#

#x=2/sqrt3# #(d^2y)/(dx^2)=0#

#x=1.5# #(d^2y)/(dx^2)=1056/15625#

Therefore there is a change in concavity so there is a point of inflexion at #x=2/sqrt3# #(2/sqrt3,9/16)#