Question #7ec7d Calculus Graphing with the Second Derivative Determining Points of Inflection for a Function 1 Answer Jim H Apr 8, 2017 Please see below. Explanation: #y = (6x)/(x^2+3)# The domain is #RR# #y' = (-6(x^2-3))/(x^2+3)# #y''=(12x(x^2-9))/(x^2+3)^3# #y'' = 0# at #x = 0#, #3# and #-3#. The denominator of #y''# is always positive and the numerator changes sign at each of its three zeros. Therefore, #y''# changes sign 3 times. Answer link Related questions How do you find the inflection points for the function #f(x)=8x+3-2 sin(x)#? How do you find the inflection point of a cubic function? How do you find the inflection point of a logistic function? What is the inflection point of #y=xe^x#? How do you find the inflection points for the function #f(x)=x^3+x#? How do you find the inflection points for the function #f(x)=x/(x-1)#? How do you find the inflection points for the function #f(x)=x/(x^2+9)#? How do you find the inflection points for the function #f(x)=xsqrt(5-x)#? How do you find the inflection points for the function #f(x)=e^sin(x)#? How do you find the inflection points for the function #f(x)=x-ln(x)#? See all questions in Determining Points of Inflection for a Function Impact of this question 1186 views around the world You can reuse this answer Creative Commons License