What are the points of inflection of #f(x)=x+sinx # on the interval #x in [0,2pi]#? Calculus Graphing with the Second Derivative Determining Points of Inflection for a Function 1 Answer Sonnhard May 27, 2018 #x_1=0# or #x_2=pi# or #x_3=2*pi# Explanation: We have #f(x)=x+sin(x)# Then #f'(x)=1+cos(x)# #f''(x)=-sin(x)# Solving #sin(x)=0# for #x#in the given interval we get #x=0# #x=pi# #x=2pi# 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 6689 views around the world You can reuse this answer Creative Commons License