How do you find the inflection points for the function #f(x)=8x+3-2 sin(x)#? Calculus Graphing with the Second Derivative Determining Points of Inflection for a Function 1 Answer Sonnhard May 26, 2018 Solve the equation #sin(x)=0# and it must be #cos(x)ne 0# Explanation: #f(x)=8x+3-2sin(x)# #f'(x)=8-2cos(x)# #f''(x)=2sin(x)# #f'''(x)=2cos(x)# Answer link Related questions 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)#? How do you find the inflection points for the function #f(x)=e^(3x)+e^(-2x)#? See all questions in Determining Points of Inflection for a Function Impact of this question 7464 views around the world You can reuse this answer Creative Commons License