How do you find the inflection points for the function #f(x)=8x+3-2 sinx#? Calculus Graphing with the Second Derivative Determining Points of Inflection for a Function 1 Answer Jim H Apr 24, 2015 #f(x)=8x+3-2 sinx# #f'(x)=8-2 cosx# #f''(x)= 2 sinx# #f''(x) =2sinx = 0# at every integer multiple of #pi# and, we know the sine changes sign at every #x# intercept, so every point: #(n pi,0)# with integer #n# is an inflection point. 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 2333 views around the world You can reuse this answer Creative Commons License