What are the extrema of #x^3-2x^2-8x#? Calculus Graphing with the First Derivative Identifying Turning Points (Local Extrema) for a Function 1 Answer Shaun Y. · Roella W. Dec 23, 2015 #[2+ 2sqrt(7)]/3 and [2- 2sqrt(7)]/3# Explanation: #f(x)=x^3 - 2x^2 - 8x# #f'(x)= 3x^2 - 4x -8# #3x^2 - 4x -8=0# since this is not factorable we use the quadratic equation #x= [2+ 2sqrt(7)]/3# Answer link Related questions How do you find the x coordinates of the turning points of the function? How do you find the turning points of a cubic function? How many turning points can a cubic function have? How do you find the coordinates of the local extrema of the function? How do you find the local extrema of a function? How many local extrema can a cubic function have? How do I find the maximum and minimum values of the function #f(x) = x - 2 sin (x)# on the... If #f(x)=(x^2+36)/(2x), 1 <=x<=12#, at what point is f(x) at a minimum? How do you find the maximum of #f(x) = 2sin(x^2)#? How do you find a local minimum of a graph using the first derivative? See all questions in Identifying Turning Points (Local Extrema) for a Function Impact of this question 1672 views around the world You can reuse this answer Creative Commons License