How do you find the local extrema for #f(x) = (x-3)^3# on (-∞, ∞)? Calculus Graphing with the First Derivative Identifying Turning Points (Local Extrema) for a Function 1 Answer Jim H Nov 11, 2016 #f# has no local extrema. Explanation: #f'(x) = 3(x-3)^2# is never undefined and is #0# only at #0#, so the only critical number is #0#. #f'(x)# is always positive for #x != 0#, so #f# is increasing on #(-oo,oo)#. 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 1139 views around the world You can reuse this answer Creative Commons License