How do you graph #f(x)=1/3x^3-6x# using the information given by the first derivative?

1 Answer
Aug 17, 2016

From #f'(x) = 0, f(x)# has turning points at #x=+-sqrt6#

Explanation:

#f(x)= 1/3x^3-6x#

#f'(x) = cancel3 x^2/cancel 3 -6#

#f(x)# has tuning points where #f'(x) = 0#
I.e. when #x^2 = 6 -> x=+-sqrt6#

From the graph below, #f(-sqrt6)# has curvature concave down #-> # a local maximum and #f(+sqrt6)# has curvature concave up #-># a local minimum.

Also, #f(x) =0# when #x(x^2/3-6)=0#
I.e. when #x=0# or #x= +-sqrt18 = +-3sqrt2#

graph{1/3x^3-6x [-25.67, 25.65, -12.83, 12.84]}