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]}