Question

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

Answer 1

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]}