Let's see, if I have time.

Critical numbers here are values of x when f'(x) = 0. They are where the graph of the function is steady; or not increasing, not decreasing; or where the tangent line is horizontal.

f'(x) = 3x^2 -4x -4

Set that to zero,

0 = 3x^2 -4x -4.

(3x +2)(x -2) = 0

x = -2/3 or 2 --------------at critical points.

2nd derivative test is about change in slope. If the slope of the tangent line changes from positive to negative, the critical point is maximum. From negative to positive f'(x), the critical point is minimum.

Or, in other words, if the 2nd derivative at the critical point is negative, then the critical point is maximum. Positive 2nd derivative means the critical point is minimum.

Or, 2nd deivative is about concavity of the graph also. If the concavity of the curve at the critical point if facing or opens downward, then that is for maximum. Concavity opening upward is for minimum.

Blah, blah, blah.

Words!

Anyway, the 2nd derivative, f''(x) = d/dx f'(x):

f'(x) = 3x^2 -4x -4

f''(x) = 6x -4

Test the critical mpoint at x = -2/3:

f''(-2/3) = 6(-2/3) -4 = -4 -4 = -8 -----negative.

Therefore, the graph has a maximum at x = -2/3 -------------***

Test the critical mpoint at x = 2:

f''(-2/3) = 6(2) -4 = 12 -4 = 8 -----positive.

Therefore, the graph has a minimum at x = 2 -------------***

Gotta go.