George C. answered • 12/30/12

Humboldt State and Georgetown graduate

From the second derivative, f" = x - 4

f' = (1/2)x^2 - 4x + c1

f'(4) = 2 = (1/2)(4)^2 - 4(4) + c1

f' = (1/2)x^2 - 4x + 10

f = (1/6)x^3 - 2x^2 + 10x + c2

f(4) = -1 = (1/6)(4)^3 - 2(4)^2 + 10(4) + c2

f = (1/6)x^3 - 2x^2 +10x - (59/3)

A cubic polynomial with no "hump" or "dip" in it. Think of a different version of plain old x^3.

Zeroes of the first derivative occur at 4 ± 2i in complex plane.

The point (4, -1) is an inflection point, foil (d).