>I'm asked to determine the relative min and/or relative max for the function:
>
>f(x)=x^3-x^2-x+2
Find the first derivative:
f'(x) = 3�x^2 - 2�x - 1
Find values for x such that f'(x) is zero.
Factoring gives x = -1/3 and x = 1
These are the POSSIBLE relative extrema.
Find the second derivative:
f"(x) = 6�x - 2
Determine the curvature at the POSSIBLE relative extrema.
f"(-1/3) = -4 -- Negative, making a relative maximum
f"(1) = 4 -- Positive, making a relative minimum.
If you wish to knew the value fo the relative extrema, you must evaluate the original function at those points, f(-1/3) and f(1).