For the function, use the second derivative test (if possible) to determine if each critical point is a minimum, maximum, or neither. If the second derivative test can't be used, say so.
f (x) = x3 – 2x2 + x
The first derivative is
f '(x) = 3x2 - 4x + 1
and the second derivative is
f "(x) = 6x – 4.
We use the first derivative to find the critical points. Happily, it factors:
The critical points are and x = 1. Now we find the (sign of) the second derivative at each critical point.
so f is concave down and has a maximum at x = ⅓.
f "(1) = 6(1) 0 – 4 = 2 > 0
so f is concave up and has a minimum at x = 1.