The Second Derivative Test is used to find the local extrema of a function (When a function rises and then falls, the local maximum is the highest point of the rise. When a function falls and then rises, the local minimum is the lowest pint of the fall).
Second Derivative Test:
Suppose that f is differentiable on an open interval containing c and that f'(c) = 0.
 If f''(c) < 0, then f has a local maximum at c.
 If f''(c) > 0, then f has a local minimum at c.
Proof:
If f'(c) = 0, then the tangent line to the graph at P(c, f(c)) is horizontal. If, in addition, f''(c) < 0, then the graph is concave downward at c, and hence there is an interval (a,b) containing c such that the graph lies below the tangent lines. It follows that f(c) is a local maximum for f (see ill. 1). Furthermore, if f''(c) > 0, then the graph is concave up at c, and therefore there is an interval (a,b) contaning c such that the graph lies above the tangent lines. It follows that f(c) is a local minimum for f (see ill. 2).
If f''(c) = 0, the second derivative test is not applicable and the first derivative test should be used.
1.(a crappy representation of an upsidedown parabola)
 P(c,f(c))
_____________ f'(c) = 0
 / \
 / \
 / \
 f''(c)<0

___________________________
c
2.(and a crappy version of a parabola)
 f''(c)>0
 \ /
 \ /
 \ /
_____\_/_____ f'(c) = 0

 P(c,f(c))

____________________________
c
Example:
f(x) = 12 + 2x^{2}  x^{4}
f'(x) = 4x  4x^{3} = 4x(1  x^{2})
so the critical numbers (where f'(x) = 0) are 0, 1, and 1. These are now plugged into the second derivative.
f''(x) = 4  12x^{2}
f''(0) = 4 > 0 so (0,12) is a local minimum
f''(1) = 8 < 0 so (1,13) is a local maximum
f''(1) = 8 < 0 so (1,13) is a local maximum