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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.

  1. If f''(c) < 0, then f has a local maximum at c.
  2. 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 upside-down 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 + 2x2 - x4
f'(x) = 4x - 4x3 = 4x(1 - x2)
so the critical numbers (where f'(x) = 0) are 0, 1, and -1. These are now plugged into the second derivative.
f''(x) = 4 - 12x2
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

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