Fermat's Theorem : If  f ( x ) contain a relative extrema at x = c & f ′ (c ) exists then x = c is a critical point of f ( x ) . Actually, it will be a critical point such that f ′ (c ) = 0 .

Note as well that we can say that f ′ (c ) = 0 since we are also supposing that  f ′ (c ) exists.

This theorem described us that there is a nice relationship between relative extrema and critical points. In fact it will let to get a list of all possible relative extrema.  As a relative extrema have to be a critical point the list of all critical points will give us a list of all possible relative extrema.

Consider the case of f ( x ) = x² .  We illustrated that this function had a relative minimum at x = 0 in various earlier examples. Hence according to Fermat's theorem x = 0 must be a critical point. The derivative of the function is,

                                                               f ′ ( x ) = 2x

Certain enough x = 0 is a critical point.

Be careful not to use wrongly this theorem.  This doesn't say that a critical point will be a relative extrema.  To illustrate this, consider the following case.

f ( x ) = x³                          f ′ ( x ) = 3x²

Clearly x = 0 is a critical point. Though we know that this function has no relative extrema of any kind.  Thus, critical points do not have to be relative extrema.

Also note as well that this theorem says nothing regarding absolute extrema.  An absolute extrema might or might not be a critical point.