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The condition that the derivative at the extremum of an uniformly differentiable function is zero.

is this necessary and sufficient or both?

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    @NateEldredge ya thanks, it clearly did2011-12-04

1 Answers 1

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Define $x^*$ to be a point where f'(x^*)=0.

Necessary

In order to show if f'(x^*)=0 is necessary for $x^*$ to be an extremum, we must show that:

If f'(x^*) \ne 0 then $x^*$ cannot be an extremum.

Intuitively, if f'(x^*) \ne 0 then it must be the case that either f'(x^*) \gt 0 or f'(x^*) \lt 0 which implies that the function $f(x)$ is either increasing or decreasing at $x^*$ respectively. Therefore, $x^*$ cannot be an extremum.

Sufficient

In order to show sufficiency we have to show that:

If f'(x^*) = 0 then $x^*$ must be an extremum.

Suppose that f'(x^*) = 0. If the point $x^*$ must be an extremum it must be the case that either:

  1. $f(x)$ is decreasing to the left of $x^*$ and increasing to the right of $x^*$

                                  or 
  2. $f(x)$ is increasing to the left of $x^*$ and decreasing to the right of $x^*$

But, f'(x^*)=0 by itself cannot tell us anything about the behavior of $f(x)$ in a local neighborhood of $x^*$. Thus, f'(x^*)=0 is not a sufficient condition for $x^*$ to be an extremum.


Note that the above arguments are informal and rely on geometrical intuition. However, they can be formalized depending on how much rigor you wish for your proof.

The above arguments only indicate that the following:

  1. $x^*$ may be a candidate for an extremum. You need additional tests (in particular the second derivative test) to check if it is in fact an extremum.

  2. $x^*$ may not be a global extremum.

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    @RileyE hmm, true. Let me fix my answer.2011-11-29