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This is a fairly simple proof. We’ll assume that  has a relative maximum to do the proof. The proof for a relative minimum is nearly
identical. So, if we assume that we
have a relative maximum at then we know that for all x
that are sufficiently close to . In particular for all h that are sufficiently close to zero (positive or negative) we
must have,
or, with a little rewrite we must have,
(1)
Now, at this point assume that and divide both sides of (1)
by h. This gives,
Because we’re assuming that we can now take the right-hand limit of both
sides of this.
We are also assuming that exists and recall that if a normal limit
exists then it must be equal to both one-sided limits. We can then say that,
If we put this together we have now shown that .
Okay, now let’s turn things around and assume that and divide both sides of (1)
by h. This gives,
Remember that because we’re assuming we’ll need to switch the inequality when we
divide by a negative number. We can
now do a similar argument as above to get that,
The difference here is that this time we’re going to be
looking at the left-hand limit since we’re assuming that . This argument shows that .
We’ve now shown that and . Then only way both of these can be true at
the same time is to have and this in turn means that must be a critical point.
As noted above, if we assume that has a relative minimum then the proof is
nearly identical and so isn’t shown here.
The main differences are simply some inequalities need to be switched.
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