Fermat's Theorem
If f(x) has a relative extrema at x = c and f′(c) exists then x = c is a critical point of f(x). Actually, this will be a critical point that f′(c) =0.
Proof
It is a fairly easy proof. We will suppose that f(x) has a relative maximum to do the proof.
The proof for a relative minimum is nearly the same. Therefore, if we suppose that we have a relative maximum at x = c after that we know that f(c) ≥ f(x) for all x which are sufficiently close to x = c.
Particularly for all h which are sufficiently close to zero may be positive or negative we must contain,
f(c) ≥ f(c + h)
or, with a little rewrite we should have,
f(c + h) - f(c) < 0 (1)
Now, here suppose that h > 0 and divide both sides of (1) with h. It provides,
(f(c + h) - f(c))/h < 0
Since we're assuming that h > 0 we can here take the right-hand limit of both sides of such.
= lim_{h}_{→}0¯ (f(c + h) - f(c))/h < lim_{h}_{→}0¯ 0 = 0
We are also assume that f′(c) exists and recall this if a general limit exists then this should be equal to both one-sided limits. We can so say that,
f′(c) = lim_{h}_{→}0¯ (f(c + h) - f(c))/h = lim_{h}_{→}0¯ (f(c + h) - f(c))/h < 0
If we place this together we have here demonstrated that, f′(c) ≤ 0 .
Fine, now let's turn things around and suppose that h < 0 provides,and divide both sides of (1) with h. It gives
(f(c + h) - f(c))/h > 0
Keep in mind that as we're assuming h < 0 we will require to switch the inequality while we divide thorugh a negative number. We can here do a same argument as above to find that,
f′(c) = lim_{h}_{→}0 (f(c + h) - f(c))/h = lim_{h}_{→}0¯ (f(c + h) - f(c))/h > lim_{h}_{→}0¯ 0 = 0
The difference now is that currently we're going to be considering at the left-hand limit as we're assuming that h < 0 . This argument illustrates that f′(c) ≥ 0 .
We've now shown that
f′(c) ≤ 0 and f′(c) ≥ 0. So only way both of such can be true at similar time is to have f′(c) = 0 and it means that x = c must be a critical point.
As considered above, if we suppose that f(x) has a relative minimum then the proof is nearly the same and therefore isn't illustraten here. The major differences are simply several inequalities require to be switched.