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.