Fermat's Theorem as relates to calculus states that if f has a local maximum or minimum at c, and if f'(c) exists, then f'(c) = 0.

```|            _
|     _     / \
|    / \   /   \
|   /   \_/
|  /
---------------
```
Clearly, at the local maximums and minimums, the tangent line is horizontal, and f'(c) = 0. Much as you'd like it to be good enough, however, most professors refuse to accept the proof by picture argument. So, the real one follows:

Suppose f has a local maximum at c. Then it follows that f(c) >= f(x) if x is sufficiently close to c. This implies that if h is sufficiently close to 0, f(c) >= f(c+h). Therefore,

```f(c) >= f(c+h)
f(c) - f(c+h) >= 0```
If h > 0,
```
f(c) - f(c+h) >= 0
h```
Taking the right-hand limit of both sides,
```lim    (f(c)-f(c+h) >= lim 0 = 0
h->0+        h          h->0+
```
Similarly, if h < 0,
```f(c) - f(c+h) <= 0
h```
Taking the left-hand limit of both sides,
```lim    (f(c)-f(c+h) <= lim 0 = 0
h->0-        h          h->0-
```
Since f'(c) exists,
```f'(c) = lim    (f(c)-f(c+h) = lim    (f(c)-f(c+h)
h->0+       h          h->0-        h```
Therefore,
```f'(c) >= 0 and f'(c) <= 0
f'(c) = 0.```
QED.

A virtually identical argument can be written down to prove that this is also true if f has a minimum at c.

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