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.
Proof by picture:
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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.