Finding extreme values of a single variable function
If a function is continuous, its rate of change is non-constant and there are intervals in which it's crescent and others where it's decrescent, a natural conclusion is to expect the rate of change to be zero somewhere. [math]\displaystyle{ f'(x) = 0 }[/math] is a necessary but insufficient condition for a point to be a local maximum or minimum. It's insufficient because there are points where [math]\displaystyle{ f'(x) = 0 }[/math] and yet that point is neither a maximum nor a minimum. The same concept applies to multivariable functions. A point can have a necessary property to be a critical point and yet fail to meet other necessary criteria to be a maximum or a minimum.
Take the function [math]\displaystyle{ f(x) = x^3 }[/math] for example. At the origin its rate of change inverts its sign, but this function is strictly crescent and doesn't have a local maximum or minimum anywhere.
Fermat's stationary points
If [math]\displaystyle{ f }[/math] is differentiable at [math]\displaystyle{ c }[/math] and [math]\displaystyle{ c \in D_f }[/math] in an open interval. A requisite for that point to be a local maximum or minimum is [math]\displaystyle{ f'(c) = 0 }[/math].
We begin with the hypothesis that [math]\displaystyle{ c }[/math] is a local maximum. Now let's consider a small step to the right and to the left of that point and call the step [math]\displaystyle{ s }[/math]:
[math]\displaystyle{ f(c) \geq f(x + s) }[/math] (with our starting assumption every step that we take to the right or to the left, from the local maximum, results in going down)
Therefore:
[math]\displaystyle{ f(x + s) - f(c) \leq 0 }[/math] (added [math]\displaystyle{ -f(c) }[/math] to both sides. Note that the inequality wasn't inverted.)
We can divide both sides by a positive constant without affecting the inequality:
[math]\displaystyle{ \frac{f(x + s) - f(c)}{s} \leq \frac{0}{s} }[/math]
We are assuming that [math]\displaystyle{ f }[/math] is differentiable at [math]\displaystyle{ c }[/math], therefore the sided limits exist and are equal to each other:
[math]\displaystyle{ f'(c) = \lim_{s \ \to \ 0^{+}} \frac{f(x + s) - f(c)}{s} \leq 0 }[/math]
With that we have shown that [math]\displaystyle{ f'(c) \leq 0 }[/math].
If we consider [math]\displaystyle{ s \lt 0 }[/math] the previous inequality inverts to being greater than or equal to zero. We calculate the same limit, but from the left [math]\displaystyle{ h \to 0^{-} }[/math]. With that we prove that [math]\displaystyle{ f'(c) \geq 0 }[/math]. With both inequalities being true we conclude that [math]\displaystyle{ f'(c) = 0 }[/math]. It makes perfect sense if we think graphically. To the right and to the left of [math]\displaystyle{ f'(c) }[/math] the derivatives have opposite signs. When the take the sided limits, both converge to the same, horizontal, tangent line.
The same reasoning follows to prove that a local minimum also has [math]\displaystyle{ f'(c) = 0 }[/math].
Note: I have a textbook that does the proof by applying the sign conservation theorem, which condenses the proof by not having to deal with positive and negative increments separately.
Links for the proofs:
