What is the foundation of differential calculus? We will discover it
from one of the most important theorem in differential calculus which
called Mean-Value theorem. In order to prove this theorem, we only need
several steps.

  • THEOREM 1—The Extreme Value Theorem
    If f f f is continuous on a closed interval [ a , b ] [a,b] [a,b], then f f f attains
    both an absolute maximum value M M M and an absolute minimum value m m m
    in [ a , b ] [a,b] [a,b].
    Note: The proof of this theorem is beyond the content of
    differential calculus.

  • THEOREM 2—The First Derivative Theorem for Local Extreme
    Values

    If f f f has a local maximum or minimum value at on interior point c c c
    of its domain, and if f ′ f^{'} f′ is defined at c c c, then
    f ′ ( c ) = 0. f^{'}(c)=0. f′(c)=0. Hint. Assume that f ′ ( c ) ≠ 0 f^{'}(c)\ne 0 f′(c)​=0.Let
    f ′ ( c ) > 0 f^{'}(c)>0 f′(c)>0. If c c c is a local maximum point, it is obvious that
    there exists another point b b b close to c c c at which f ( b ) > f ( c ) f(b)>f(c) f(b)>f(c).
    Similarly, we can prove other situations.

  • THEOREM 3—Rolle’s Theorem
    Suppose that y = f ( x ) y=f(x) y=f(x) is continuous over the closed interval [ a , b ] [a,b] [a,b]
    and differentiable at every point of its interior ( a , b ) (a,b) (a,b). If
    f ( a ) = f ( b ) f(a)=f(b) f(a)=f(b), then there is at least one number c c c in ( a , b ) (a,b) (a,b) at
    which f ′ ( c ) = 0 f^{'}(c)=0 f′(c)=0
    Hint. Using the THEOREM 2, this theorem can be easily solved.

  • THEOREM 4—The Mean Value Theorem
    Suppose y = f ( x ) y=f(x) y=f(x) is continuous over a closed interval [ a , b ] [a,b] [a,b] and
    differentiable on the inverval’s interior ( a , b ) (a,b) (a,b). Then there is at
    least one point c c c in ( a , b ) (a,b) (a,b) at wich
    f ( b ) − f ( a ) b − a = f ′ ( c ) \frac{f(b)-f(a)}{b-a}=f^{'}(c) b−af(b)−f(a)​=f′(c)

    Hint. With the auxiliary function
    g ( x ) = f ( x ) − f ( a ) − f ( b ) − f ( a ) b − a ( x − a ) g(x)=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a) g(x)=f(x)−f(a)−b−af(b)−f(a)​(x−a) , this theorem is a
    direct result of rolle’s theorem by noticing that g ( a ) = g ( b ) = 0 g(a)=g(b)=0 g(a)=g(b)=0.

Conclusion
We shall notice that the theorem 1 is the origination of the proof of
the other theorems. It is so important that we cannot ignore its
correctness. However, no differential calculus teacher tells you the
proof of this thorem, because it is really hard. As you can imagine,
this is exactly the field that mathematical analysis will deal with.
From grabbing those essential concepts behind theorem 1, the door of
mathematical analysis will open to you.

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