Unizor Derivatives - Constant Functions

Notes to a video lecture on http://www.unizor.com

Derivatives - Constant Function

The following statement is obvious.
If a smooth function f(x), defined on segment [a,b] (including endpoints), is constant, that is if
∀ x∈[a,b]: f(x)=f(a)=f(b),
then its derivative at any inner point of this interval equals to zero:
∀ x∈(a,b): f'(x) = 0

What is more interesting is the converse theorem.

Theorem 1

If a smooth function f(x), defined on segment [a,b] (including endpoints), has a derivative at any inner point equaled to zero, that is if
∀ x∈(a,b): f'(x) = 0,
then this function is constant on this segment, that is
∀ x∈[a,b]: f(x)=f(a)=f(b).

Proof

Let's choose any point x inside interval (a,b).
Now use Lagrange Theorem for our function f(x) on an segment [a, x] that starts at point a and ends at point x.
This theorem states that there exists a point x0∈(a, x) such that
f'(x0) = [f(x)−f(a)] / (x−a)
Since we know that the derivative of function f(x) on an interval (a,b) is zero at any point, we conclude that
0 = [f(x)−f(a)] / (x−a)
from which follows
f(x)=f(a).
Recall that point x was chosen as any point in interval (a,b). It means that f(x)=f(a) is true for any inner point of this interval.
Since function f(x) is smooth (which, in particular, implies continuity), the values at the end of this interval are also the same.
Hence, our function is constant on segment [a,b].

End of proof.

Theorem 2

If two smooth functions f(x) and g(x), defined on segment [a,b] (including endpoints), have equal derivatives, that is if
∀ x∈(a,b): f'(x) = g'(x)
then these functions are different only by a constant on this segment, that is
∃c:∀ x∈[a,b]: f(x)=g(x)+c.

Proof

Consider a new function
h(x)=f(x)−g(x).
Since derivatives of f(x) and g(x) are equal to each other, derivative of h(x) equals to zero as well:
h'(x) = f'(x)−g'(x)=0.
Now use the theorem above that states that if a smooth function h(x) has derivative equaled to zero at any inner point of a segment [a,b], then it is constant on this segment, that is
h(x) = c, where c=h(a)=h(b)
Therefore,
f(x)−g(x)=c for any x∈[a,b].

End of proof.

Important corollary
If we are given a derivative of some function and we have guessed an original function, from which this derivative was taken, we can say that any other function with the same derivative differs from the one we have guessed by a constant, and there are no other functions with this derivative.
For example, we can guess that, if derivative is f I(x)=x², then original function might be f(x)=x³/3. Now, based on the above theorem, we can state that an expression f(x)=x³/3+c describes all the functions that have derivative f I(x)=x², where c - any real number, and no other function with this derivative exists.