Mean value theorem
Implications

Theorem 1: Assume that f {\displaystyle f} is a continuous, real-valued function, defined on an arbitrary interval I {\displaystyle I} of the real line. If the derivative of f {\displaystyle f} at every interior point of the interval I {\displaystyle I} exists and is zero, then f {\displaystyle f} is constant in the interior.

Proof: Assume the derivative of f {\displaystyle f} at every interior point of the interval I {\displaystyle I} exists and is zero. Let (a, b) {\displaystyle (a,b)} be an arbitrary open interval in I {\displaystyle I}. By the mean value theorem, there exists a point c {\displaystyle c} in (a, b) {\displaystyle (a,b)} such that

This implies that f (a) = f (b) {\displaystyle f(a)=f(b)}. Thus, f {\displaystyle f} is constant on the interior of I {\displaystyle I} and thus is constant on I {\displaystyle I} by continuity. (See below for a multivariable version of this result.)

Remarks:

Theorem 2: If f ′ (x) = g ′ (x) {\displaystyle f'(x)=g'(x)} for all x {\displaystyle x} in an interval (a, b) {\displaystyle (a,b)} of the domain of these functions, then f − g {\displaystyle f-g} is constant, i.e. f = g + c {\displaystyle f=g+c} where c {\displaystyle c} is a constant on (a, b) {\displaystyle (a,b)}.

Proof: Let F (x) = f (x) − g (x) {\displaystyle F(x)=f(x)-g(x)}, then F ′ (x) = f ′ (x) − g ′ (x) = 0 {\displaystyle F'(x)=f'(x)-g'(x)=0} on the interval (a, b) {\displaystyle (a,b)}, so the above theorem 1 tells that F (x) = f (x) − g (x) {\displaystyle F(x)=f(x)-g(x)} is a constant c {\displaystyle c} or f = g + c {\displaystyle f=g+c}.

Theorem 3: If F {\displaystyle F} is an antiderivative of f {\displaystyle f} on an interval I {\displaystyle I}, then the most general antiderivative of f {\displaystyle f} on I {\displaystyle I} is F (x) + c {\displaystyle F(x)+c} where c {\displaystyle c} is a constant.

Proof: It directly follows from the theorem 2 above.
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