User:Shaesthetic/sandbox

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

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


 * $$0=f'(c)=\frac{f(b)-f(a)}{b-a}.$$

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

Remarks: Theorem 2: If $$f'(x) = g'(x)$$ for all $$x$$ in an interval $$(a, b)$$ of the domain of these functions, then $$f - g$$ is constant, i.e. $$f = g + c$$ where $$c$$ is a constant on $$(a, b)$$.
 * Only continuity of $$f$$, not differentiability, is needed at the endpoints of the interval $$I$$. No hypothesis of continuity needs to be stated if $$I$$ is an open interval, since the existence of a derivative at a point implies the continuity at this point.  (See the section continuity and differentiability of the article derivative.)
 * The differentiability of $$f$$ can be relaxed to one-sided differentiability, a proof is given in the article on semi-differentiability.

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

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

Proof: It directly follows from the theorem 2 above.