One of the most fundamental theorems in mathematical analysis is the mean value theorem. Geometrically, the theorem says that somewhere between points A and B on a differentiable curve there is at least one tangent line parallel to the secant line AB.

Letâ€™s discover together this great theorem and give it some applications. Here will use the concept of differential functions.

**Statement and applications of the mean value theorem**

The mean value theorem is used to prove certain regularities of differentiable functions. In fact, it is used to prove that a function is norm continuity; to demonstrate that a function is a Lipschitz function, etc.

**Theorem:** Let $a$ and $b$ be real numbers and le $f$ be a real-valued function continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. Then there exists a real number $c\in (a,b)$ such that $f(b)-f(a)=f'(c)(b-a)$.

**Some applications:** Show that for any real numbers $x,y$, we have $|\sin(x)-\sin(y)|\le |x-y|$ and $|\arctan(x)-\arctan(y)|\le |x-y|$. In fact, without loss of generality, we can assume that $x<y$. The functions $t\mapsto \sin(t)$ and $t\mapsto \arctan(t)$ are continuous and differentiable on $\mathbb{R}$. Thus, according to the above theorem, there exists $c_1\in (x,y)$ and $c_2\in (x,y)$ such that $\sin(x)-\sin(y)=\cos(c_1)(x-y)$ and $\arctan(x)-\arctan(y)=\frac{1}{1+c_2^2}(x-y)$. This is because $\sin'(t)=\cos(t)$ and $\arctan'(t)=\frac{1}{1+t^2}$ for any $t\in\mathbb{R}$. Now we take the absolute and use the fact that $|\cos(c_1)|\le 1$ and $\frac{1}{1+c_2^2}\le 1$ to ends the proof.

**The vectorial version of the mean value theorem**

Consider a normed vector space $(E,\|\cdot\|)$. We deal with vector-valued functions of the form $f:[a,b]\to E$. Then the mean value theorem for such functions is somehow different. In fact, instead of equality, we have inequality. More precisely, we have the following result.

**Theorem:** Assume that a function $f:[a,b]\to E$ is continue on $[a,b]$ and differential on $(a,b)$. On the other hand, assume that there exists a constant $M>0$ such that $\|f'(t)\|\le M$ for any $t\in (a,b)$. Then $$ \|f(b)-f(a)\|\le M |a-b|.$$