Theorem 1. Let \(f:[a,b]\to\mathbb{R}\) be differentiable.
Then there exists some \(\hat{x}\in(a,b)\) such that \[f(b)-f(a)=f'(\hat{x})\cdot
(b-a).\]

Before the proof is presented, we give some graphical interpretation:
A division of the above equation by \(b-a\) gives \[\frac{f(b)-f(a)}{b-a}=f'(\hat{x}).\]
The quantity on the left hand side is equal to the slope of the secant
of \(f\) through \(a\) and \(b\), whereas \(f'(\hat{x})\) corresponds to the slope
of tangent of \(f\) at \(\hat{x}\). Therefore, the secant of \(f\) through \(a\) and \(b\) is parallel to a tangent of \(f\).

Proof: Consider the function \(F:[a,b]\to\mathbb{R}\) with \[F(x):=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}\cdot(x-a).\]
Then we have \(F(a)=F(b)=0\). By
Rolle’s Theorem, we get that there exists some \(\hat{x}\in(a,b)\) with \[0=F'(\hat{x})=f'(\hat{x})-\frac{f(b)-f(a)}{b-a}\]
and thus \[f'(\hat{x})=\frac{f(b)-f(a)}{b-a}.\]\(\Box\)

The mean value theorem leads us to determine monotonicity properties
of a function by means of its derivative.

Theorem 2. Let \(f:[a,b]\to\mathbb{R}\) be a differentiable
function. Then the following holds true.

If \(f'(x)>0\) for
all \(x\in(a,b)\), then \(f\) is strictly monotonically
increasing.

If \(f'(x)<0\) for
all \(x\in(a,b)\), then \(f\) is strictly monotonically
decreasing.

If \(f'(x)\geq0\) for
all \(x\in(a,b)\), then \(f\) is monotonically
increasing.

If \(f'(x)\leq0\) for
all \(x\in(a,b)\), then \(f\) is monotonically
decreasing.

Proof: (i) By the mean value theorem we have
that for \(x_1,x_2\in(a,b)\) with \(x_1<x_2\), there exists some \(\hat{x}\in(x_1,x_2)\) with \[f(x_2)-f(x_1)=f'(\hat{x})\cdot(x_2-x_1)>0.\]
The results (ii)-(iv) can be proven analogously.\(\Box\)

_{The data for the interactive network on this webpage was generated with pntfx Copyright Fabian Gabel and Julian Großmann. pntfx is licensed under the MIT license. Visualization of the network uses the open-source graph theory library Cytoscape.js licensed under the MIT license.
}