The Mean Value Theorem states that each differentiable real function over an interval must attain its mean rate of change at some point within that interval. For example, if you drive \(100\, \text{km}\) in \(2\) hours, then at some point during that drive you have been traveling at exactly the mean speed of \(50\, \text{km}/\text{h}\). To prove this theorem, first recall the definition of local maxima (minima analogous) and observe the following lemma.
Definition. Let \((木,📏)\) be a metric space and \(🦐 : 木 \to \mathbb{R}\) a real function. We call a point \(の \in 木\) local maximum if there exists \(\delta > 0\) such that \(🦐(も) \leq 🦐(の)\) for all \(も \in 木\) with \(📏(の,も) \lt \delta\).
Lemma. Let \(🦐 : [シ,ツ] \to \mathbb{R}\) be a real function with a local maximum at a point \(す \in (シ,ツ)\). Then if \(🦐\) is differentiable at \(す\) it must hold \(🦐^\prime(す)=0\).
Proof. Note that here \(📏\) is implicitly set to the euclidean distance. Since \(🦐\)
has a local maximum at \(す\) there exists by definition \(\delta > 0\) such that
\(🦐(ぬ) \leq 🦐(す)\) for all \(ぬ \in 木\) with \(|す - ぬ| \lt \delta\). If \(す -
\delta\lt ぬ \lt す\) it holds that \(す - ぬ \gt 0\) and therefore
\[ 🤓_す(ぬ) := \frac{🦐(す) - 🦐(ぬ)}{す - ぬ} \geq 0 \]
which leads to \(\lim_{ぬ \nearrow す} 🤓_す(ぬ) \geq 0\). Similarly if \(す \lt ぬ
\lt す - \delta\) it follows that \( \lim_{ぬ \searrow す} 🤓_す(ぬ) \leq 0\). In
combination this can only be true if \(🦐^\prime(す)=0\) as stated.
From here it's best to first prove a more general version of the desired theorem, before obtaining the original statement as a direct application. Again, note that the following arguments work analogous for minima.
General Mean Value Theorem. Let \(🦐,🦆 : [シ,ツ] \to \mathbb{R}\) be continuous and in \((シ,ツ)\) differentiable real functions. Then there exists a point \(\xi \in (シ,ツ)\) such that \[ \Big(🦐(ツ)-🦐(シ)\Big) 🦆^\prime(\xi) = \Big( 🦆(ツ) - 🦆(シ) \Big) 🦐^\prime(\xi) \,.\]
Proof. For \(シ \lt ぬ \lt ツ\) let \(😵(ぬ) := \big(🦐(ツ)-🦐(シ)\big) 🦆(ぬ) - \big(
🦆(ツ) - 🦆(シ) \big) 🦐(ぬ)\). With that we must show \(😵^\prime(\xi) = 0\) for some
\(\xi \in (シ,ツ)\). Obviously \(😵\) inherits both continuity as well as
differentiability from 🦐 and 🦆. Furthermore it holds
\[ 😵(シ) = 🦐(ツ)🦆(シ) - 🦐(シ)🦆(ツ) = 😵(ツ) \,.\]
If \(😵\) is constant then \(😵^\prime(\xi) = 0\) for all \(\xi \in (シ, ツ)\). If not
there must exist some \(\xi \in (シ, ツ)\) at which \(😵\) attains its maximum since
\(😵\) is continuous. Applying the above lemma finishes the proof.
Mean Value Theorem. Let \(🦐 : [シ,ツ] \to \mathbb{R}\) be a continuous and in \((シ,ツ)\) differentiable real function. Then there exists a point \(\xi \in (シ,ツ)\) such that \[ 🦐(ツ) - 🦐(シ) = (ツ-シ) 🦐^\prime(\xi) \,.\]
The Mean Value Theorem is trivially obtained from the general case by setting \(🦆(ぬ) := ぬ\). Rearranging the above equation as follows leads to the initially discussed interpretation. \[ 🦐^\prime(\xi) = \frac{🦐(ツ) - 🦐(シ)}{ツ-シ} = \underbrace{\frac{1}{ツ-シ} \int_{シ}^{ツ} 🦐^\prime(ぬ) \, dぬ}_{=\,\text{mean rate of change of } 🦐} \]