Rolle's Theorem

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Proof: We assume that \(x_0\) is a local maximum (the case of minimum is shown analogously). Let \(U\) be a neighbourhood of \(x_0\) with \(U\subset I\) and \(f(x_0)=\max\{f(x):\;x\in U\}\). Let \[f(x)=f(x_0)+(x-x_0)\cdot\Delta_{f,x_0}(x).\] Assume that \(f'(x_0)=\Delta_{f,x_0}(x_0)>0\). Since \(\Delta_{f,x_0}\) is continuous in \(x_0\), then there exists a neighbourhood \(V\subset U\) of \(x_0\) such that \(\Delta_{f,x_0}(x)>0\) for all \(x\in V\). Then for all \(x_1\in V\) with \(x_1>x_0\) holds \(f(x_1)=f(x_0)+(x_1-x_0)\cdot\Delta_{f,x_0}(x_1)>f(x_0)\). This is a contradiction.
On the other hand, assume that \(f'(x_0)=\Delta_{f,x_0}(x_0)<0\). Since \(\Delta_{f,x_0}\) is continuous in \(x_0\), then there exists a neighbourhood \(V\subset U\) such that \(\Delta_{f,x_0}(x)<0\) for all \(x\in V\). Then for all \(x_1\in V\) with \(x_1<x_0\) holds \(f(x_1)=f(x_0)+(x_1-x_0)\cdot\Delta_{f,x_0}(x_1)>f(x_0)\). This is also a contradiction.\(\Box\)

As a consequence, we will formulate the following result stating that derivatives of functions with equal boundary conditions have at least one zero.

Proof: If \(f\) is constant, the statement is clear (since then \(f'(x)=0\) for all \(x\in(a,b)\)). If \(f\) is not constant, consider the maximum and the minimum of \(f\) on \([a,b]\) (we know by the theorem about compactness of images under continuout functions the that they exist). So, let \(x_-,x_+\in[a,b]\) such that \[f(x_+)=\max\{f(x)\;:\;x\in[a,b]\},\qquad f(x_-)=\min\{f(x)\;:\;x\in[a,b]\}.\] Then we have that \(x_+\in(a,b)\) or \(x_-\in(a,b)\) since, otherwise, \(f(x_+)=f(x_-)\) (constant).
Then \(f'(x_-)=0\) or \(f'(x_+)=0\).\(\Box\)

As a corollary, we have that for a function \(f\) differentiable in some interval \(I\), the following holds: Between two zeros of \(f\), there always exists some point \(x_0\) with \(f'(x_0)=0\).