Rolle's Theorem/Proof 2
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Theorem
Let $f$ be a real function which is:
- continuous on the closed interval $\closedint a b$
and:
- differentiable on the open interval $\openint a b$.
Let $\map f a = \map f b$.
Then:
- $\exists \xi \in \openint a b: \map {f'} \xi = 0$
Proof
First take the case where:
- $\forall x \in \openint a b: \map f x = 0$
Then:
- $\forall x \in \openint a b: \map {f'} x = 0$
Otherwise:
- $\exists c \in \openint a b: \map f c \ne 0$
Let $\map f c > 0$.
Then there exists an absolute maximum at a point $\xi \in \openint a b$.
Hence:
\(\ds \dfrac {\map f {\xi + h} - \map f \xi} h\) | \(\le\) | \(\ds 0\) | for $\xi < \xi + h < b$ | |||||||||||
\(\ds \dfrac {\map f {\xi + h} - \map f \xi} h\) | \(\ge\) | \(\ds 0\) | for $a < \xi + h < \xi$ |
As $h \to 0$, we see that both of the above approach $\map {f'} \xi$, which is then both non-negative and non-positive.
That is:
- $\map {f'} \xi = 0$
Similarly, let $\map f c < 0$.
Then there exists an absolute minimum at a point $\xi \in \openint a b$.
Hence:
\(\ds \dfrac {\map f {\xi + h} - \map f \xi} h\) | \(\ge\) | \(\ds 0\) | for $\xi < \xi + h < b$ | |||||||||||
\(\ds \dfrac {\map f {\xi + h} - \map f \xi} h\) | \(\le\) | \(\ds 0\) | for $a < \xi + h < \xi$ |
Again, as $h \to 0$, we see that both of the above approach $\map {f'} \xi$, which is then both non-negative and non-positive.
That is:
- $\map {f'} \xi = 0$
Hence the result.
$\blacksquare$
Sources
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 2$. Functions of One Variable: $2.3$ Rolle's Theorem
- 1996: H. Jerome Keisler and Joel Robbin: Mathematical Logic and Computability ... (previous) ... (next): $\S 1.12$: Valid Arguments: Proposition $1.12.3$