📂Calculus

Proof of Rolle's Theorem in Calculus

Theorem¹

If the function $f(x)$ is continuous at $[a,b]$ and differentiable at $(a,b)$ and if $f(a)=f(b)$, then there exists at least one $c$ in $(a,b)$ that satisfies $f ' (c)=0$.

Description

In high school courses, it is introduced only as an auxiliary lemma to prove the mean value theorem and is not used at all otherwise. However, beyond the high school level, it is sometimes used as an auxiliary lemma. Although the mean value theorem is more general, when there is no need to use complex forms like $\displaystyle f '(c) = {{f(b) - f(a)} \over {b - a}}$, it makes the proof more concise.

Proof

Strategy: Apply Fermat’s theorem by dividing it into two cases: when $f(x)$ is a constant function and when it is not.

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• Case 1. $f(x)$ is a constant function

  Since $f ' (x)=0$, there exists at least one $c$ in $(a,b)$ that satisfies $f ' (c)=0$.

• Case 2. When $f(x)$ is not a constant function

  Since $f(x)$ has a maximum or minimum and is differentiable in $(a,b)$, there exists a critical point $c$ satsifying $f ' (c)$.

  Fermat’s Theorem

  If function $f(x)$ has a maximum or minimum at $x=c$ and $f ' (c)$ exists, then $f ' (c) = 0$

  The critical point $c$ must satisfy $f ' (c) = 0$ by Fermat’s theorem.

Therefore, whether $f(x)$ is a constant function or not, there exists at least one $c$ in $(a,b)$ that satisfies $f ' (c)=0$.

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