Question

rolles thm. let \\(f: a,b \to \mathbb{r}\\) continuous, differentiable on \\((a,b)\\). if \\(f(a) = f(b)\\), then \\(\exists\\ c \in (a,b)\\) with \\(f(c) = 0\\).
pf. if \\(f(a) = f(b)\\), then \\(f\\) is constant on \\(a,b\\) or not.
if constant, then \\(f(c) = 0\\ \forall\\ c \in (a,b)\\) and thus the result holds.
if not constant, then as \\(f\\) continuous over compact domain \\(a,b\\), the extreme values of \\(f\\) are attained on \\(a,b\\). let \\(c\\) be where \\(f\\) attains a global min, \\(d\\) for global max. it must be at least one of \\(c, d\\) not equal to \\(a\\) or \\(b\\), otherwise i.e. \\(c=a\\) and \\(d=b\\) we have \\(f(b) = f(a) = f(c) \le f(x) \le f(d) = f(b)\\) contradicts \\(f\\) nonconstant on \\(a,b\\).
thus at least one of \\(c\\) or \\(d\\) in \\((a,b)\\), by another thm \\(f(c) = 0\\) or \\(f(d) = 0\\) at this point as global (local) extrema.

Explanation:

State Rolle's Theorem

Rolle's Theorem states: Let \(f: [a, b] \to \mathbb{R}\) be continuous on \([a, b]\) and differentiable on \((a, b)\). If \(f(a) = f(b)\), then there exists some \(c \in (a, b)\) such that \(f'(c) = 0\).

Case 1: Constant Function

If \(f\) is constant on \([a, b]\), then for any \(c \in (a, b)\), the derivative \(f'(c) = 0\). Thus, the theorem holds.

Case 2: Non-constant Function

If \(f\) is not constant, then by the Extreme Value Theorem, since \([a, b]\) is compact and \(f\) is continuous, \(f\) attains its global maximum and global minimum on \([a, b]\). Let \(c\) be the point where \(f\) attains its global minimum, and \(d\) be the point where \(f\) attains its global maximum.

Locate Local Extremum in Open Interval

Since \(f\) is non-constant and \(f(a) = f(b)\), at least one of the extremum points \(c\) or \(d\) must lie in the open interval \((a, b)\). If both occurred at the endpoints, then \(f(x)\) would be bounded between \(f(a)\) and \(f(b)\), forcing \(f\) to be constant, which is a contradiction.

Apply Fermat's Theorem

Let \(x_0 \in (a, b)\) be this interior extremum point (either \(c\) or \(d\)). Since \(f\) is differentiable on \((a, b)\) and has a local extremum at \(x_0\), by Fermat's Theorem on Stationary Points, we must have \(f'(x_0) = 0\).

Answer:

The handwritten text provides a proof of Rolle's Theorem:

1. Theorem Statement: If \(f: [a, b] \to \mathbb{R}\) is continuous on \([a, b]\), differentiable on \((a, b)\), and \(f(a) = f(b)\), then there exists some \(c \in (a, b)\) such that \(f'(c) = 0\).
2. Proof Structure:

• Constant Case: If \(f\) is constant, \(f'(c) = 0\) for all \(c \in (a, b)\).
• Non-constant Case: By the Extreme Value Theorem, \(f\) attains a global maximum and minimum on \([a, b]\). Since \(f\) is non-constant and \(f(a) = f(b)\), at least one of these extrema must occur at an interior point \(x_0 \in (a, b)\). By Fermat's Theorem, the derivative at this interior local extremum is zero: \(f'(x_0) = 0\).