[Understanding Analysis by Stephen Abbott] - [Chapter 5 The Derivatives] - [5.3 The Mean Value Theorem]

Theorem 5.3.1 (Rolle's Theorem). 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 a point \(c \in (a, b)\) where \(f^\prime(c) = 0\)

Theorem 5.3.2 (Mean Value Theorem). If \(f : [a, b] \to \mathbb{R}\) is continuous on \([a, b]\) and differentiable on \((a, b)\), then there exists a point \(c \in (a, b)\) where $$ f^\prime(c) = \frac{f(b) - f(a)}{b - a} $$

Corollary 5.3.3. if \(g : A \to \mathbb{R}\) is differentiable on an interval \(A\) and satisfies \(g^\prime(x) = 0\) for all \(x \in A\), then \(g(x) = k\) for some constant \(k \in \mathbb{R}\)

Corollary 5.3.4. If \(f\) and \(g\) are differentiable functions on an interval \(A\) and satisfies \(f^\prime(x) = g^\prime(x)\) for all \(x \in A\), then \(f(x) = g(x) + k\) for some constant \(k \in \mathbb{R}\)

Theorem 5.3.5 (Generalized Mean Value Theorem). If \(f\) and \(f\) are continuous on the closed interval \([a, b]\) and differentiable on the open interval \((a, b)\), then there exists a point \(c \in (a, b)\) where $$ [f(b) - f(a)]g^\prime(c) = [g(b) - g(a)]f^\prime(c) $$ if \(g^\prime\) is never zero on \((a, b)\), then the conclusion can be stated as $$ \frac{f^\prime(c)}{g^\prime(c)} = \frac{f(b) - f(a)}{g(b) - g(a)} $$

Theorem 5.3.6 (L'Hospital's Rule: 0/0 case). Let \(f\) and \(g\) be continuous on an interval containing \(\alpha\), and assume \(f\) and \(g\) are differential on this interval with the possible exception of the point \(\alpha\). If \(f(\alpha) = g(\alpha) = 0\) and \(g^\prime \neq 0\) for all \(x \neq \alpha\), then $$ \lim_{x \to a}\frac{f^\prime(x)}{g^\prime(x)} = L \text{ implies } \lim_{x \to a}\frac{f(x)}{g(x)} = L $$

Definition 5.3.7. Given \(g : A \to \mathbb{R}\) and a limit point \(c\) of \(A\), we say that \(limi_{x \to c}g(x) \to \infty\) if, for every \(M > 0\), there exists a \(\delta > 0\) such that whenever \(0 < |x - c| < \delta\) it follows that \(g(x) \geqslant M\).

Theorem 5.3.8 (L'Hospital's Rule: \(\infty/\infty\) casne). Assume \(f\) and \(g\) are differentiable on \((a, b)\) and that \(g^\prime(x) \neq 0\) for all \(x \in (a, b)\). If \(\lim_{x \to \alpha}g(x) = \infty\) or \(-\infty\), then $$ \lim_{x \to a}\frac{f^\prime(x)}{g^\prime(x)} = L \text{ implies } \lim_{x \to a}\frac{f(x)}{g(x)} = L $$