Definition 1 (Differentiability) \(\text{Let} \hspace{1.5mm} f: [a, b] \rightarrow \mathbb{R}. \hspace{1.5mm} \text{For} \hspace{1.5mm} p \in [a, b], \hspace{0.5mm} f \hspace{1.5mm} \text{is said to be differentiable at} \hspace{1.5mm} p \hspace{1.5mm} \text{if}\)

\[\begin{equation} \large \lim_{x \to p} \frac{f(x) - f(p)}{x - p} \end{equation}\]

 

\(\text{exists. Moreover, if that limit —which is denoted by} \hspace{1.5mm} f'(p)\text{— exists for every} \hspace{1.5mm} p \hspace{1.5mm} \text{belonging to a subset} \hspace{1.5mm} E \subseteq [a, b] \hspace{1.5mm} \text{then} \hspace{1.5mm} f \hspace{1.5mm} \text{is} \\ \text{said to be differentiable at} \hspace{1.5mm} E.\)

 

 

Theorem 1 (Differentiability implies continuity) \(\text{Let} \hspace{1.5mm} f: [a, b] \rightarrow \mathbb{R}. \hspace{1.5mm} \text{If} \hspace{1.5mm} f \hspace{1.5mm} \text{is differentiable at a point} \hspace{1.5mm} p \in [a, b] \hspace{1.5mm} \text{then} \hspace{1.5mm} f \hspace{1.5mm} \text{is continuous at} \hspace{1.5mm} p.\)

 

Proof. Perhaps a proof by contrapositive will be easier to pull off. Assume \(f\) is not continuous at \(p\). That is

\[\begin{equation} \exists \epsilon > 0 \hspace{1.5mm} \text{such that} \hspace{1.5mm} \forall \delta > 0, \hspace{0.5mm} \exists x \in N_{\delta}(p) \cap [a, b] \hspace{1.5mm} \text{such that} \hspace{1.5mm} f(x) \notin N_{\epsilon}(f(p)). \end{equation}\]

 

From here we deduce that either

\[\begin{equation} f(x) - f(p) \geq \epsilon \quad \text{or} \quad f(x) - f(p) \leq - \epsilon. \end{equation}\]

 

On the other hand, \(x\) satisfies \(-\delta < x - p < \delta\). Thus, either

\[\begin{equation} \frac{1}{x - p} > \frac{1}{\delta} \quad \text{or} \quad \frac{1}{x - p} < -\frac{1}{\delta}. \end{equation}\]

 

Whatever the case is, we conclude that either

\[\begin{equation} \frac{f(x) - f(p)}{x - p} > \frac{\epsilon}{\delta} \quad \text{or} \quad \frac{f(x) - f(p)}{x - p} < -\frac{\epsilon}{\delta} \end{equation}\]

 

and that is true for an arbitrarily small \(\boldsymbol{\delta}\). Therefore, as we consider ever narrower neighborhoods around \(p\) we will be certain to find a value \(x\) such that the expression \(\large \frac{f(x) - f(p)}{x - p}\) gets ever farther from the origin of the real line. It refuses to be contained within any finite-radius neighborhood of a real number \(f'(p)\). And thus we conclude such a number \(f'(p)\) does not exist —hence, \(f\) is not differentiable at \(p\)—.

 

 

Theorem 2 (Sign of the derivative) \(\text{Let} \hspace{1.5mm} f: [a, b] \rightarrow \mathbb{R} \hspace{1.5mm} \text{be differentiable at} \hspace{1.5mm} p \in [a, b].\)

\(\text{If} \hspace{1.5mm} f'(p) > 0 \hspace{1.5mm} \text{then} \hspace{1.5mm} \exists \delta > 0 \hspace{1.5mm} \text{such that}\)

 

\(\text{Similarly, if} \hspace{1.5mm} f'(p) < 0 \hspace{1.5mm} \text{then} \hspace{1.5mm} \exists \delta > 0 \hspace{1.5mm} \text{such that}\)

 

Proof. Assume \(f\) is differentiable at \(p\). That is, there exists a number \(f'(p)\) such that

\[\begin{equation} \forall \epsilon > 0, \hspace{0.5mm} \exists \delta > 0 \hspace{1.5mm} \text{such that} \hspace{1.5mm} \forall x \in \big(N_{\delta} \cap [a, b]\big) \setminus \{p\}, \hspace{0.5mm} \frac{f(x) - f(p)}{x - p} \in N_{\epsilon}(f'(p)) \tag{1} \end{equation}\]

 

Since the statement above is true for every \(\epsilon > 0\) then let us assume \(f'(p) > 0\) and consider \(\epsilon = f'(p)\). Therefore, every \(x\) belonging to \(\big(N_{\delta} \cap [a, b]\big) \setminus \{p\}\) satisfies

\[\begin{align} -f'(p) &< \frac{f(x) - f(p)}{x - p} - f'(p) < f'(p) \\ \\ &\overbrace{0 < \frac{f(x) - f(p)}{x - p}}^{\large \text{This is what} \\ \large \text{interests us}} < 2f'(p) \end{align}\]

 

Consider \(\boldsymbol{p - \delta < x < p}\). Equivalently: \(-\delta < \underline{x - p < 0}\). Now, since \(\large \frac{f(x) - f(p)}{x - p} \normalsize > 0\) and \(x - p < 0\) we conclude that the product between these numbers is negative. That is:

\[\begin{align} f(x) - f(p) &< 0 \\ \boldsymbol{f(x)} &\boldsymbol{< f(p)} \end{align}\]

 

Now consider \(\boldsymbol{p < x < p + \delta}\). Equivalently: \(\underline{0 < x - p} < \delta\). In this case the product between \(\large \frac{f(x) - f(p)}{x - p}\) and \(x - p\) is positive. That is:

\[\begin{align} f(x) - f(p) &> 0 \\ \boldsymbol{f(x)} &\boldsymbol{> f(p)} \end{align}\]

 

For the next two propositions we need to assume \(f'(p) < 0\). Again, since (1) holds true, it particularly holds true for \(\epsilon = -f'(p)\). Then

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