Suppose $f : (a, b) \to \R$ and $x \in (a, b)$. Show that $f$ is differentiable at $x$ if and only if there is a function $E : (a, b) \to \R$ continuous at $x$ and a constant $C \in \R$ so that $E(x) = 0$ and for all $t \in (a, b)$, \[f(t) = f(x) + C(t-x) + E(t)(t-x).\] Moreover, show that in this case $C = f'(x)$.
Give an example of continuous functions $f, g : \R \to \R$ so that:
Let $S, C : \R \to [-1, 1]$ be differentiable functions with the following properties:
Let \begin{align*} f : \R &\longrightarrow \R \\ t &\longmapsto \begin{cases} 0 & \text{ if } t = 0 \\ t^2S\paren{\frac1t} & \text{ if } t \neq 0.\end{cases} \end{align*}
Suppose that $f : (a, b) \to \R$ is differentiable with $f'(x) \gt 0$ for all $x \in (a, b)$.
Suppose that $f : \R \to \R$ is differentiable, and $f'(0) \gt 0$.
Here are some further problems to think about out of interest. You do not need to attempt them, nor should you submit them with the assignment.
Suppose that $f, g : \R \to \R$ are such that $f(0) = g(0)$ and $f'(x) \lt g'(x)$ for all $x \in \R$. Prove that $\abs{f(x)} \leq \abs{g(x)}$ for all $x \in \R$, with equality only when $x = 0$.
Suppose $f : \R \to \R$. Define a function \begin{align*} F : \R^2 \setminus \set{(x, x) \mid x \in \R} &\longrightarrow \R \\ (x, y) &\longmapsto \frac{f(x) - f(y)}{x-y}. \end{align*} Under what conditions does $F$ extend continuously to $\R^2$, in the sense of Assignment 8 Question 4?
(While this certainly seems to have something to do with differentiability, it is a little subtle. Notice, for example, that taking a derivative corresponds to holding one variable fixed and letting the other tend to it, or approaching the missing diagonal horizontally or vertically only; meanwhile continuity requires everything in an open ball to be close to the correct value, not just along the two axes. (It just now occurs to me how "axis" and "axe" both have the plural "axes".))