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Due: November 22nd, 2019
Math 104 Assignment 10
A criterion for differentiability
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 $C = f'(x)$.
Some counterexamples of converses
Give an example of continuous functions $f, g : \R \to \R$ so that:
-
$f+g$ is differentiable but $f$ is not.
-
$fg$ is differentiable, $f$ is not, and $g(x) \gt 0$ for all $x \in \R$.
-
$f\circ g$ and $g$ are differentiable but $f$ is not.
-
$f\circ g$ and $f$ are differentiable but $g$ is not.
More derivatives
Recall that $\sin$ and $\cos$ are differntiable functions $\R\to[-1,1]$ with the properties that $\sin'(x) = \cos(x)$, $\cos'(x) = -\sin(x)$, $\sin(x) \gt 0$ for $x \in (0, \pi)$, $\cos(x) = \sin(x + \frac\pi2)$ (from which the other translations follow, e.g. $\sin(x) = \sin(x+2\pi)$, $\cos(x) = \cos(x+2\pi)$, ...), $\sin(0) = 0$, and $\cos(0) = 1$.
(In fact, $\sin$ is the only differentiable function on $\R$ with $\sin'' = -\sin$, $\sin(0) = 0$, and $\sin'(0) = 1$, but showing this is difficult, and defining $\sin$ as the unique function with these properties makes it rather difficult to show basic properties such as periodicity.)
Let
\begin{align*}
f : \R &\longrightarrow \R \\
t &\longmapsto \begin{cases} 0 & \text{ if } t = 0 \\ t^2\sin\paren{\frac1t} & \text{ if } t \neq 0.\end{cases}
\end{align*}
- Show that $f$ is differentiable on $\R$, and find its derivative.
- Show that $f'$ is discontinuous at $0$.
Inverses
Suppose that $f : (a, b) \to \R$ is differentiable with $f'(x) \gt 0$ for all $x \in (a, b)$.
- Prove that $f$ is strictly monotonically increasing, i.e., if $a \lt x \lt y \lt b$ then $f(x) \lt f(y)$.
- Note that $f$ is one-to-one, and so has an inverse function $g : f((a, b)) \to (a, b)$.
Show that $g$ is continuous.
- Show that for all $x \in (a, b)$,
\[g'(f(x)) = \frac1{f'(x)}.\]
Positive derivative at a point
Show that there is a differentiable function $f : \R \to \R$ with $f'(0) \gt 0$, but so that there is no $\delta\gt0$ with $f$ monotonically increasing on $(-\delta, \delta)$.
(I.e., there is no $\delta \gt 0$ so that for all $x, y\in\R$ with $-\delta \lt x \lt y \lt \delta$ we have $f(x) \leq f(y)$.)
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.
Differentiability and monotony
Does there exist a differentiable function $f : \R\to\R$ which is not monotonic on any open interval?
(This is not an easy question to answer.)