\chapter{The Derivative} \label{der:chapter}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\section{The derivative}
\label{sec:der}

%mbxINTROSUBSECTION

\sectionnotes{1 lecture}

The idea of a derivative is the following.
If the graph of a function looks locally like a straight line,
then we can talk about the slope of this line.
The slope tells us the rate at which 
the value of the function is changing at that particular point.
Of course, we are leaving out any function that has corners or
discontinuities.  
Let us be precise.

\subsection{Definition and basic properties}

\begin{defn}
Let $I$ be an interval, let
$f \colon I \to \R$ be a function, and let $c \in I$.  If 
the limit
\begin{equation*}
L \coloneqq \lim_{x \to c} \frac{f(x)-f(c)}{x-c} 
\end{equation*}
exists, then we say $f$ is
\emph{\myindex{differentiable}}\index{function!differentiable} at
$c$, we call $L$ the \emph{\myindex{derivative}} of $f$ at $c$,
and we write $f'(c) \coloneqq L$.\glsadd{not:derivative}

\medskip

If $f$ is differentiable at all $c \in I$, then we simply say that
$f$ is \emph{differentiable}, and then we obtain a function
$f' \colon I \to \R$.
The derivative is sometimes written as $\frac{df}{dx}$ or
$\frac{d}{dx}\bigl( f(x) \bigr)$.

\medskip

The expression $\frac{f(x)-f(c)}{x-c}$ is called the
\emph{\myindex{difference quotient}}.
\end{defn}

The graphical interpretation of the derivative is  depicted in
\figureref{derivfig}.  The left-hand plot gives the line through
$\bigl(c,f(c)\bigr)$
and $\bigl(x,f(x)\bigr)$ with slope
$\frac{f(x)-f(c)}{x-c}$, that is,
the \emph{\myindex{secant line}}.  When we take the limit as $x$ goes to $c$,
we get the right-hand plot, where we see
that the derivative of the function
at the point $c$ is the slope of the line tangent to the graph of $f$
at the point $\bigl(c,f(c)\bigr)$.

\begin{myfigureht}
\myincludepdft{deriv_derivd}{%
Two diagrams of a plot of a function f.  On the left,
the points c and x are marked on the horizontal axis.
The corresponding points on the graph of f are marked
and through them is drawn a line (a secant line).  It is labeled
with slope equals f of x -f of c the entire thing divided
by quantity x minus c.
On the right, the same function is drawn but only the point c
is marked.  A line that is tangent to the graph of f
through the corresponding point is drawn and labeled
with slope equals f prime of c.}
\caption{Graphical interpretation of the derivative.\label{derivfig}}
\end{myfigureht}

We allow $I$ to be a closed interval and we allow
$c$ to be an endpoint of $I$.  Some calculus books do not allow $c$ to be an
endpoint of an interval, but all the theory still works by allowing it, and
it makes our work easier.

\begin{example}
Let $f(x) \coloneqq x^2$ be defined on the whole real line.  Let $c \in \R$ be arbitrary.  We find that if
$x \neq c$,
\begin{equation*}
\frac{x^2-c^2}{x-c} =
\frac{(x+c)(x-c)}{x-c} =
(x+c) .
\end{equation*}
Therefore,
\begin{equation*}
f'(c) = 
\lim_{x\to c} \frac{x^2-c^2}{x-c} =
\lim_{x\to c} (x+c) = 2c.
\end{equation*}
\end{example}

\begin{example}
Let $f(x) \coloneqq ax + b$ for numbers $a, b \in \R$.
Let $c \in \R$ be arbitrary.
For $x \neq c$,
\begin{equation*}
\frac{f(x)-f(c)}{x-c} =
\frac{a(x-c)}{x-c} = a .
\end{equation*}
Therefore,
\begin{equation*}
f'(c) =
\lim_{x\to c} 
\frac{f(x)-f(c)}{x-c} =
\lim_{x\to c} 
a = a.
\end{equation*}
In fact, every differentiable function \myquote{infinitesimally} behaves like
the affine function $ax + b$.  You can guess many results
and formulas for derivatives if you work them out for affine functions
first.
\end{example}

\begin{example}
The function $f(x) \coloneqq \sqrt{x}$ is differentiable for $x > 0$.  To see this
fact, fix $c > 0$,
and suppose $x \neq c$ and $x > 0$.  Compute
\begin{equation*}
\frac{\sqrt{x}-\sqrt{c}}{x-c}
=
\frac{\sqrt{x}-\sqrt{c}}{(\sqrt{x}-\sqrt{c})(\sqrt{x}+\sqrt{c})}
=
\frac{1}{\sqrt{x}+\sqrt{c}} .
\end{equation*}
Therefore,
\begin{equation*}
f'(c) =
\lim_{x\to c}
\frac{\sqrt{x}-\sqrt{c}}{x-c}
=
\lim_{x\to c}
\frac{1}{\sqrt{x}+\sqrt{c}}
=
\frac{1}{2\sqrt{c}} .
\end{equation*}
\end{example}

\begin{example}
The function $f(x) \coloneqq \sabs{x}$ is not differentiable
at the origin.  When $x > 0$,
\begin{equation*}
\frac{\sabs{x}-\sabs{0}}{x-0} =
\frac{x-0}{x-0} = 1 .
\end{equation*}
When $x < 0$,
\begin{equation*}
\frac{\sabs{x}-\sabs{0}}{x-0} =
\frac{-x-0}{x-0} = -1 .
\end{equation*}
\end{example}

A famous example of Weierstrass shows that there exists a continuous
function that is not differentiable at \emph{any} point.  The construction
of this function is beyond the scope of this chapter.  On the other hand,
a differentiable function is always continuous.

\begin{prop}
If $f \colon I \to \R$ is differentiable at $c \in I$,
then it is continuous at $c$.
\end{prop}

\begin{proof}
We know the limits
\begin{equation*}
\lim_{x\to c}\frac{f(x)-f(c)}{x-c} = f'(c)
\qquad
\text{and}
\qquad
\lim_{x\to c}(x-c) = 0
\end{equation*}
exist.  Furthermore,
\begin{equation*}
f(x)-f(c) = 
\left( \frac{f(x)-f(c)}{x-c} \right) (x-c) .
\end{equation*}
Therefore, the limit of $f(x)-f(c)$ exists and
\begin{equation*}
\lim_{x\to c} \bigl( f(x)-f(c) \bigr) =
\left(\lim_{x\to c} \frac{f(x)-f(c)}{x-c} \right)
\left(\lim_{x\to c} (x-c) \right) =
f'(c) \cdot 0  = 0.
\end{equation*}
Hence, $\lim\limits_{x\to c} f(x) = f(c)$, and $f$ is continuous at $c$.
\end{proof}

An important property of the derivative is linearity.  The
derivative is the approximation of a function by a straight line.
The slope of a line through two points changes linearly when the
$y$-coordinates are changed linearly.  Taking the limit,
it makes sense that the derivative is linear.

\begin{prop}[Linearity]
\index{linearity of the derivative}
Let $I$ be an interval, let
$f \colon I \to \R$ and $g \colon I \to \R$ be differentiable at $c \in I$,
and let $\alpha \in \R$.
\begin{enumerate}[(i)]
\item
Define $h \colon I \to \R$ by $h(x) \coloneqq \alpha f(x)$.  Then
$h$ is differentiable at $c$ and
$h'(c) = \alpha f'(c)$.
\item
Define $h \colon I \to \R$ by $h(x) \coloneqq  f(x) + g(x)$.  Then
$h$ is differentiable at $c$ and
$h'(c) =  f'(c) + g'(c)$.
\end{enumerate}
\end{prop}

\begin{proof}
First, let $h(x) \coloneqq \alpha f(x)$.
For $x \in I$, $x \neq c$,
\begin{equation*}
\frac{h(x)-h(c)}{x-c} =
\frac{\alpha f(x) - \alpha f(c)}{x-c}
=
\alpha \frac{f(x) - f(c)}{x-c} .
\end{equation*}
The limit as $x$ goes to $c$ exists on the right-hand side
by \corref{falg:cor}.  We get
\begin{equation*}
\lim_{x\to c}\frac{h(x)-h(c)}{x-c} =
\alpha \lim_{x\to c} \frac{f(x) - f(c)}{x-c} .
\end{equation*}
Therefore, $h$ is differentiable at $c$,
and the derivative is computed as given.

Next, define $h(x) \coloneqq f(x)+g(x)$.
For $x \in I$, $x \neq c$, we have
\begin{equation*}
\frac{h(x)-h(c)}{x-c} =
\frac{\bigl(f(x) + g(x)\bigr) - \bigl(f(c) + g(c)\bigr)}{x-c}
=
\frac{f(x) - f(c)}{x-c}
+
\frac{g(x) - g(c)}{x-c} .
\end{equation*}
The limit as $x$ goes to $c$ exists on the right-hand side
by \corref{falg:cor}.  We get
\begin{equation*}
\lim_{x\to c}\frac{h(x)-h(c)}{x-c} =
\lim_{x\to c} \frac{f(x) - f(c)}{x-c}
+
\lim_{x\to c}\frac{g(x) - g(c)}{x-c} .
\end{equation*}
Therefore, $h$ is differentiable at $c$,
and the derivative is computed as given.
\end{proof}

It is not true that the derivative of a product of two functions is
the product of the derivatives.
Instead, we get what is called the \emph{product rule}
or the \emph{\myindex{Leibniz rule}}%
\footnote{Named for the German mathematician
\href{https://en.wikipedia.org/wiki/Leibniz}{Gottfried Wilhelm Leibniz}
(1646--1716).}.

\begin{prop}[Product rule]\index{product rule}
Let $I$ be an interval, let
$f \colon I \to \R$ and $g \colon I \to \R$ be 
functions differentiable at $c$.  If $h \colon I \to \R$
is defined by
\begin{equation*}
h(x) \coloneqq f(x) g(x) ,
\end{equation*}
then $h$ is differentiable at $c$ and
\begin{equation*}
h'(c) = f(c) g'(c) + f'(c) g(c) .
\end{equation*}
\end{prop}

The proof of the product rule is left as an exercise.  The key to the proof is 
the identity
$f(x) g(x) - f(c) g(c) =
f(x)\bigl( g(x) - g(c) \bigr)
+ \bigl( f(x) - f(c) \bigr) g(c)$,
which is illustrated in \figureref{figprodrule}.
\begin{myfigureht}
\myincludepdft{figprodrule}{%
A diagram of a rectangular area in the plane, divided
into 3 rectangles.  The horizontal side runs from 0 to
f of x, and the vertical side runs from 0 to g of x.
The largest white area is a rectangle that runs horizontally
from 0 to some f of c, which is slightly less than f of x,
and vertically from 0 to some g of c, which is slightly less than
g of x.  This white area is labeled as f of c times g of c.
On top, there is a thin lightly shaded rectangle running
on the horizontal side from 0 to f of x, and on the vertical
side from g of c to g of x.  It is labeled 
as f of x times the quantity g of x minus g of c.
Finally, there is a thin darkly shaded rectangle
running horizontally from f of c to f of x and vertically
from 0 to g of c.  It is labeled as f of x minus f of c
the entire thing times g of c.}
\caption{The idea of product rule.  The area of the entire rectangle
$f(x)g(x)$ differs from the area of the white rectangle $f(c)g(c)$
by the area of the lightly shaded rectangle
$f(x)\bigl( g(x) - g(c) \bigr)$ plus the darker rectangle
$\bigl( f(x) - f(c) \bigr) g(c)$.
In other words, roughly,  $\Delta (f \cdot g)
= f \cdot \Delta g + \Delta f \cdot g$.\label{figprodrule}}
\end{myfigureht}



\begin{prop}[Quotient rule]\index{quotient rule}
Let $I$ be an interval, let
$f \colon I \to \R$ and $g \colon I \to \R$ be differentiable at $c$
and $g(x) \neq 0$ for all $x \in I$.
If $h \colon I \to \R$
is defined by
\begin{equation*}
h(x) \coloneqq \frac{f(x)}{g(x)},
\end{equation*}
then $h$ is differentiable at $c$ and
\begin{equation*}
h'(c) = \frac{f'(c) g(c) - f(c) g'(c)}{{\bigl(g(c)\bigr)}^2} .
\end{equation*}
\end{prop}

Again, the proof is left as an exercise.

\subsection{Chain rule}

More complicated functions are often obtained by composition,
which is differentiated via the chain rule.  The rule also tells us
how a derivative changes if we change variables.

\begin{prop}[Chain rule]
\index{chain rule}
Let $I_1, I_2$ be intervals, let
$g \colon I_1 \to I_2$ be differentiable at $c \in I_1$,
and
$f \colon I_2 \to \R$ be differentiable at $g(c)$.
If $h \colon I_1 \to \R$
is defined by
\begin{equation*}
h(x) \coloneqq (f \circ g) (x) = f\bigl(g(x)\bigr) ,
\end{equation*}
then $h$ is differentiable at $c$ and
\begin{equation*}
h'(c) = f'\bigl(g(c)\bigr)g'(c) .
\end{equation*}
\end{prop}

\begin{proof}
Let $d \coloneqq g(c)$.  Define
$u \colon I_2 \to \R$ and $v \colon I_1 \to \R$ by
\begin{equation*}
u(y) \coloneqq
\begin{cases}
 \frac{f(y) - f(d)}{y-d}  & \text{if } y \neq d, \\
 f'(d)                    & \text{if } y = d,
\end{cases}
\qquad
v(x) \coloneqq
\begin{cases}
 \frac{g(x) - g(c)}{x-c} & \text{if } x \neq c, \\
 g'(c)                   & \text{if } x = c.
\end{cases}
\end{equation*}
Because $f$ is differentiable at $d = g(c)$, we find that
$u$ is continuous at $d$.  Similarly, $v$ is continuous at $c$.
For any $x$ and $y$,
\begin{equation*}
f(y)-f(d) = u(y) (y-d)
\qquad \text{and} \qquad
g(x)-g(c) = v(x) (x-c) .
\end{equation*}
Plug in to obtain
\begin{equation*}
h(x)-h(c)
=
f\bigl(g(x)\bigr)-f\bigl(g(c)\bigr)
=
u\bigl( g(x) \bigr) \bigl(g(x)-g(c)\bigr)
=
u\bigl( g(x) \bigr) \bigl(v(x) (x-c)\bigr) .
\end{equation*}
Therefore, if $x \neq c$,
\begin{equation} \label{eq:chainruleeq}
\frac{h(x)-h(c)}{x-c}
=
u\bigl( g(x) \bigr) v(x) .
\end{equation}
By continuity of $u$ and $v$ at $d$ and $c$ respectively, we find
$\lim_{y \to d} u(y)
= f'(d) = f'\bigl(g(c)\bigr)$ and
$\lim_{x \to c} v(x) = g'(c)$.
The function $g$ is continuous at $c$, and so $\lim_{x \to c} g(x) = g(c)$.
Hence, the limit of
the right-hand side of \eqref{eq:chainruleeq}
as $x$ goes to $c$
exists and is equal to $f'\bigl(g(c)\bigr) g'(c)$.  Thus $h$
is differentiable at $c$ and $h'(c) = f'\bigl(g(c)\bigr)g'(c)$.
\end{proof}

\subsection{Exercises}

\begin{exercise}
Prove the product rule.
Hint: Prove and use
$f(x) g(x) - f(c) g(c) = f(x)\bigl( g(x) - g(c) \bigr) + \bigl( f(x) -
f(c) \bigr) g(c)$.
\end{exercise}

\begin{exercise}
Prove the quotient rule.  Hint: You can do this directly, but it may be
easier to find the derivative of $\nicefrac{1}{x}$ and then use
the chain rule and the product rule.
\end{exercise}

\begin{exercise} \label{exercise:diffofxn}
For $n \in \Z$,
prove that $x^n$ is differentiable and find the derivative,
unless, of course, $n < 0$ and $x=0$.
Hint: Use the product rule.
\end{exercise}

\begin{exercise}
Prove that a polynomial is differentiable, and find the derivative.
Hint: Use the previous exercise.
\end{exercise}

\begin{exercise}
Define $f \colon \R \to \R$ by
\begin{equation*}
f(x) \coloneqq
\begin{cases}
x^2 & \text{if } x \in \Q,\\
0 & \text{otherwise.}
\end{cases}
\end{equation*}
Prove that $f$ is differentiable at $0$, but discontinuous at all points
except $0$.
\end{exercise}

\begin{exercise}
Assume the inequality $\babs{x-\sin(x)} \leq x^2$.  Prove that $\sin$ is
differentiable at $0$, and find the derivative at $0$.
\end{exercise}

\begin{exercise}
Using the previous exercise, prove that $\sin$ is differentiable at all $x$
and that the derivative is $\cos(x)$.  Hint: Use the sum-to-product
trigonometric identity as we did before.
\end{exercise}

\begin{exercise}
Let $f \colon I \to \R$ be differentiable.  For $n \in \Z$, let $f^n$
be the function defined by $f^n(x) \coloneqq {\bigl( f(x) \bigr)}^n$.  If
$n < 0$, assume $f(x) \neq 0$ for all $x \in I$.  Prove that
$(f^n)'(x) = n {\bigl(f(x) \bigr)}^{n-1} f'(x)$.
\end{exercise}

\begin{exercise}
Suppose $f \colon \R \to \R$ is a differentiable
Lipschitz continuous function.
Prove that $f'$ is a bounded function.
\end{exercise}

\begin{exercise} \label{exercise:inversederformula}
Let $I_1, I_2$ be intervals.
Let $f \colon I_1 \to I_2$ be a bijective function and $g \colon I_2 \to I_1$
be the inverse.  Suppose that both $f$ is differentiable at $c \in I_1$ and
$f'(c) \neq 0$ and $g$ is differentiable at $f(c)$.  Use the chain rule
to find a formula for $g'\bigl(f(c)\bigr)$ (in terms of $f'(c)$).
\end{exercise}

\begin{exercise} \label{exercise:bndmuldiff}
Suppose $f \colon I \to \R$ is bounded, $g \colon I \to
\R$ is differentiable at $c \in I$, and $g(c) = g'(c) = 0$.  Show
that $h(x) \coloneqq f(x) g(x)$ is differentiable at $c$.  Hint: You
cannot apply the product rule.
\end{exercise}

\begin{exercise} \label{exercise:diffsqueeze}
Suppose $f \colon I \to \R$, 
$g \colon I \to \R$, and
$h \colon I \to \R$, are functions.  Suppose $c \in I$ is such that
$f(c) = g(c) = h(c)$, $g$ and $h$ are differentiable at $c$,
and $g'(c) = h'(c)$.  Furthermore, suppose $h(x) \leq f(x) \leq g(x)$ for
all $x \in I$.  Prove $f$ is differentiable at $c$ and $f'(c) = g'(c) =
h'(c)$.
\end{exercise}

\begin{exercise}
\leavevmode
\begin{enumerate}[a)]
\item
Suppose $f \colon (-1,1) \to \R$ is a function such that $f(x) = x h(x)$ for
a bounded function $h \colon (-1,1) \to \R$.
Show that $g(x) \coloneqq {\bigl( f(x) \bigr)}^2$ is
differentiable at the origin and $g'(0) = 0$.
\item
Find an example of a
continuous function $f \colon (-1,1) \to \R$ with $f(0) = 0$, but such
that $g(x) \coloneqq {\bigl( f(x) \bigr)}^2$ is not differentiable at the origin.
\end{enumerate}
\end{exercise}

\begin{exercise}
Suppose $f \colon I \to \R$ is differentiable at $c \in I$.
Prove that there exist numbers $a$ and $b$ with the property that
for every $\epsilon > 0$, there is a $\delta > 0$, such that
$\babs{a+b(x-c) - f(x)} \leq \epsilon \sabs{x-c}$, whenever $x \in I$ and
$\sabs{x-c} < \delta$.
In other words, show that
there exists a function $g \colon I \to \R$
such that $\lim_{x\to c} g(x) = 0$ and
$\babs{a+b(x-c) - f(x)} = g(x) \sabs{x-c}$.
\end{exercise}

\begin{exercise} \label{exercise:simpleLHopital}
Prove the following simple version of \myindex{L'H\^opital's rule}\index{L'Hospital's rule}.
Suppose 
$f \colon (a,b) \to \R$ and $g \colon (a,b) \to \R$ are differentiable
functions
whose derivatives $f'$ and $g'$ are continuous functions.
Suppose that at $c \in (a,b)$, $f(c) = 0$, $g(c)=0$,
$g'(x) \neq 0$ for all $x \in (a,b)$, and
$g(x) \neq 0$ whenever $x \neq c$.   Note
that the limit of $\nicefrac{f'(x)}{g'(x)}$ as $x$ goes to $c$ exists.  Show that
\begin{equation*}
\lim_{x \to c} \frac{f(x)}{g(x)} = 
\lim_{x \to c} \frac{f'(x)}{g'(x)} .
\end{equation*}
\end{exercise}

\begin{exercise}
Suppose $f \colon (a,b) \to \R$ is differentiable at $c \in (a,b)$,
$f(c)=0$, and $f'(c) > 0$.
Prove that there is a $\delta > 0$ such that $f(x) < 0$
whenever $c-\delta < x < c$ and
$f(x) > 0$ 
whenever $c < x < c+\delta$.
\end{exercise}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\sectionnewpage
\section{Mean value theorem}
\label{sec:mvt}

%mbxINTROSUBSECTION

\sectionnotes{2 lectures (some applications may be skipped)}

\subsection{Relative minima and maxima}

We previously talked about absolute maxima and minima.  These are the tallest peaks and
the lowest valleys in the entire mountain range.  What
about peaks of individual mountains and bottoms of individual valleys?
The derivative, being a local concept, is like walking around in a fog; it
cannot tell you if you are on the highest peak, but it can tell you whether you are
at the top of some peak.

\begin{defn}
Let $S \subset \R$ be a set and
let $f \colon S \to \R$ be a function.  The function $f$ is said to have
a \emph{\myindex{relative maximum}}\index{maximum!relative}
at $c \in S$ if there exists a $\delta>0$
such that for all $x \in S$ where $\sabs{x-c} < \delta$,
we have $f(x) \leq f(c)$.
The definition of
\emph{\myindex{relative minimum}}\index{minimum!relative}
is analogous.
\end{defn}

\begin{lemma}\label{relminmax:lemma}
Suppose $f \colon (a,b) \to \R$ is differentiable at $c \in (a,b)$,
and $f$ has
a relative minimum or a relative maximum at $c$.  Then
$f'(c) = 0$.
\end{lemma}

\begin{proof}
Suppose $c$ is a 
relative maximum of $f$.  That is, there is a $\delta > 0$ such
that for every $x \in (a,b)$ where
$\sabs{x-c} < \delta$, we have $f(x)-f(c) \leq 0$.
Consider the difference
quotient.  If $c < x < c+\delta$, then
\begin{equation*}
\frac{f(x)-f(c)}{x-c} \leq 0 ,
\end{equation*}
and if $c-\delta < y < c$, then
\begin{equation*}
\frac{f(y)-f(c)}{y-c} \geq 0 .
\end{equation*}
See \figureref{fig:critpt} for an illustration.
\begin{myfigureht}
\myincludegraphics{critpt}{%
A graph of a function f is shown.
Three points are marked on the horizontal axis, from left
to right, y, c, and x.  The function has a maximum
at the point c.  The points corresponding to x, c, and y are
marked on the graph.
Through the points corresponding
to y and c is a secant line that slopes upwards and is
marked with slope equals f of y minus f of c the whole thing
over quantity y-c is greater than or equal to 0.
Through the points corresponding
to c and x is a secant line that slopes downwards and is
marked with slope equals f of x minus f of c the whole thing
over quantity x-c is less than or equal to 0.}
\caption{Slopes of secants at a relative maximum.\label{fig:critpt}}
\end{myfigureht}

As $a < c < b$, there exist
sequences $\{ x_n \}_{n=1}^\infty$ and
$\{ y_n \}_{n=1}^\infty$ in $(a,b)$
such that $c < x_n < c+\delta$ and
$c-\delta < y_n < c$ for all $n \in \N$, and such that
$\lim_{n\to\infty} x_n = \lim_{n\to\infty} y_n = c$.
Since $f$
is differentiable at $c$,
\begin{equation*}
0 \geq \lim_{n\to\infty} \frac{f(x_n)-f(c)}{x_n-c} 
=
f'(c)
=
\lim_{n\to\infty} \frac{f(y_n)-f(c)}{y_n-c} \geq 0.
\end{equation*}
We are done with a maximum.
For a minimum, consider
the function $-f$.
\end{proof}

For a differentiable function, a point where 
$f'(c) = 0$ is called a \emph{\myindex{critical point}}.  When $f$ is not
differentiable at some points,
it is common to also say that $c$ is a critical point
if $f'(c)$ does not exist.
The theorem says that a relative minimum or maximum at an interior point
of an interval must be a critical point.
As you remember from calculus, one finds minima and maxima of a function
by finding all the critical points together with the endpoints of
the interval and simply checking at which of these points
the function is largest or smallest.

\subsection{Rolle's theorem}

Suppose a function has the same value at both endpoints of an interval.
Intuitively, it ought to attain a minimum or a maximum in the interior of the
interval.
Then at such a minimum or a maximum, the derivative should be zero.
See \figureref{rollefig} for the geometric idea.
This is the content of Rolle's theorem%
\footnote{Named after the French mathematician
\href{https://en.wikipedia.org/wiki/Michel_Rolle}{Michel Rolle}
(1652--1719).}.

\begin{myfigureht}
\myincludegraphics{rollefig}{%
A graph of a function is drawn on the interval from a to b.
The function is zero at both a and b.  The function goes up to
a maximum,
then down to a minimum, then back up.  The maximum of the function occurs
at a point marked c.  A horizontal tangent line is drawn on the
corresponding point on the graph.  Note that it is parallel with
the horizontal axis.}
\caption{Point where the tangent line is horizontal, that is $f'(c) =
0$.\label{rollefig}}
\end{myfigureht}

\begin{thm}[Rolle] \label{thm:rolle}
\index{Rolle's theorem}
Let $f \colon [a,b] \to \R$ be a continuous function
differentiable on $(a,b)$ such that $f(a) = f(b)$.
Then there exists a $c \in (a,b)$ such that $f'(c) = 0$.
\end{thm}

\begin{proof}
As $f$ is continuous on $[a,b]$, it attains an absolute minimum and an
absolute 
maximum in $[a,b]$.  We wish to apply \lemmaref{relminmax:lemma}, and
so we need to find some $c \in (a,b)$ where $f$ attains a minimum or a
maximum.
Write $K \coloneqq f(a) = f(b)$.
If there exists an $x$ such that $f(x) > K$, then the absolute
maximum is larger than $K$ and hence occurs at some $c \in (a,b)$, and
therefore $f'(c) = 0$.  On the other hand, if there exists an $x$
such that $f(x) < K$, then the absolute minimum occurs at some
$c \in (a,b)$, and so $f'(c) = 0$.  If there is no $x$ such that
$f(x) > K$ or
$f(x) < K$, then $f(x) = K$ for all $x$ and then
$f'(x) = 0$ for all $x \in [a,b]$, so any $c \in (a,b)$ works.
\end{proof}

It is absolutely necessary that the derivative exists for all $x
\in (a,b)$.  Consider the function $f(x) \coloneqq \sabs{x}$ on $[-1,1]$.
Clearly $f(-1) = f(1)$, but there is no point $c$ where $f'(c) = 0$.

\subsection{Mean value theorem}

We extend \hyperref[thm:rolle]{Rolle's theorem}
to functions that attain different
values at the endpoints.

\begin{thm}[Mean value theorem] \label{thm:mvt}
\index{mean value theorem}
Let $f \colon [a,b] \to \R$ be a continuous function
differentiable on $(a,b)$.  Then there exists a point $c \in (a,b)$
such that
\begin{equation*}
f(b)-f(a) = f'(c)(b-a) .
\end{equation*}
\end{thm}

For a geometric interpretation of the mean value theorem, see
\figureref{mvtfig}.  The idea is that the value $\frac{f(b)-f(a)}{b-a}$
is the slope of the line between the points $\bigl(a,f(a)\bigr)$
and $\bigl(b,f(b)\bigr)$.
Then $c$ is the point such that $f'(c) = \frac{f(b)-f(a)}{b-a}$, that 
is, the tangent line at the point $\bigl(c,f(c)\bigr)$ has the same slope as the
line between $\bigl(a,f(a)\bigr)$ and $\bigl(b,f(b)\bigr)$.
The name comes from the fact that the slope of the secant line
is the mean value of the derivative, so the average derivative is achieved
in the interior of the interval.

The theorem follows from \hyperref[thm:rolle]{Rolle's theorem}
by subtracting from $f$ the affine linear function
with the same values at $a$ and $b$ as $f$.
The graph of this affine linear function is the 
secant line---the straight line through
$\bigl(a,f(a)\bigr)$ and $\bigl(b,f(b)\bigr)$---and
its derivative is
$\frac{f(b)-f(a)}{b-a}$.
Then we are looking for a point where this
difference has zero derivative.

\begin{myfigureht}
\myincludegraphics{mvtfig}{%
A graph of a function is drawn on the interval from a to b.
The function looks similar as before but it is no longer
zero at the endpoints, the left endpoint is negative
and the right is positive.  A straight line through the endpoints
of the curve is drawn, that is, it is a line through the point
(a, f of a) and (b, f of b).  At the point corresponding to 
c on the horizontal axis the tangent line to the function is
parallel to the line through the endpoints.}
\caption{Graphical interpretation of the mean value theorem.\label{mvtfig}}
\end{myfigureht}


\begin{proof}
Define the
function $g \colon [a,b] \to \R$ by
\begin{equation*}
g(x) \coloneqq
f(x)-
\left( f(b)+\frac{f(b)-f(a)}{b-a}(x-b) \right)
=
f(x)-
f(b)-\frac{f(b)-f(a)}{b-a}(x-b) .
\end{equation*}
The function $g$ is differentiable on $(a,b)$,
continuous on $[a,b]$, such that $g(a) = 0$ and $g(b) = 0$.  Thus there exists
a
$c \in (a,b)$ such that $g'(c) = 0$, that is,
\begin{equation*}
0 = g'(c) = f'(c)-\frac{f(b)-f(a)}{b-a} .
\end{equation*}
In other words,
$f(b)-f(a) = f'(c)(b-a)$.
\end{proof}

The proof generalizes.  By considering
$g(x) \coloneqq
\bigl(f(x)-f(b)\bigr)
\bigl(\varphi(b)-\varphi(a)\bigr)
-
\bigl(f(b)-f(a)\bigr)
\bigl(\varphi(x)-\varphi(b)\bigr)$,
one can prove the following version.  We leave the proof as an exercise.

\begin{thm}[Cauchy's mean value theorem] \label{thm:cauchymvt}
\index{Cauchy's mean value theorem}
Let $f \colon [a,b] \to \R$ and $\varphi \colon [a,b] \to \R$ be continuous
functions
differentiable on $(a,b)$.  Then there exists a point $c \in (a,b)$
such that
\begin{equation*}
\bigl(f(b)-f(a)\bigr)\varphi'(c) = f'(c)\bigl(\varphi(b)-\varphi(a)\bigr) .
\end{equation*}
\end{thm}

The mean value theorem has the distinction of being one of the few theorems
cited
in court.  That is, when police measure the speed of cars by aircraft, or
via cameras reading license plates, they 
measure the time the car takes to go between two points.
The mean value theorem then
says that the car must have somewhere attained the speed you get by dividing the
difference in distance by the difference in time.



\subsection{Applications}

Let us look at a few applications of the mean value theorem.
The applications show the typical use of the theorem, which is
to get rid of a limit by finding the right sort of points where the
derivative is not just close to some difference quotient, but actually
equal to it.
First, we solve our very first differential equation.

\begin{prop} \label{prop:derzeroconst}
Let $I$ be an interval and
let $f \colon I \to \R$ be a differentiable function such that $f'(x) = 0$
for all $x \in I$.
Then $f$ is constant.
\end{prop}

\begin{proof}
Take arbitrary $x,y \in I$ with $x < y$.
As $I$ is an interval, $[x,y] \subset I$.
Then $f$ restricted to $[x,y]$ satisfies the hypotheses
of the \hyperref[thm:mvt]{mean value theorem}.
Therefore, there is a $c \in (x,y)$ such that
\begin{equation*}
f(y)-f(x) = f'(c)(y-x).
\end{equation*}
As $f'(c) = 0$, we have $f(y) = f(x)$.  Hence,
the function is constant.
\end{proof}

Now that we know what it means for the function to stay constant, we look
at increasing and decreasing functions.
We say $f \colon I \to \R$ is \emph{\myindex{increasing}}
(resp.\  \emph{\myindex{strictly increasing}}) if
$x < y$ implies $f(x) \leq f(y)$ (resp.\ $f(x) < f(y)$).
We define
\emph{\myindex{decreasing}} and
\emph{\myindex{strictly decreasing}} in the same way by switching the
inequalities for $f$.

\begin{prop} \label{incdecdiffprop}
Let $I$ be an interval and
let $f \colon I \to \R$ be a differentiable function.
%\begin{enumerate}[(i),itemsep=0.5\itemsep,parsep=0.5\parsep,topsep=0.5\topsep,partopsep=0.5\partopsep]
\begin{enumerate}[(i)]
\item $f$ is increasing if and only if $f'(x) \geq 0$ for all $x \in I$.
\item $f$ is decreasing if and only if $f'(x) \leq 0$ for all $x \in I$.
\end{enumerate}
\end{prop}

\begin{proof}
Let us prove the first item.  Suppose $f$ is increasing.
For all $x,c \in I$ with $x \neq c$,
\begin{equation*}
\frac{f(x)-f(c)}{x-c} \geq 0 .
\end{equation*}
Taking a limit as $x$ goes to $c$, we see that $f'(c) \geq 0$.

For the other direction, suppose $f'(x) \geq 0$ for all $x \in I$.
Take any $x, y \in I$ where $x < y$, and note that $[x,y] \subset I$.
By the \hyperref[thm:mvt]{mean value theorem}, there is some $c \in (x,y)$ such that
\begin{equation*}
f(y)-f(x) = f'(c)(y-x) .
\end{equation*}
As $f'(c) \geq 0$ and $y-x > 0$, we have $f(y) - f(x) \geq 0$ or $f(x) \leq
f(y)$, and so
$f$ is increasing.

We leave the second item, decreasing $f$, to the reader as an exercise.
\end{proof}

A similar but weaker statement is true for strictly increasing and
decreasing functions.

\begin{prop} \label{incdecdiffstrictprop}
Let $I$ be an interval and
let $f \colon I \to \R$ be a differentiable function.
\begin{enumerate}[(i)]
\item
\label{incdecdiffstrictprop:i}
If $f'(x) > 0$ for all $x \in I$, then
$f$ is strictly increasing.
\item
\label{incdecdiffstrictprop:ii}
If $f'(x) < 0$ for all $x \in I$,
then $f$ is strictly decreasing.
\end{enumerate}
\end{prop}

The proof of
\ref{incdecdiffstrictprop:i}
is left as an exercise.
Then \ref{incdecdiffstrictprop:ii}
follows from 
\ref{incdecdiffstrictprop:i} by considering $-f$
instead.
The converse of this proposition is not true.  The function
$f(x) \coloneqq x^3$ is strictly increasing, but $f'(0) = 0$.

\medskip

Another application of the \hyperref[thm:mvt]{mean value theorem} is the following result about
the location of extrema, sometimes called the
\emph{\myindex{first derivative test}}.
The result is stated for an absolute minimum and maximum.
To apply it to find relative minima
and maxima, restrict $f$ to an interval $(c-\delta,c+\delta)$.

\begin{prop} \label{firstderminmaxtest}
Let $f \colon (a,b) \to \R$ be continuous.  Let $c \in (a,b)$
and suppose
$f$ is differentiable on $(a,c)$ and $(c,b)$.
\begin{enumerate}[(i)]
\item If $f'(x) \leq 0$ whenever $x \in (a,c)$ and
 $f'(x) \geq 0$ whenever $x \in (c,b)$, then $f$ has an absolute minimum 
at $c$.
\item If $f'(x) \geq 0$ whenever $x \in (a,c)$ and
 $f'(x) \leq 0$ whenever $x \in (c,b)$, then $f$ has an absolute maximum
at $c$.
\end{enumerate}
\end{prop}

\begin{proof}
We prove the first item and leave the second to the reader.
Take $x \in (a,c)$
and a sequence $\{ y_n\}_{n=1}^\infty$ such that $x < y_n < c$ for all $n$
and $\lim_{n\to\infty} y_n = c$.
By the preceding proposition,
$f$ is decreasing on $(a,c)$ so $f(x) \geq f(y_n)$ for all $n$.
As $f$ is
continuous at $c$, we take the limit to get
$f(x) \geq f(c)$.

Similarly, take $x \in (c,b)$
and $\{ y_n\}_{n=1}^\infty$ a sequence such that $c < y_n < x$ and
$\lim_{n\to\infty} y_n = c$.
The function is increasing on $(c,b)$ so $f(x) \geq f(y_n)$ for all $n$.
By continuity of $f$, we get
$f(x) \geq f(c)$.  Thus $f(x) \geq f(c)$ for all
$x \in (a,b)$.
\end{proof}

The converse of the proposition does not hold.  See
\exampleref{baddifffunc:example} below.

\medskip

Another often used application of the mean value theorem you have possibly
seen in calculus is the following result on differentiability at the
end points of an interval.  The proof is
\exerciseref{exercise:endpointderivative}.

\begin{prop} \label{prop:endpointderivative}
\leavevmode
\begin{enumerate}[(i)]
\item
Suppose $f \colon [a,b) \to \R$ is continuous, differentiable in $(a,b)$,
and $\lim_{x \to a} f'(x) = L$.  Then $f$ is differentiable at $a$ and
$f'(a) = L$.
\item
Suppose $f \colon (a,b] \to \R$ is continuous, differentiable in $(a,b)$,
and $\lim_{x \to b} f'(x) = L$.  Then $f$ is differentiable at $b$ and
$f'(b) = L$.
\end{enumerate}
\end{prop}

In fact, using the extension result \propref{context:prop}, you do not need to assume
that $f$ is defined at the end point.  See \exerciseref{exercise:extendboundedder}.

\subsection{Continuity of derivatives and the intermediate value theorem}

Derivatives of functions satisfy an
intermediate value property.

\begin{thm}[Darboux] \label{thm:darboux} \index{Darboux's theorem}
Let $f \colon [a,b] \to \R$ be differentiable.  Suppose $y \in \R$ is such
that $f'(a) < y < f'(b)$ or
$f'(a) > y > f'(b)$.  Then there exists a $c \in (a,b)$ such that $f'(c) =
y$.
\end{thm}

The proof follows by letting $g$ be the difference of $f$
and a linear function with derivative $y$.
The new function $g$ reduces the problem
to the case $y=0$, where $g'(a) > 0 > g'(b)$.  That is, $g$ is increasing at $a$ and
decreasing at $b$, so it must attain a maximum inside $(a,b)$,
where the derivative is zero.  See \figureref{darbouxthmfig}.

\begin{myfigureht}
\myincludegraphics{darbouxthmfig}{%
A graph of a function with three points, a, c, b, in this order,
marked on the horizontal axis.  Tangent lines are drawn at
the corresponding points on the graph.  From left to right,
the first line slopes up and is labeled g prime of a is greater than 0.
The second line is horizontal and is labeled g prime of c equals 0.
The third line slopes down and is labeled g prime of b is less than 0.}
\caption{Idea of the proof of Darboux theorem.\label{darbouxthmfig}}
\end{myfigureht}

\begin{proof}
Suppose 
$f'(a) < y < f'(b)$.
Define
\begin{equation*}
g(x) \coloneqq yx - f(x) .
\end{equation*}
The function $g$ is continuous on $[a,b]$, and so $g$ attains a maximum at some $c \in
[a,b]$.

The function $g$ is also differentiable on $[a,b]$.
Compute $g'(x) = y-f'(x)$.  Thus $g'(a) > 0$.  As the derivative is
the limit of difference quotients and is positive, there must be some
difference quotient that is positive.  That is, there must exist
an $x > a$ such that
\begin{equation*}
\frac{g(x)-g(a)}{x-a} > 0 ,
\end{equation*}
or $g(x) > g(a)$.  Thus $g$
cannot possibly have a maximum at $a$.  Similarly, as $g'(b) < 0$,
we find an $x < b$ (a different $x$) such that
$\frac{g(x)-g(b)}{x-b} < 0$ or that $g(x) > g(b)$, thus
$g$ cannot possibly have a maximum at $b$.
Therefore, $c \in (a,b)$,
and \lemmaref{relminmax:lemma} applies: As $g$ attains a maximum
at $c$, we find $g'(c) = 0$
and so $f'(c) = y$.

Similarly, if $f'(a) > y > f'(b)$, consider $g(x) \coloneqq f(x)- yx$.
\end{proof}

We have seen already that
there exist discontinuous functions that have the
intermediate value property.  While it is hard to imagine at first, there
also
exist functions that are differentiable everywhere and the derivative is not
continuous.

\begin{example} \label{baddifffunc:example}
Let $f \colon \R \to \R$ be the function defined by
\begin{equation*}
f(x) \coloneqq
\begin{cases}
{\bigl( x \sin(\nicefrac{1}{x}) \bigr)}^2 & \text{if } x \neq 0, \\
0 & \text{if } x = 0.
\end{cases}
\end{equation*}
We claim that $f$ is differentiable everywhere, but
$f' \colon \R \to \R$ is not continuous at
the origin.  Furthermore, $f$ has a minimum at $0$, but the derivative
changes sign infinitely often near the origin.
See \figureref{fig:nonc1diff}.
\begin{myfigureht}
\myincludepdft{nonc1diff_full}{%
On the left is a graph of a function that oscillates between
the x-axis and the function x squared, whose graph is given
as a dashed line.
On the right is a function that oscillates faster and faster
as it gets close to the origin, looking somewhat like the graph
of sine of 1 over x, however the entire graph appears sheared a bit
so that it goes upwards as well.}
\caption{A function with a discontinuous derivative. The function $f$ is on the left
and $f'$ is on the right.  Notice that $f(x) \leq x^2$ on the left graph.\label{fig:nonc1diff}}
\end{myfigureht}

Proof: It is immediate from the definition that $f$ has an absolute
minimum at $0$; we know $f(x) \geq 0$ for all $x$ and $f(0) = 0$.

For $x \neq 0$,
$f$ is differentiable
and
the derivative 
is $2 \sin (\nicefrac{1}{x}) \bigl( x \sin (\nicefrac{1}{x}) -
\cos(\nicefrac{1}{x}) \bigr)$.
As an exercise, show that for $x_n = \frac{4}{(8n+1)\pi}$,
we have
$\lim_{n\to\infty} f'(x_n) = -1$, and for
$y_n = \frac{4}{(8n+3)\pi}$, we have
$\lim_{n\to\infty} f'(y_n) = 1$.  So $f'$
cannot be continuous at $0$
no matter what $f'(0)$ is.

Let us show that $f$ is differentiable at $0$ and $f'(0)=0$.
For $x \neq 0$,
\begin{equation*}
\abs{\frac{f(x)-f(0)}{x-0} - 0}
=
\abs{\frac{x^2 \sin^2(\nicefrac{1}{x})}{x}}
=
\babs{x \sin^2(\nicefrac{1}{x})}
\leq
\sabs{x} .
\end{equation*}
And, of course, as $x$ tends to zero, $\sabs{x}$ tends to zero, and hence
$\abs{\frac{f(x)-f(0)}{x-0} - 0}$ goes to zero.  Therefore, $f$
is differentiable at 0 and the derivative at 0 is 0.
A key point in the calculation above
is that $\babs{f(x)} \leq x^2$,
see also Exercises \ref{exercise:bndmuldiff} and
\ref{exercise:diffsqueeze}.
\end{example}

It is sometimes useful to assume the derivative of a differentiable
function is continuous.  If $f \colon I \to \R$ is differentiable and
the derivative $f'$ is continuous on $I$, then we say $f$ is
\emph{\myindex{continuously
differentiable}}\index{differentiable!continuously}.  It is common to
write $C^1(I)$ for the set of continuously differentiable functions defined
on $I$.

\subsection{Exercises}

\begin{exercise}
Finish the proof of \propref{incdecdiffprop}.
\end{exercise}

\begin{exercise}
Finish the proof of \propref{firstderminmaxtest}.
\end{exercise}

\begin{exercise} \label{exercise:boundeddermeanslip}
Suppose $f \colon \R \to \R$ is a differentiable
function such that $f'$ is a bounded function.  Prove
that $f$ is a Lipschitz continuous function.
\end{exercise}

\begin{exercise}
\pagebreak[1]
Suppose $f \colon [a,b] \to \R$ is differentiable and $c \in [a,b]$.
Show there exists a sequence $\{ x_n \}_{n=1}^\infty$ converging to $c$, $x_n
\neq c$ for all $n$, such that
\begin{equation*}
f'(c) = \lim_{n\to \infty} f'(x_n).
\avoidbreak
\end{equation*}
Do note this does \emph{not} imply that $f'$ is continuous (why?).
\end{exercise}

\begin{exercise}
Suppose $f \colon \R \to \R$ is a function such that
$\babs{f(x)-f(y)} \leq \sabs{x-y}^2$ for all $x$ and $y$.  Show that
$f(x) = C$ for some constant $C$.  Hint: Show that $f$ is differentiable
at all points and compute the derivative.
\end{exercise}

\begin{exercise} \label{exercise:posderincr}
Finish the proof of \propref{incdecdiffstrictprop}.  That is,
suppose $I$ is an interval and
$f \colon I \to \R$ is a differentiable function such that
$f'(x) > 0$ for all $x \in I$.
Show that $f$ is strictly increasing.
\end{exercise}

\begin{exercise}
Suppose $f \colon (a,b) \to \R$ is a differentiable function
such that
$f'(x) \neq 0$ for all $x \in (a,b)$.  Suppose there
exists
a point $c \in (a,b)$ such that $f'(c) > 0$.
Prove $f'(x) > 0$ for all $x \in (a,b)$.
\end{exercise}

\begin{exercise} \label{exercise:samediffconst}
Suppose $f \colon (a,b) \to \R$ and $g \colon (a,b) \to \R$ are
differentiable functions such that $f'(x) = g'(x)$ for all $x \in (a,b)$,
then show that there exists a constant $C$ such that $f(x) = g(x) + C$.
\end{exercise}

\begin{exercise}
Prove the following version of \myindex{L'H\^opital's rule}\index{L'Hospital's rule}.
Suppose 
$f \colon (a,b) \to \R$ and $g \colon (a,b) \to \R$ are differentiable
functions and $c \in (a,b)$.  Suppose that $f(c) = 0$, $g(c)=0$,
$g'(x) \neq 0$ when $x \neq c$, and
that the limit of $\nicefrac{f'(x)}{g'(x)}$ as $x$ goes to $c$ exists.  Show that
\begin{equation*}
\lim_{x \to c} \frac{f(x)}{g(x)} = 
\lim_{x \to c} \frac{f'(x)}{g'(x)} .
\end{equation*}
Compare to \exerciseref{exercise:simpleLHopital}.
Note: Before you do anything else, prove that $g(x) \neq 0$ when $x \neq c$.
\end{exercise}

\begin{exercise}
Let $f \colon (a,b) \to \R$ be an unbounded differentiable function.  Show
$f' \colon (a,b) \to \R$ is unbounded.
Remark: It is important that $(a,b)$ is a bounded interval.
\end{exercise}

\begin{exercise}
Prove the theorem Rolle actually proved in 1691:
\emph{If $f$ is a polynomial,
$f'(a) = f'(b) = 0$ for some $a < b$,
and there is no $c \in (a,b)$ such that $f'(c) = 0$,
then there is at most one root of $f$ in $(a,b)$,
that is at most one $x \in (a,b)$ such that $f(x) = 0$.}
In other words, between any two consecutive roots of $f'$ is at most one
root of $f$.
Hint: Suppose there are two roots and see what happens.
\end{exercise}

\begin{exercise}
Suppose $a,b \in \R$ and $f \colon \R \to \R$ is differentiable,
$f'(x) = a$ for all $x$, and $f(0) = b$.  Find $f$ and prove that 
it is the unique differentiable function with this property.
\end{exercise}

\begin{exercise} \label{exercise:endpointderivative}
\leavevmode
\begin{enumerate}[a)]
\item
Prove \propref{prop:endpointderivative}.
\item 
Suppose $f \colon (a,b) \to \R$ is continuous, and suppose $f$ is differentiable everywhere except at $c \in
(a,b)$ and $\lim_{x \to c} f'(x) = L$.
Prove that $f$ is differentiable at
$c$ and $f'(c) = L$.
\end{enumerate}
\end{exercise}

\begin{exercise} \label{exercise:extendboundedder}
Suppose $f \colon (0,1) \to \R$ is differentiable and $f'$
is bounded.
\begin{enumerate}[a)]
\item
Show that there exists a continuous function $g \colon [0,1) \to \R$
such that $f(x) = g(x)$ for all $x \neq 0$.\\
Hint: \propref{context:prop} and
\exerciseref{exercise:boundeddermeanslip}.
\item
Find an example where the $g$ is not differentiable at $x=0$.
\\
Hint: Consider something based on $\sin(\ln x)$,
and assume you know basic properties of
$\sin$ and $\ln$ from calculus.
\item
Instead of assuming that $f'$ is bounded, assume that $\lim_{x \to 0} f'(x)
= L$.  Prove that not only does $g$ exist but it is differentiable at $0$
and $g'(0) = L$.
\end{enumerate}
\end{exercise}

\begin{exercise}
Prove \thmref{thm:cauchymvt}.
\end{exercise}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\sectionnewpage
\section{Taylor's theorem}
\label{sec:taylor}

%mbxINTROSUBSECTION

\sectionnotes{less than a lecture (optional section)}

\subsection{Derivatives of higher orders}

When $f \colon I \to \R$ is differentiable, we obtain a function
$f' \colon I \to \R$.  The function
$f'$ is called the \emph{\myindex{first derivative}} of $f$.
If $f'$ is differentiable, we denote by
$f'' \colon I \to \R$ the derivative of $f'$.  The function $f''$
is called the \emph{\myindex{second derivative}} of $f$.
\glsadd{not:secondthirdfourthder}
We similarly obtain
$f'''$, $f''''$, and so on.
With a larger number of derivatives
the notation would get out of hand; we denote
by $f^{(n)}$ the
\glsadd{not:nthder}
\emph{$n$th derivative}\index{nth derivative@$n$th derivative} of $f$.
%
When $f$ possesses $n$ derivatives, we say $f$ is
\emph{$n$ times differentiable}\index{n times differentiable@$n$ times
differentiable}\index{differentiable!n times@$n$ times}.

\subsection{Taylor's theorem}

Taylor's theorem%
\footnote{Named for the English mathematician
\href{https://en.wikipedia.org/wiki/Brook_Taylor}{Brook Taylor}
(1685--1731).
It was found by
the Scottish mathematician
\href{https://en.wikipedia.org/wiki/James_Gregory_(mathematician)}{James Gregory}
(1638--1675).  The statement we give
is due to
\href{https://en.wikipedia.org/wiki/Lagrange}{Joseph-Louis Lagrange}
(1736--1813).} (at least the version we give here)
is a generalization of the \hyperref[thm:mvt]{mean value theorem}.
Mean value theorem says that up to a small error $f(x)$ for $x$ near $x_0$ can be
approximated by $f(x_0)$:
\begin{equation*}
f(x) = f(x_0) + f'(c)(x-x_0),
\end{equation*}
where the \myquote{error} $f'(c)(x-x_0)$
is measured in terms of the first derivative
at some point $c$ between $x$ and $x_0$.
Taylor's theorem generalizes this result to higher derivatives.
It tells us that up to a small error, any $n$
times differentiable function can be approximated at a point $x_0$
by a polynomial of degree $n$.  The
error of this approximation goes to zero faster than ${(x-x_0)}^{n}$ as
$x$ goes to $x_0$.
To see why this is a good approximation, notice that for a big $n$, 
${(x-x_0)}^n$ is very small in a small interval around $x_0$.
We will require one more (so $n+1$) derivative to write the error
as we do in the mean value theorem.

\begin{defn}
For an $n$ times differentiable function $f$ defined near a point $x_0 \in \R$, define the
$n$th order \emph{\myindex{Taylor polynomial}}%
\index{nth order Taylor polynomial@$n$th order Taylor polynomial}
for $f$ at $x_0$ as
\begin{equation*}
\begin{split}
P_n^{x_0}(x)
& \coloneqq
\sum_{k=0}^n
\frac{f^{(k)}(x_0)}{k!}{(x-x_0)}^k
\\
& =
f(x_0)
+ f'(x_0)(x-x_0)
+ \frac{f''(x_0)}{2}{(x-x_0)}^2
+ \frac{f^{(3)}(x_0)}{6}{(x-x_0)}^3
\\
& \qquad
+ \cdots
+ \frac{f^{(n)}(x_0)}{n!}{(x-x_0)}^n .
\end{split}
\end{equation*}
\end{defn}

See \figureref{fig:taylorsin} for
the odd-degree Taylor polynomials for the sine function at $x_0=0$.
The even-degree terms are all zero, as even derivatives 
of sine are again sines, which are zero at the origin.
\begin{myfigureht}
\myincludegraphics{taylorsin}{%
The graph of the sine function is given.  Also several
approximations are given.  First a straight line that is
tangent at the origin is labeled as y equals P sub 1 super 0 of x.
Second a graph of a cubic that is tangent to the graph of sine
at the origin and somewhat approximates it nearby is labeled
as y equals P sub 3 super 0 of x.
Similarly with degree 5 and degree 7 approximations.  The higher the
degree the further away from the origin is the graph reasonably close 
to the sine function, for example the degree 7 approximation seems
relatively close between two points where the sine
is zero around the origin, that is, plus and minus pi.}
\caption{The odd degree Taylor polynomials for the sine
function.\label{fig:taylorsin}}
\end{myfigureht}

The statement of Taylor's theorem that we give includes the
\hyperref[thm:mvt]{mean value theorem}, which can be thought of as
Taylor's theorem for $n=0$.

\begin{thm}[Taylor]\index{Taylor's theorem} \label{thm:taylor}
Suppose $f \colon [a,b] \to \R$ is a function with $n$ continuous
derivatives on $[a,b]$ and such that $f^{(n+1)}$ exists on $(a,b)$.
Given distinct points $x_0$ and $x$ in $[a,b]$,
there is a point $c$ between $x_0$
and $x$ such that
\begin{equation*}
f(x)=P_{n}^{x_0}(x)+\frac{f^{(n+1)}(c)}{(n+1)!}{(x-x_0)}^{n+1} .
\end{equation*}
\end{thm}

The term $R_n^{x_0}(x)\coloneqq
f(x)-P_n^{x_0}(x) =
\frac{f^{(n+1)}(c)}{(n+1)!}{(x-x_0)}^{n+1}$ is called the
\emph{remainder term}\index{remainder term in Taylor's formula}.  This
form is called the
\emph{\myindex{Lagrange form}} of the remainder.  There are other ways
to write the remainder term, but we skip those.  Note that $c$ depends on
both $x$ and $x_0$.

\begin{proof}
Let $M_{x,x_0}$ be the number (depending on $x$ and $x_0$) solving the equation
\begin{equation*}
f(x)=P_{n}^{x_0}(x)+M_{x,x_0}{(x-x_0)}^{n+1} .
\end{equation*}
Define a function $g(s)$ by
\begin{equation*}
g(s) \coloneqq f(s)-P_n^{x_0}(s)-M_{x,x_0}{(s-x_0)}^{n+1} .
\end{equation*}
We compute
the $k$th derivative at $x_0$ of the Taylor polynomial
${(P_n^{x_0})}^{(k)}(x_0) = f^{(k)}(x_0)$ for
$k=0,1,2,\ldots,n$ (the zeroth derivative of a function is the function
itself).  Therefore,
\begin{equation*}
g(x_0) = g'(x_0) = g''(x_0) = \cdots = g^{(n)}(x_0) = 0 .
\end{equation*}
In particular, $g(x_0) = 0$.
We also have $g(x) = 0$.
By the
\hyperref[thm:mvt]{mean value theorem},
there exists an $x_1$ between $x_0$ and $x$ such that $g'(x_1) = 0$.
Applying the \hyperref[thm:mvt]{mean value theorem}
to $g'$, we obtain that there exists
$x_2$ between $x_0$ and $x_1$ (and therefore between $x_0$ and $x$)
such that $g''(x_2) = 0$.  We repeat the
argument $n+1$ times to obtain a number $x_{n+1}$ between $x_0$ and $x_n$
(and therefore between $x_0$ and $x$) such that $g^{(n+1)}(x_{n+1}) = 0$.

Let $c \coloneqq x_{n+1}$.
We compute the $(n+1)$th derivative of $g$ to find
\begin{equation*}
g^{(n+1)}(s) = f^{(n+1)}(s)-(n+1)!\,M_{x,x_0} .
\end{equation*}
Plugging in $c$ for $s$, we obtain $M_{x,x_0} = \frac{f^{(n+1)}(c)}{(n+1)!}$, and
we are done.
\end{proof}

In the proof, we found
${(P_n^{x_0})}^{(k)}(x_0) = f^{(k)}(x_0)$ for $k=0,1,2,\ldots,n$.
Therefore, the Taylor polynomial has the same derivatives as $f$ at $x_0$
up to the $n$th derivative.  That is why the Taylor polynomial is
a good approximation to $f$.
Notice how in \figureref{fig:taylorsin} the Taylor polynomials are
reasonably good approximations to the sine near $x=0$.

We do not necessarily get good approximations
by the Taylor polynomial everywhere.
Consider expanding the function
$f(x) \coloneqq \frac{x}{1-x}$ around 0,
for $x < 1$, we get the graphs in
\figureref{fig:taylorgeom}.  The dotted lines are the first, second, and
third degree approximations.
The dashed line is the 20th degree polynomial.
The approximation does seem to get
better as the degree rises for $x > -1$.
For $x < -1$, it in fact gets visibly worse.
The polynomials
are the partial sums of the geometric series $\sum_{n=1}^\infty x^n$,
and the series only converges on $(-1,1)$.
See the discussion of power series in
\sectionref{sec:moreonseries}.

\begin{myfigureht}
\myincludegraphics{taylorgeom}{%
In bold line is a graph of a function on the interval
from minus 2 to 1, that comes in flat from the right from below
the x-axis, crosses the x-axis at the origin, and then turns sharply
upward leaving the picture before we get to 1.
The graphs of the first three Taylor polynomials are shown in
dotted line.  That is,
a straight line, a quadratic, and a cubic.
They all seem to be 
somewhat reasonable approximation very near the origin
but get worse as we get further.
A dashed line signifying the degree 20 approximation is drawn and
from some point slightly more than minus 1 until where the bold
line leaves the picture on the right, we cannot tell the difference
between it and the bold line.  However, right around minus 1
if we move from right to left, it shoots upwards leaving the picture,
so for x less than minus 1 the approximation would be ludicrously worse
than the small degree approximations (which are not good already).}
\caption{The function $\frac{x}{1-x}$, and the Taylor polynomials
$P_1^0$, $P_2^0$, $P_3^0$ (all dotted), and the polynomial $P_{20}^0$
(dashed).\label{fig:taylorgeom}}
\end{myfigureht}

If $f$ is \emph{\myindex{infinitely
differentiable}}\index{differentiable!infinitely},
that is, if $f$ can be
differentiated any number of times, then 
we define the \emph{\myindex{Taylor series}}:
\begin{equation*}
%T^{x_0}(x)
%\coloneqq
\sum_{k=0}^\infty
\frac{f^{(k)}(x_0)}{k!}{(x-x_0)}^k .
\end{equation*}
There is no guarantee that this series converges for any
$x \neq x_0$.  Even where it does converge, there is no guarantee
that it converges to the function $f$.  Functions $f$
whose Taylor series at every point $x_0$
converges to $f$ in some open interval containing $x_0$
are called
\emph{analytic functions}\index{analytic function}.
Many functions one tends to see in practice are analytic.
See \exerciseref{exercise:nonanalytic}, for an example of a non-analytic
function.

\medskip

The definition of derivative says that
a function is
differentiable if it
is locally approximated by a line.
We mention in passing that there exists a converse to Taylor's
theorem,
which we will neither state nor prove,
saying that if a function is
locally approximated in a certain way by a polynomial of degree $d$, then it
has $d$ derivatives.

\medskip

Taylor's theorem gives a quick proof of a version of
the second derivative test.
By a \emph{\myindex{strict relative minimum}}\index{minimum!strict relative}
of $f$ at $c$,
we mean that there exists a $\delta > 0$ such that $f(x) > f(c)$ for
all $x \in (c-\delta,c+\delta)$ where $x\neq c$.
A \emph{\myindex{strict relative maximum}}\index{maximum!strict relative}
is defined similarly.
Continuity of the second derivative is not needed, but the proof is more
difficult and is left as an exercise.  The proof also generalizes
into the $n$th derivative test, which is also left as
an exercise.

\begin{prop}[Second derivative test]\index{second derivative test}
Suppose $f \colon (a,b) \to \R$ is twice continuously differentiable,
$x_0 \in (a,b)$, $f'(x_0) = 0$ and $f''(x_0) > 0$.  Then $f$ has a strict relative
minimum at $x_0$.
\end{prop}

\begin{proof}
As $f''$ is continuous, there exists a $\delta > 0$
such that $f''(c) > 0$ for all $c \in (x_0-\delta,x_0+\delta)$,
see \exerciseref{exercise:positivecontneigh}.
Take $x \in (x_0-\delta,x_0+\delta)$, $x \neq x_0$.
Taylor's theorem says that for some $c$ between $x_0$ and $x$,
\begin{equation*}
f(x) 
=
f(x_0) + f'(x_0) (x-x_0) +
\frac{f''(c)}{2}{(x-x_0)}^{2} 
=
f(x_0) + \frac{f''(c)}{2}{(x-x_0)}^{2}  .
\end{equation*}
As $f''(c) > 0$ and ${(x-x_0)}^2 > 0$, we have $f(x) > f(x_0)$.
\end{proof}

\subsection{Exercises}

\begin{exercise}
Compute the $n$th Taylor polynomial at $0$ for the exponential function.
\end{exercise}

\begin{exercise}
Suppose $p$ is a polynomial of degree $d$.  Given $x_0 \in \R$,
show that
the $d$th Taylor polynomial for $p$ at $x_0$ is equal to $p$.
\end{exercise}

\begin{exercise}
Let $f(x) \coloneqq \sabs{x}^3$.  Compute $f'(x)$ and $f''(x)$ for all $x$,
but show that $f^{(3)}(0)$ does not exist.
\end{exercise}

\begin{exercise}
Suppose $f \colon \R \to \R$ has $n$ continuous derivatives.  Show
that for every $x_0 \in \R$,
there exist polynomials $P$ and $Q$ of degree $n$ and 
an $\epsilon > 0$ such that $P(x) \leq f(x) \leq Q(x)$ for all $x \in
[x_0,x_0+\epsilon]$  and
$Q(x)-P(x) = \lambda {(x-x_0)}^n$ for some $\lambda \geq 0$.
\end{exercise}

\begin{exercise}
If $f \colon [a,b] \to \R$ has $n+1$ continuous derivatives\footnote{%
The same statement holds if we only require $n$ derivatives,
but it is quite a bit harder to prove.
}
and $x_0 \in [a,b]$,
prove
$\lim\limits_{x\to x_0} \frac{R_n^{x_0}(x)}{{(x-x_0)}^n} = 0$.
\end{exercise}

\begin{exercise}
Suppose $f \colon [a,b] \to \R$ has $n+1$ continuous derivatives
and $x_0 \in (a,b)$.
Prove: $f^{(k)}(x_0) = 0$ for all $k = 0, 1, 2, \ldots, n$
if and only if $\lim\limits_{x\to x_0} \frac{f(x)}{{(x-x_0)}^{n+1}}$ exists.
\end{exercise}

\begin{exercise}
Suppose $a,b,c \in \R$ and $f \colon \R \to \R$ is twice differentiable,
$f''(x) = a$ for all $x$, $f'(0) = b$, and $f(0) = c$.  Find $f$ and prove that 
it is the unique differentiable function with this property.
\end{exercise}

\begin{exercise}[Challenging]
Show that a simple converse to Taylor's theorem does not hold.
Find a function $f \colon \R \to \R$ with no second derivative at $x=0$ such that
$\babs{f(x)} \leq \babs{x^3}$, that is, $f$ goes to zero at 0 faster than $x^2$, and
while $f'(0)$ exists, $f''(0)$ does not.
\end{exercise}

\begin{exercise} \label{exercise:extendboundedder2}
Suppose $f \colon (0,1) \to \R$ is differentiable and $f''$
is bounded.
\begin{enumerate}[a)]
\item
Show that there exists a once differentiable function $g \colon [0,1) \to \R$
such that $f(x) = g(x)$ for all $x \neq 0$.  Hint: 
See
\exerciseref{exercise:extendboundedder}.
\item
Find an example where the $g$ is not twice differentiable at $x=0$.
\end{enumerate}
\end{exercise}

\begin{exercise}
Prove the
\emph{$n$th derivative test}\index{nth derivative test@$n$th derivative test}.
Suppose $n \in \N$,
$x_0 \in (a,b)$, and $f \colon (a,b) \to \R$ is $n$ times continuously
differentiable, with $f^{(k)}(x_0) = 0$ for $k=1,2,\ldots,n-1$, and
$f^{(n)}(x_0)
\neq 0$.
Prove:
\begin{enumerate}[a)]
\item
If $n$ is odd, then $f$ has neither a relative minimum,
nor a maximum at $x_0$.
\item
If $n$ is even, then $f$ has a strict relative minimum at $x_0$ if
$f^{(n)}(x_0) > 0$
and a strict relative maximum at $x_0$ if $f^{(n)}(x_0) < 0$.
\end{enumerate}
\end{exercise}

\begin{exercise}
Prove the more general version of the second derivative test.
Suppose $f \colon (a,b) \to \R$ is differentiable and $x_0 \in (a,b)$
is such that, $f'(x_0) = 0$, $f''(x_0)$ exists, and $f''(x_0) > 0$.
Prove that $f$ has a strict relative
minimum at $x_0$.  Hint: Consider the limit definition of $f''(x_0)$.
\end{exercise}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\sectionnewpage
\section{Inverse function theorem}
\label{sec:ift}

%mbxINTROSUBSECTION

\sectionnotes{less than 1 lecture (optional section, needed for
\sectionref{sec:logandexp}, requires 
\sectionref{sec:monotonefunc})}

\subsection{Inverse function theorem}

We start with a simple example.  Consider the function $f(x) \coloneqq a x$ for a
number $a \neq 0$.  Then $f \colon \R \to \R$ is bijective, and the inverse
is $f^{-1}(y) = \frac{1}{a} y$.  In particular, $f'(x) = a$ and 
$(f^{-1})'(y) = \frac{1}{a}$.  As differentiable functions are
\myquote{infinitesimally like} linear functions, we expect the same
sort of behavior from the inverse of a differentiable function.
The main idea of differentiating inverse functions is the following lemma.

\begin{lemma} \label{lemma:ift}
Let $I,J \subset \R$ be intervals.
If $f \colon I \to J$ is strictly monotone (hence one-to-one),
onto ($f(I) = J$),
differentiable at $x_0 \in I$, and $f'(x_0) \neq 0$,
then the inverse 
$f^{-1}$ is differentiable at $y_0 = f(x_0)$ and
\begin{equation*}
(f^{-1})'(y_0) = \frac{1}{f'\bigl( f^{-1}(y_0) \bigr)} = \frac{1}{f'(x_0)} .
\end{equation*}
If $f$ is continuously differentiable and $f'$ is never zero, then $f^{-1}$
is continuously differentiable.
\end{lemma}

\begin{proof}
By \propref{prop:invcont}, $f$ has a continuous inverse.  For convenience,
call the inverse $g \colon J \to I$.
Let $x_0,y_0$ be as in the statement.  For $x \in I$, write $y \coloneqq f(x)$.
If $x \neq x_0$, and so $y \neq y_0$, we find
\begin{equation*}
\frac{g(y)-g(y_0)}{y-y_0} =
\frac{g\bigl(f(x)\bigr)-g\bigl(f(x_0)\bigr)}{f(x)-f(x_0)} =
\frac{x-x_0}{f(x)-f(x_0)} .
\end{equation*}
See \figureref{inversefig} for the geometric idea.
\begin{myfigureht}
\myincludepdft{inversefigAB}{%
Two diagrams.
On the left, a function f that looks like the square root
is shown in the first quadrant.  On the horizontal axis
there is a point labeled x equals g of y.  A dashed
line goes vertically up from this point until it hits the corresponding
point on the graph of f.  Another dashed line, now horizontal goes left
until it hits the y axis and a point marked f of x equals y.
Another point x sub 0 equals g of y sub 0 is also marked
a bit to the right of the first.  The same setup with the
dashed lines is repeated except instead of x it is x sub 0 and
instead of y it is y sub 0.
A secant line is drawn through the two points on the graph
and it is labeled as having slope
f of x minus f of x sub 0, the whole thing over quantity
x minus x sub 0
and that equals
y minus y sub 0, the whole thing over quantity
g of y minus g of y sub 0.
The right hand diagram is the same thing but the picture
is simply reflected across the x equals y line, so the x-axis
becomes the y-axis and vice-versa, and the graph now looks more
like x squared.  The two expressions of the tangent
line are the reciprocals.}
\caption{Interpretation of the derivative of the inverse
function.\label{inversefig}}
\end{myfigureht}

Let
\begin{equation*}
Q(x) \coloneqq
\begin{cases}
\frac{x-x_0}{f(x)-f(x_0)} & \text{if } x \neq x_0, \\
\frac{1}{f'(x_0)}         & \text{if } x = x_0 \quad \text{(notice that }
f'(x_0) \neq 0 \text{)}.
\end{cases}
\end{equation*}
As $f$ is differentiable at $x_0$, 
\begin{equation*}
\lim_{x \to x_0} Q(x) =
\lim_{x \to x_0} 
\frac{x-x_0}{f(x)-f(x_0)} 
=
\frac{1}{f'(x_0)}
=
Q(x_0) ,
\end{equation*}
that is, $Q$ is continuous at $x_0$.
As $g(y)$ is continuous at $y_0$,
the composition $Q\bigl(g(y)\bigr) = \frac{g(y)-g(y_0)}{y-y_0}$
is continuous at $y_0$ by
\propref{prop:compositioncont}.
Therefore,
\begin{equation*}
\frac{1}{f'\bigl(g(y_0)\bigr)}
= Q\bigl(g(y_0)\bigr)
= \lim_{y \to y_0} Q\bigl(g(y)\bigr)
= \lim_{y \to y_0} \frac{g(y)-g(y_0)}{y-y_0} .
\end{equation*}
So $g$ is differentiable at $y_0$, and $g'(y_0) =
\frac{1}{f'\left(\vphantom{1^1_1}g(y_0)\right)}$.

If $f'$ is continuous and nonzero at all $x \in I$,
then the lemma applies at all $x \in I$.  As $g$ is also
continuous (it is differentiable), the derivative $g'(y) =
\frac{1}{f'\left(\vphantom{1^1_1}g(y)\right)}$ must be continuous.
\end{proof}

What is usually called the inverse function theorem is the following result.

\begin{thm}[Inverse function theorem]\index{inverse function theorem}
Let $f \colon (a,b) \to \R$ be a continuously differentiable function,
$x_0 \in (a,b)$ a point where $f'(x_0) \neq 0$.  Then there exists
an open interval $I \subset (a,b)$ with $x_0 \in I$, the
restriction $f|_{I}$ is injective with a continuously differentiable inverse
$g \colon J \to I$ defined on an interval $J \coloneqq f(I)$,
and
\begin{equation*}
g'(y) = \frac{1}{f'\bigl( g(y) \bigr)} \qquad \text{for all } y \in J.
\end{equation*}
\end{thm}

\begin{proof}
Without loss of generality, suppose $f'(x_0) > 0$.  As $f'$ is
continuous, there must exist an open interval $I = (x_0-\delta,x_0+\delta)$ 
such that $f'(x) > 0$ for all $x \in I$.  See
\exerciseref{exercise:positivecontneigh}.

By \propref{incdecdiffstrictprop}, $f$ is strictly increasing
on $I$, and hence the restriction $f|_{I}$ is bijective onto $J \coloneqq f(I)$.
As $f$ is continuous,
\corref{cor:continterval}
(or directly via the
\hyperref[IVT:thm]{intermediate value theorem})
implies that
$f(I)$ is an interval.
Now apply \lemmaref{lemma:ift}.
\end{proof}

In \exampleref{example:sqrt2}, we saw how difficult an
endeavor was proving the existence of $\sqrt{2}$ without any
tools.
With the \hyperref[IVT:thm]{intermediate value theorem},
the existence of roots is almost trivial, and
with the machinery of this section, we will prove
far more than mere existence.

\begin{cor}
Given $n \in \N$ and $x \geq 0$, there exists a unique 
number $y \geq 0$ (denoted $x^{1/n} \coloneqq y$), such that $y^n = x$.  Furthermore,
the function $g \colon (0,\infty) \to (0,\infty)$ defined by
$g(x) \coloneqq x^{1/n}$ is continuously differentiable and
\begin{equation*}
g'(x) = \frac{1}{nx^{(n-1)/n}} = \frac{1}{n} \, x^{(1-n)/n} ,
\end{equation*}
using the convention $x^{m/n} \coloneqq {(x^{1/n})}^{m}$.
\end{cor}

\begin{proof}
For $x=0$, the existence of a unique root is trivial.

Let $f \colon (0,\infty) \to (0,\infty)$ be defined by $f(y) \coloneqq y^n$.
The function $f$ is continuously differentiable,
and $f'(y) = ny^{n-1}$, see \exerciseref{exercise:diffofxn}.
For $y > 0$, the derivative $f'$ is strictly positive,
and so again by \propref{incdecdiffstrictprop}, $f$ is strictly
increasing (this can also be proved directly) and hence injective.
Suppose $M$ and $\epsilon$ are such that
$M > 1$ and $1 > \epsilon > 0$.
Then
$f(M) = M^n \geq M$ and
$f(\epsilon) = \epsilon^n \leq \epsilon$.
For every $x$ with $\epsilon < x < M$,
we have, by the
\hyperref[IVT:thm]{intermediate value theorem}, that $x \in
f\bigl( [\epsilon,M] \bigr) \subset
f\bigl( (0,\infty) \bigr)$.  As $M$ and $\epsilon$ were arbitrary, $f$ is onto
$(0,\infty)$, and hence $f$ is bijective.
Let $g$ be the inverse of $f$, and we obtain
the existence and uniqueness of positive
$n$th roots.  \lemmaref{lemma:ift} says that $g$ has a continuous
derivative and $g'(x) =
\frac{1}{f'\left(\vphantom{1^1_1}g(x)\right)} = \frac{1}{n {(x^{1/n})}^{n-1}}$.
\end{proof}

\begin{example}
The corollary provides a good example of where the inverse function theorem
gives us an interval smaller than $(a,b)$.  Take $f \colon \R \to \R$
defined by $f(x) \coloneqq x^2$.  Then $f'(x_0) \neq 0$
as long as $x_0 \neq 0$.  If $x_0 > 0$, we can take $I=(0,\infty)$, but
no larger.
\end{example}

\begin{example}
Another useful example is $f(x) \coloneqq x^3$.  The function $f \colon \R \to \R$ is
one-to-one and onto, so $f^{-1}(y) = y^{1/3}$ exists on the entire real
line, including zero and negative $y$.  The function $f$ has
a continuous derivative, but $f^{-1}$ has no derivative at the origin.  The
point is that $f'(0) = 0$.  See \figureref{cubecuberootfig} for a graph.
Notice the vertical tangent on the cube root at the origin.
See also \exerciseref{exercise:oddroot}.
\begin{myfigureht}
\myincludegraphics{cubecuberoot}{%
The graph y equals the cube root of x in bold line
and the graph of y equals x cubed in dashed line.
We note that while the graph of x cubed is tangent to the x-axis
at the origin, the cube root is a reflection across the x equals y
line and is therefore tangent to the vertical y-axis at the origin.}
\caption{Graphs of $x^3$ and $x^{1/3}$.\label{cubecuberootfig}}
\end{myfigureht}
\end{example}


\subsection{Exercises}

\begin{exercise}
Suppose $f \colon \R \to \R$ is continuously differentiable and
$f'(x) > 0$ for all $x$.  Show that $f$ is invertible on the interval $J =
f(\R)$, the inverse is continuously differentiable, and ${(f^{-1})}'(y) >
0$ for all $y \in f(\R)$.
\end{exercise}

\begin{exercise}
\pagebreak[2]
Suppose $I,J$ are intervals and a monotone onto $f \colon I \to J$ has an inverse $g \colon J \to I$.
Suppose you already know that both $f$ and $g$ are differentiable
everywhere and $f'$ is never zero.  Using the chain rule but not
\lemmaref{lemma:ift}, prove the
formula $g'(y) = \frac{1}{f'\left(\vphantom{1_1^1}g(y)\right)}$. % \bigl( \bigr) are overly big here!
Remark: This exercise is the same as \exerciseref{exercise:inversederformula},
no need to do it again if you have
solved it already.
\end{exercise}

\begin{exercise}
\pagebreak[2]
Let $n\in \N$ be even.
Prove that every $x > 0$ has a unique negative $n$th root.
That is, there exists a negative number $y$ such that $y^n = x$.
Compute the derivative
of the function $g(x) \coloneqq y$.
\end{exercise}

\begin{exercise} \label{exercise:oddroot}
Let $n \in \N$ be odd and $n \geq 3$.
Prove that every $x$ has a unique $n$th root.
That is, there exists a number $y$ such that $y^n = x$.  Prove that
the function defined by $g(x) \coloneqq y$ is differentiable except at $x=0$
and compute the derivative.  Prove that $g$ is not differentiable at $x=0$.
\end{exercise}

\begin{exercise}[requires \sectionref{sec:taylor}]
Show that if in the inverse function theorem $f$ has $k$ continuous
derivatives, then the inverse function $g$ also has $k$ continuous
derivatives.
\end{exercise}

\begin{exercise}
Let $f(x) \coloneqq x + 2 x^2 \sin(\nicefrac{1}{x})$ for $x \neq 0$ and
$f(0) \coloneqq 0$.  Show that $f$ is differentiable at all $x$, that $f'(0) > 0$,
but that $f$ is not invertible
on any open interval containing the origin.
\end{exercise}

\begin{exercise}
\leavevmode
\begin{enumerate}[a)]
\item
Let $f \colon \R \to \R$ be a continuously differentiable function
and $k > 0$ be a number such that $f'(x) \geq k$ for all $x \in \R$.
Show $f$ is one-to-one and onto, and has a continuously differentiable
inverse $f^{-1} \colon \R \to \R$.
\item
Find an example $f \colon \R \to \R$
where $f'(x) > 0$
for all $x$, but $f$ is not onto.
\end{enumerate}
\end{exercise}

\begin{exercise}
Suppose $I,J$ are intervals and a monotone onto $f \colon I \to J$ has an inverse $g \colon J \to I$.
Suppose $x \in I$ and $y \coloneqq f(x) \in J$, and that $g$ is differentiable at
$y$.  Prove:
\begin{enumerate}[a)]
\item
If $g'(y) \neq 0$, then $f$ is differentiable at $x$.
\item
If $g'(y) = 0$, then $f$ is not differentiable at $x$.
\end{enumerate}
\end{exercise}