Created the Real Analysis Theorems and Definitions packet
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\chapter{Differentiation}
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\subimport{./}{the-derivative.tex}
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\subimport{./}{the-mean-value-theorem.tex}
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\subimport{./}{lhopitals-rules.tex}
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\subimport{./}{taylors-theorem.tex}
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\section{L'Hopital's Rules}
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\begin{theorem}
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Let $f$ and $g$ be defined on $[a,b]$, let $f(a)=g(a)=0$, and let $g(x) \neq 0$ for $a < x < b$. If $f$ and $g$ are differentiable at $a$ and if $g'(a) \neq 0$, then the limit of $f/g$ at $a$ exists and is equal to $f'(a)/g'(a)$. Thus
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\[\lim\limits_{x\to a^+} \frac{f(x)}{g(x)} = \frac{f'(a)}{g'(a)}\]
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\end{theorem}
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\begin{theorem}[\textbf{Cauchy Mean Value Theorem}]
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Let $f$ and $g$ be continuous on $[a,b]$ and differentiable on $(a,b)$, and assume that $g'(x) \neq 0$ for all $x$ in $(a,b)$. Then there exists $c$ in $(a,b)$ such that
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\[\frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)}\]
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\end{theorem}
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\begin{theorem}[\textbf{L'Hopital's Rule, I}]
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Let $-\infty \leq a < b \leq \infty$ and let $f,g$ be differentiable on $(a,b)$ such that $g'(x) \neq 0$ for all $x \in (a,b)$. Suppose that
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\[\lim\limits_{x\to a^+} f(x) = 0 = \lim\limits_{x\to a^+} g(x)\]
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\begin{enumerate}
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\item If $\lim\limits_{x\to a^+} \frac{f'(x)}{g'(x)}=L \in \R$, then $\lim\limits_{x\to a^+} \frac{f(x)}{g(x)}=L$.
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\item If $\lim\limits_{x\to a^+} \frac{f'(x)}{g'(x)}=L \in \{-\infty, \infty\}$, then $\lim\limits_{x\to a^+} \frac{f(x)}{g(x)}=L$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}[\textbf{L'Hopital's Rule, II}]
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Let $-\infty \leq a < b \leq \infty$ and let $f,g$ be differentiable on $(a,b)$ such that $g'(x) \neq 0$ for all $x \in (a,b)$. Suppose that
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\[\lim\limits_{x\to a^+} g(x) = \pm \infty\]
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\begin{enumerate}
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\item If $\lim\limits_{x\to a^+} \frac{f'(x)}{g'(x)}=L \in \R$, then $\lim\limits_{x\to a^+} \frac{f(x)}{g(x)}=L$.
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\item If $\lim\limits_{x\to a^+} \frac{f'(x)}{g'(x)}=L \in \{-\infty, \infty\}$, then $\lim\limits_{x\to a^+} \frac{f(x)}{g(x)}=L$.
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\end{enumerate}
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\end{theorem}
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\section{Taylor's Theorem}
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\begin{theorem}[\textbf{Taylor's Theorem}]
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Let $n \in \N$, let $I:=[a,b]$, and let $f:I \rightarrow \R$ be such that $f$ and its derivative $f', f'', \dots, f^{(n)}$ are continuous on $I$ and that $f^{(n+1)}$ exists on $(a,b)$. If $x_0\in I$, then for any $x$ in $I$ there exists a point $c$ between $x$ and $x_0$ such that
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\[f(x) = f(x_0) + f'(x_0)(x-x_0) + \frac{f''(x_0)}{2!}(x-x_0)^2\]
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\[+ \dots + \frac{f^{(n)}(x_0)}{n!}(x-x_0)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1}\]
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\end{theorem}
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\begin{theorem}
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Let $I$ be an interval, let $x_0$ be an interior point of $I$, and let $n \geq 2$. Suppose that the derivatives $f', f'', \dots, f^{(n)}$ exist and are continuous in a neighborhood of $x_0$ and that $f'(x_0) = \dots = f^{(n-1)(x_0)}$, but $f^{(n)}(x_0) \neq 0$.
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\begin{enumerate}
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\item If $n$ is even and $f^{(n)}(x_0) >0$, then $f$ has a relative minimum at $x_0$.
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\item If $n$ is even and $f^{(n)}<0$, then $f$ has a relative maximum at $x_0$.
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\item If $n$ is odd, then $f$ has neither a relative minimum nor relative maximum at $x_0$.
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\end{enumerate}
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\end{theorem}
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\begin{definition}
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Let $I \subseteq \R$ be an interval. A function $f:I \rightarrow \R$ is said to be \textbf{convex} on $I$ if for any $t$ satisfying $0 \leq t \leq 1$ and any points $x_1, x_2$ in $I$, we have
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\[f((1-t)x_1+tx_2) \leq (1-t)f(x_1)+tf(x_2).\]
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\end{definition}
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\begin{theorem}
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Let $I$ be an open interval and let $f: I \rightarrow \R$ have a second derivative on $I$. Then $f$ is a convex function on $I$ if and only if $f''(x) \geq 0$ for all $x \in I$.
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\end{theorem}
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\begin{theorem}[\textbf{Newton's Method}]
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Let $I:=[a,b]$ and let $f:I \rightarrow \R$ be twice differentiable on $I$. Suppose that $f(a)f(b) < 0$ and that there are constants $m,M$ such that $|f'(x)| \geq m > 0$ and $|f''(x)| \leq M$ for $x \in I$ and let $K:=M/2m$. Then there exists a subinterval $I^*$ containing a zero $r$ of $f$ such that for any $x_1 \in I^*$ the sequence $(x_n)$ defined by
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\[|x_{n+1}-r|\leq x_n - \frac{f(x_n)}{f'(x_n)}\ \ \ \ \text{for all}\ \ \ \ n \in \N,\]
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belongs to $I^*$ and $(x_n)$ converges to $r$. Moreover
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\[|x_{n+1}-r| \leq K|x_n-r|^2\ \ \ \ \text{for all}\ \ \ \ n \in \N.\]
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\end{theorem}
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\section{The Derivative}
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\begin{definition}
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Let $I \subseteq \R$ be an interval, let $f:I \rightarrow \R$, and let $ c \in I$. We say that a real number $L$ is the \textbf{derivative of $f$ at $c$} if given any $\varepsilon > 0$ there exists $\delta (\varepsilon) > 0$ such that if $x \in I$ satisfies $0 < |x-c|<\delta (\varepsilon)$, then
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\[\abs{\frac{f(x)-f(c)}{x-c}-L}<\varepsilon.\]
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In this case we say that $f$ is \textbf{differentiable} at $c$, and we write $f'(c)$ for $L$. In other words, the derivative of $f$ at $c$ is given by the limit
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\[f'(c) = \lim\limits_{x\to c} \frac{f(x)-f(c)}{x-c}\]
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provided this limit exists. (We allow the possibility that $c$ may be the endpoint of the interval.)
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\end{definition}
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\begin{theorem}
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If $f:I \rightarrow \R$ has a derivative at $c \in I$, then $f$ is continuous at $c$.
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\end{theorem}
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\begin{theorem}
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Let $I \subseteq \R$ be an interval, let $c \in I$ , and let $f:I \rightarrow \R$ and $g:I \rightarrow \R$ be functions that are differentiable at $c$. Then:
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\begin{enumerate}
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\item If $\alpha \in \R$, then the function $\alpha f$ is differentiable at $c$, and \[(\alpha f)'(c) = \alpha f'(c)\]
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\item The function $f+g$ is differentiable at $c$, and
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\[(f+g)'(c) = f'(c)+g'(c)\]
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\item (Product Rule) The function $fg$ is differentiable at $c$, and
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\[(fg)'(c) = f'(c)g(c) + f(c)g'(c).\]
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\item (Quotient Rule) If $g(c) \neq 0$, then the function $f/g$ is differentiable at $c$, and
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\[\left( \frac{f}{g}\right)'(c) = \frac{f'(c)g(c)-f(c)g'(c)}{(g(c))^2}\]
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\end{enumerate}
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\end{theorem}
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\begin{corollary}
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If $f_1, f_2, \dots, f_n$ are functions on an interval $I$ to $\R$ that are differentiable at $c \in I$, then:
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\begin{enumerate}
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\item The function $f_1 + f_2 + \dots + f_n$ is differentiable at $c$ and
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\[(f_1 + f_2 + \dots + f_n)'(c) = f_1'(c) + f_2'(c) + \dots + f_n'(c)\]
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\item The function $f_1f_2 \dots f_n$ is differentiable at $c$ and
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\[(f_1f_2 \dots f_n)'(c) = f_1'(c)f_2(c) \dots f_n(c)+f_1(c)f_2'(c) \dots f_n(c) + \dots + f_1(c)f_2(c) \dots f_n'(c).\]
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An important special case of the extended product rule occurs if the functions are equal, that is, $f_1 = f_2 = \dots = f_n = f$. Then the above becomes
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\[(f^n)'(c) = n(f(c))^{n-1}f'(c)\]
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\end{enumerate}
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\end{corollary}
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\begin{theorem}[\textbf{Carathéodory's Theorem}]
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Let $f$ be defined on an interval $I$ containing the point $c$. Then $f$ is differentiable at $c$ if and only if there exists a function $\varphi$ on $I$ that is continuous at $c$ and satisfies
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\[f(x)-f(c)=\varphi (x)(x-c)\ \ \ \ \text{for}\ \ \ \ x \in I\]
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In this case, we have $\varphi (c)=f'(c)$.
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\end{theorem}
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\begin{theorem}[\textbf{Chain Rule}]
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Let $I, J$ be intervals in $\R$, let $g:I \rightarrow \R$ and $f:J \rightarrow \R$ be functions such that $f(J) \subseteq I$, and let $c \in J$. If $f$ is differentiable at $c$ and if $g$ is differentiable at $f(c)$, then the composite function $g \circ f$ is differentiable at $c$ and
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\[(g \circ f)'(c) = g'(f(c)) \cdot f'(c).\]
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\end{theorem}
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\begin{theorem}
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Let $I$ be an interval in $\R$ and let $f:I \rightarrow \R$ be strictly monotone and continuous on $I$. Let $J:=f(I)$ and let $g:J \rightarrow \R$ be the strictly monotone and continuous function inverse to $f$. If $f$ is differentiable at $c \in I$ and $f'(c) \neq0$, then $g$ is differentiable at $d:=f(c)$ and
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\[g'(d)=\frac{1}{f'(c)}=\frac{1}{f'(g(d))}\]
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\end{theorem}
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\begin{theorem}
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Let $I$ be an interval and let $f:I \rightarrow \R$ be strictly monotone on $I$. Let $J:= f(I)$ and let $g:J \rightarrow \R$ be the function inverse to $f$. If $f$ is differentiable on $I$ and $f'(x) \neq 0$ for $x \in I$, then $g$ is differentiable on $J$ and
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\[g' = \frac{1}{f' \circ g}\]
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\end{theorem}
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\section{The Mean Value Theorem}
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\begin{theorem}[\textbf{Interior Extremum Theorem}]
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Let $c$ be an interior point of of the interval $I$ at which $f:I \rightarrow \R$ has a relative extremum. If the derivative of $f$ at $c$ exists, then $f'(c)=0$.
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\end{theorem}
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\begin{corollary}
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Let $f:I \rightarrow \R$ be continuous on an interval $I$ and suppose that $f$ has a relative extremum at an interior point $c$ of $I$. Then either the derivative of $f$ at $c$ does not exist, or it is equal to zero.
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\end{corollary}
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\begin{theorem}[\textbf{Rolle's Theorem}]
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Suppose that $f$ is continuous on a closed interval $I:= [a,b]$, that the derivative $f'$ exists at every point of the open interval $(a,b)$, and that $f(a)=f(b)=0$. Then there exists at least one point $c$ in $(a,b)$ such that $f'(c)=0$.
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\end{theorem}
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\begin{theorem}[\textbf{Mean Value Theorem}]
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Suppose that $f$ is continuous on a closed interval $I:=[a,b]$, and that $f$ has a derivative in the open interval $(a,b)$. Then there exists at least one point $c$ in $(a,b)$ such that
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\[f(b)-f(a)=f'(c)(b-a)\]
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\end{theorem}
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\begin{theorem}
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Suppose that $f$ is continuous on the closed interval $I:= [a,b]$, that $f$ is differentiable on the open interval $(a,b)$, and that $f'(x)=0$ for $x \in (a,b)$. Then $f$ is constant on $I$.
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\end{theorem}
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\begin{corollary}
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Suppose that $f$ and $g$ are continuous on $I:=[a,b]$, that they are differentiable on $(a,b)$, and that $f'(x)=g'(x)$ for all $x \in (a,b)$. Then there exists a constant $C$ such that $f = g+C$ on $I$.
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\end{corollary}
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\begin{theorem}
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Let $f:I \rightarrow \R$ be differentiable on the interval $I$. Then:
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\begin{enumerate}
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\item $f$ is increasing on $I$ if and only if $f'(x) \geq 0$ for all $x \in I$.
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\item $f$ is decreasing on $I$ if and only if $f'(x) \leq 0$ for all $x \in I$.
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\end{enumerate}
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\end{theorem}
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\begin{theorem}[\textbf{First Derivative Test for Extrema}]
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Let $f$ be continuous on the interval $I:=[a,b]$ and let $c$ be an interior point of $I$. Assume that $f$ is differentiable on $(a,c)$ and $(c,b)$. Then:
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\begin{enumerate}
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\item If there is a neighborhood $(c-\delta, c+\delta)\subseteq I$ such that $f'(x) \geq 0$ for $c-\delta< x < c$ and $f'(x) \leq 0$ for $c < x < c + \delta$, then $f$ has a relative maximum at $c$.
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\item If there is a neighborhood $(c-\delta, c+\delta) \subseteq I$ such that $f'(x) \leq 0$ for $c-\delta < x < c$ and $f'(x) \geq 0$ for $c < x < c+\delta$, then $f$ has a relative maximum at $c$.
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\end{enumerate}
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\end{theorem}
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\begin{lemma}
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Let $I \subseteq \R$ be an interval, let $f:I \rightarrow \R$, let $c \in I$, and assume that $f$ has a derivative at $c$. Then:
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\begin{enumerate}
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\item If $f'(c) >0$, then there is a number $\delta > 0$ such that $f(x) > f(c)$ for $x \in I$ such that $c < x < c+ \delta$.
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\item If $f'(c)<0$, then there is a number $\delta >0$ such that $f(x) > f(c)$ for $x \in I$ such that $c-\delta < x < c$.
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\end{enumerate}
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\end{lemma}
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\begin{theorem}[\textbf{Darboux's Theorem}]
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If $f$ is differentiable on $I = [a,b]$ and if $k$ is a number between $f'(a)$ and $f'(b)$, then there is at least one point $c$ in $(a,b)$ such that $f'(c)=k$.
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\end{theorem}
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