\documentclass[12pt]{article} \usepackage{amsfonts} %new commands % reference for equation labels \newcommand{\eq}[1]{(\ref{#1})} % abbreviations for equation environments \newcommand{\be}{\begin{equation}} \newcommand{\ee}{\end{equation}} \newcommand{\bea}{\begin{eqnarray}} \newcommand{\eea}{\end{eqnarray}} \newcommand{\beann}{\begin{eqnarray*}} \newcommand{\eeann}{\end{eqnarray*}} % abbreviation for reals \def\Cx{\mathbb{C}} \def\Nl{\mathbb{N}} \def\Ra{\mathbb{Q}} \def\Rl{\mathbb{R}} \def\Ts{\mathbb{T}} \def\Ir{\mathbb{Z}} \def\DD{\mathcal{D}} % abbreviations for \varepsilon and \varphi \newcommand{\eps}{\varepsilon} \newcommand{\vphi}{\varphi} \pagestyle{empty} \begin{document} \centerline{\textbf{Sample Questions: Solutions}} \centerline{\textbf{Midterm II}} \centerline{\textbf{Math 127B}. \textbf{Winter, 2005}} \bigskip\bigskip \noindent \textbf{1.} For each of the following statements, say if it is true or false. (No explanation is required.) \smallskip\noindent (a) If $f$ is differentiable and $f^\prime > 0$, then $f$ is strictly increasing. \smallskip\noindent (b) If $f$ is strictly increasing and differentiable, then $f^\prime > 0$. \smallskip\noindent (c) If $f$ is the sum of a convergent Taylor series in an open interval containing the origin, then $f$ is infinitely differentiable. \smallskip\noindent (d) If $f$ is infinitely differentiable in an open interval containing the origin, then the Taylor series of $f$ converges. \smallskip\noindent (e) There exists $0 0$ is $f$ differentiable at $0$? Is $f$ differentiable at $x\ne 0$? \bigskip\noindent \textbf{Solution.} \begin{itemize} \item A function $f:(a,b)\to \Rl$ is differentiable at $x_0\in (a,b)$ with derivative $f^\prime(x_0)$ if the following limit exists \[ f^\prime(x_0) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}. \] \item This function is differentiable at $0$ if and only if $a > 1$. This follows because \beann &&\frac{f(x) - f(0)}{x-0} = 0 \qquad\mbox{if $x$ is rational and nonzero},\\ &&\frac{f(x) - f(0)}{x-0} = \frac{|x|^a}{x} \qquad\mbox{if $x$ is irrational}, \eeann and \[ \lim_{x\to 0}\frac{|x|^a}{x} = 0 \] if and only if $a > 1$. \item The function $f$ is discontinuous at every $x\ne 0$, therefore it is not differentiable at any $x\ne 0$. \end{itemize} \newpage \noindent \textbf{3.} State Taylor's theorem. Prove that \[ \log(1+x) < x \] for all $x > 0$. \bigskip\noindent \textbf{Solution.} \begin{itemize} \item If $f : (a,b) \to \Rl$ is an $n$-times differentiable function on an open interval $(a,b)$ containing the origin, then for any $x\in (a,b)$ there exists a $y$ between $0$ and $x$ such that \[ f(x) - \sum_{k=0}^{n-1} \frac{f^{(k)}(0)}{k!} x^k = \frac{f^{(n)}(y)}{n!} x^n. \] \item First proof. The function $f(x) = \log(1+x)$ is infinitely differentiable in $(-1,\infty)$, and \[ f^\prime(x) = \frac{1}{1+x},\quad f^{\prime\prime}(x) = -\frac{1}{(1+x)^2}. \] Since $f(0) = 0$ and $f^\prime(0) = 1$, the Taylor polynomial of $f$ of degree $1$ is $x$. If $x > 0$, then using Taylor's theorem with $n=2$ we find that there exists $0 < y < x$ such that \[ f(x) - x = -\frac{1}{2!(1+y)^2}x^2 < 0, \] so \[ \log(1+x) < x. \] \item Second proof. For any $x > 0$, the function $g(x) = x -\log(1+x)$ is continuous on the closed interval $[0,x]$ and differentiable in the open interval $(0,x)$. By the mean value theorem, there exists $0 < y < x$ such that \[ g(x) - g(0) = g^\prime(y) (x-0). \] Since $g(0) = 0$ and \[ g^\prime(y) = 1 - \frac{1}{1+y} > 0\qquad \mbox{for $y > 0$}, \] we conclude that \[ g(x) > 0 \qquad\mbox{for $x > 0$} \] which proves the result. \end{itemize} \newpage \noindent \textbf{4.} Carefully state a version of L'Hospital's rule that applies to the following limit. Use it to prove that the limit exists, and find its value: \[ \lim_{x\to 0} \frac{1-\cos x}{x^2}. \] \bigskip\noindent \textbf{Solution.} \begin{itemize} \item Suppose that $f$, $g$ are functions that are differentiable in an open interval containing $x_0$, with $g^\prime(x)$ nonzero in the interval, and \[ \lim_{x\to x_0} f(x) = 0,\qquad \lim_{x\to x_0} g(x) = 0. \] If the limit \[ \lim_{x\to x_0} \frac{f^\prime(x)}{g^\prime(x)} = L \] exists, then so does the limit \[ \lim_{x\to x_0} \frac{f(x)}{g(x)} = L. \] \item The functions $f(x) = \sin x$ and $g(x) = 2x$ satisfy the hypotheses of L'Hospital's theorem on any interval containing the origin, and \[ \lim_{x\to 0} \frac{f^\prime(x)}{g^\prime(x)} = \lim_{x\to 0} \frac{\cos x}{2} =\frac{1}{2} \] since $\cos x$ is continuous at $x=0$. Hence \[ \lim_{x\to 0} \frac{\sin x}{2x} =\frac{1}{2} \] Applying L'Hospital's theorem again to the functions $F(x) = 1-\cos x$ and $G(x) = x^2$, which also satisfy its hypotheses, with \[ \lim_{x\to 0} \frac{F^\prime(x)}{G^\prime(x)} =\frac{1}{2} \] we conclude that \[ \lim_{x\to 0} \frac{F(x)}{G(x)} = \frac{1}{2}, \] which proves the result. \end{itemize} \newpage \noindent \textbf{5.} Define the hyperbolic sine \[ \sinh x = \frac{e^x - e^{-x}}{2}. \] Prove that $\sinh x$ is strictly increasing on $\Rl$ and hence has an inverse. Prove that the inverse is differentiable and compute its derivative. \bigskip\noindent \textbf{Solution.} \begin{itemize} \item The functions \[ \cosh x = \frac{e^x + e^{-x}}{2},\qquad \sinh x = \frac{e^x - e^{-x}}{2}, \] satisfy the following easily verified identities: \beann &&(\cosh x)^\prime = \sinh x,\quad (\sinh x)^\prime = \cosh x,\\ &&\cosh^2 x - \sinh^2 x = 1. \eeann \item Since $(\sinh x)^\prime = \cosh x > 0$ on $\Rl$, the function $\sinh x$ is strictly increasing, and therefore invertible on its range. Moreover, since $e^x \to \infty$ and $e^{-x} \to 0$ as $x\to \infty$, we have $\sinh x \to \infty$ as $x\to \infty$; similarly $\sinh x \to -\infty$ as $x\to -\infty$. Thus, the range of $\sinh : \Rl \to \Rl$ is all of $\Rl$, and $\sinh^{-1} : \Rl \to \Rl$ \item Since $\sinh x$ is differentiable on $\Rl$ and $(\sinh x)^\prime \ne 0$, the inverse function is differentiable on $\Rl$, and \[ (\sinh^{-1})^\prime(\sinh x ) = \frac{1}{(\sinh x)^\prime} = \frac{1}{\cosh x}. \] Setting $y=\sinh x$, and writing \[ \cosh x = \sqrt{1 + \sinh^2 x} = \sqrt{1 + y^2}, \] we conclude that \[ (\sinh^{-1})^\prime(y) = \frac{1}{\sqrt{1 + y^2}}. \] \end{itemize} \newpage \noindent \textbf{6.} A function $f$ has a jump discontinuity at $x_0$ if both the left and right limits \[ \lim_{x\to x_0^+} f(x), \qquad \lim_{x\to x_0^-} f(x) \] exist but have different values. Suppose that $f : (a,b) \to \Rl$ is differentiable in $(a,b)$. Prove that $f^\prime$ does not have a jump discontinuity in $(a,b)$. \bigskip\noindent \textbf{Solution.} \begin{itemize} \item Suppose, for contradiction, that the limits \[ L_\pm = \lim_{x\to x_0^\pm} f^\prime(x) \] exist and are distinct. Let \[ \epsilon = \frac{L_+ - L_-}{3}. \] Then there exists $\delta > 0$ such that \[ |f^\prime(x) - L_+| < \epsilon\quad\mbox{for $x_0 < x < x_0+\delta$} \] and \[ |f^\prime(x) - L_-| < \epsilon\quad\mbox{ for $x_0-\delta < x < x_0$.} \] For definiteness, suppose that $L_+ > L_-$ (otherwise, we can consider $-f$ instead of $f$). It follows that $f^\prime(x)$ cannot take values in the interval $[L_- + \epsilon, L_+ - \epsilon]$ of width $\epsilon > 0$ when $0 < |x-x_0| < \delta$. Since $f^\prime(x) > L_+ - \epsilon$ for $x > x_0$ and $f^\prime(x) < L_- + \epsilon$ for $x < x_0$, this contradicts the intermediate value property of the derivative, whatever the value of $f^\prime(x_0)$. \end{itemize} \end{document}