From 1231c438847fa3cc5c2eee200c9fa76451abc1cd Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Wed, 16 Sep 2015 14:32:19 +0200
Subject: Start assignment 3

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+\documentclass[10pt,a4paper]{article}
+
+\usepackage[utf8]{inputenc}
+\usepackage[margin=2cm]{geometry}
+
+\usepackage{enumitem}
+\setenumerate[1]{label=\arabic*.}
+\setenumerate[2]{label=(\alph*)}
+
+% textcomp package is not available everywhere, and we only need the Copyright symbol
+% taken from http://tex.stackexchange.com/a/1677/23992
+\DeclareTextCommandDefault{\textregistered}{\textcircled{\check@mathfonts\fontsize\sf@size\z@\math@fontsfalse\selectfont R}}
+
+\usepackage{fancyhdr}
+\renewcommand{\headrulewidth}{0pt}
+\renewcommand{\footrulewidth}{0pt}
+\fancyhead{}
+\fancyfoot[C]{Copyright {\textcopyright} 2015 Camil Staps}
+\pagestyle{fancy}
+
+\usepackage{amsmath}
+\usepackage{amsfonts}
+
+\parindent0pt
+
+\title{Calculus en Kansrekenen - assignment 3}
+\author{Camil Staps\\\small{s4498062, group Bram}}
+
+\begin{document}
+
+\maketitle
+\thispagestyle{fancy}
+
+\begin{enumerate}
+    \item \begin{enumerate}
+            \item The domain of $\arcsin$ is the range of $\sin$, so $[-1,1]$.
+            \item Filling in $x=\sin1$ gives us $x=(2k+\frac12)\cdot\pi$ with $k\in\mathbb Z$. Therefore, $\arcsin(1) = \frac\pi2$; we fill in $k=0$ which is the only $k$ s.t. $(2k+\frac12)\pi \in [-\frac\pi2,\frac\pi2]$.
+
+                Similarly, $x=\sin0$ gives us $x=k\cdot\pi$ with $k\in\mathbb Z$. We fill in $k=0$, so $\arcsin(0) = 0$.
+
+                Finally, we know $x=\sin\frac{\sqrt3}2$ gives $x=(2k+\frac13)\cdot\pi$ or $x=(2k+\frac23)\cdot\pi$ with $k\in\mathbb Z$. The only possible $x$ would then be $(2k+\frac13)\pi$ with $k=0$, so $\arcsin\frac{\sqrt3}2 = \frac\pi3$.
+            \item \begin{align*}
+                    f'(x) &= \left(\arcsin\frac{2x}{1-x}\right)' \\
+                          &= \frac1{1-\left(\frac{2x}{1-x}\right)^2} \cdot \left(\frac{2x}{1-x}\right)' \\
+                    %todo
+                \end{align*}
+        \end{enumerate}
+
+    \item \begin{enumerate}
+            \item We know $\lim_{x\to\infty}\ln(2015x) = \infty = \lim_{x\to\infty} x^3$, so let's try to apply l'Hôpital:
+                \begin{align*}
+                    (\ln(2015x))' &= (\ln 2015 + \ln x)' = \frac1x.\\
+                    (x^3)'        &= 3x^2.\\
+                    \lim\limits_{x\to\infty} \frac{\frac1x}{3x^2} &= \lim\limits_{x\to\infty} \frac1{3x^3} = 0.
+                \end{align*}
+
+                Since we saw that both numerator and denominator are differentiable and that the fraction of their derivatives has a limit for $x\to\infty$, this limit is equal to the limit of the original fraction.
+
+            \item We know $\lim_{a\to-3}\sin(a\pi) = 0 = \lim_{a\to-3}a^2-9$, so let's try to apply l'Hôpital:
+                \begin{align*}
+                    (\sin(a\pi))' &= \cos(a\pi)\cdot\pi. \\
+                    (a^2-9)'      &= 2a.\\
+                    \lim\limits_{a\to-3}\frac{\pi\cdot\cos(a\pi)}{2a} &= \frac\pi6.
+                \end{align*}
+
+                Since we saw that both numerator and denominator are differentiable and that the fraction of their derivatives has a limit for $a\to-3$, this limit is equal to the limit of the original fraction.
+
+            \item We know $\lim_{x\to-\infty}e^{3-x} = \infty = \lim_{x\to-\infty}7x^2$, so let's try to apply l'Hôpital:
+                \begin{align*}
+                    \left(e^{3-x}\right)' &= e^{3-x}\cdot-1 = -e^{3-x}.\\
+                    (7x^2)'               &= 14x.\\
+                \end{align*}
+
+                Now we know $\lim_{x\to-\infty} -e^{3-x} = -\infty = \lim_{x\to-\infty} 14x$, so to find $\lim_{x\to-\infty}\frac{-e^{3-x}}{14x}$ we can again try to apply l'Hôpital:
+                \begin{align*}
+                    \left(-e^{3-x}\right)' &= e^{3-x}.\\
+                    (14x)'                 &= 14.\\
+                    \lim\limits_{x\to-\infty}\frac{e^{3-x}}{14} &= \infty.
+                \end{align*}
+                
+                And since we saw that in both steps numerator and denominator are differentiable, and that the fractions of their derivatives has a limit for $x\to-\infty$, this limit is equal to the limit of the original fraction.
+        \end{enumerate}
+
+    \item \begin{enumerate}
+            \item $f(x) = \log_32 + \log_3x$. Then $f'(x) = \frac1{x\ln3}$. Then $f''(x) = -\frac{\ln3}{x^2(\ln3)^2} = -\frac1{x^2\ln3}$. Finally $f'''(x) = -\frac{-2\ln3\cdot x}{x^4\cdot(\ln3)^2} = \frac2{x^3\ln3}$.
+            \item %todo
+        \end{enumerate}
+
+    \item \begin{enumerate}
+            \item $x-3$, $x+1$ and $x^2$ are all from $\mathbb R$ to $\mathbb R$. Multiplication is from $\mathbb R\times\mathbb R$ to $\mathbb R$. Therefore, $f : \mathbb R \to \mathbb R$. 
+            \item To find the roots we equate $f(x)$ with zero and solve for $x$. That gives $x=-1$ or $x=3$, so we have $(-1,0)$ and $(3,0)$ as roots of $f$.
+
+                To find the y-intercept, we fill in $x=0$ in $f(x)$: $(0+1)^2(0-3)=-3$. This gives $(0,-3)$.
+            \item %todo
+            \item $f'(x) = \left((x+1)^2(x-3)\right)' = \left(x^3-x^2-5x-3\right)' = 3x^2 - 2x - 5$.
+
+                $f''(x) = \left(3x^2 - 2x - 5\right)' = 6x - 2$.
+            \item %todo
+            \item %todo
+            \item %todo
+            \item %todo
+        \end{enumerate}
+
+    \item %todo
+
+    \item %todo
+
+\end{enumerate}
+
+\end{document}
+
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