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+\documentclass[10pt,a4paper]{article}
+
+\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 1}
+\author{Camil Staps\\\small{s4498062}}
+
+\begin{document}
+
+\maketitle
+\thispagestyle{fancy}
+
+\begin{enumerate}
+    \item \begin{enumerate}[label=\arabic*.]
+            \item \begin{align*}
+                    x - x^3 &= 0 \\
+                    x &= x^3 \qquad\text{We see $x=0$ is a solution, continuing with $x\neq0$\dots}\\
+                    1 &= x^2
+                \end{align*}
+
+                This gives $x=\sqrt1=1$ and $x=-\sqrt1=-1$ as solutions, yielding $-1,0,1$ as possible values for x.
+            \item \begin{align*}
+                    x - x^3 &> 0\\
+                    x &> x^3 \qquad\text{We see $x=0$ is no solution, so we can assume $x\neq0$.}\\
+                    1 &> x^2
+                \end{align*}
+
+                We know $1=x^2$ gives $x=-1$ or $x=1$, so this inequality gives intuitively $-1>x$ or $1 > x > 0$. This gives $x\in(-\infty,-1)\cup(0,1)$.
+        \end{enumerate}
+
+    \item For particular values of $a,b$ the function $y=a+(b-a)x$ is a linear function, hence it has extremes at $x=0$ and $x=1$. This gives the extremes $y=a+(b-a)\cdot0=a$ and $y=a+(b-a)\cdot1=b$. Therefore, $y$ runs through $(a,b)$. Note that strictly we calculated the extremes on $[0,1]$ rather than $(0,1)$ but that this does not matter for a linear function and that we corrected this by giving the range $(a,b)$ rather than $[a,b]$.
+
+    \item \begin{enumerate}
+            \item $f(-x) = 3(-x)-(-x)^3 = -3x+x^3 = -\left(3x-x^3\right) = -f(x)$, hence $f$ is \textbf{odd}.
+            \item $f(-x) = \sqrt[3]{(1-(-x))^2} + \sqrt[3]{(1+(-x))^2} = \sqrt[3]{(1+x)^2} + \sqrt[3]{(1-x)^2} = \sqrt[3]{(1-x)^2} + \sqrt[3]{(1+x)^2} = f(x)$, hence $f$ is \textbf{even}.
+        \end{enumerate}
+
+    \item We find $f^{-1}(x) = \frac{b-dx}{cx-a}$, since:
+        
+        \begin{align*}
+            y &= \frac{ax+b}{cx+d}\\
+            (cx + d) y &= ax + b\\
+            cxy + dy &= ax + b\\
+            cxy - ax &= b - dy\\
+            x(cy-a) &= b - dy\\
+            x &= \frac{b-dy}{cy-a}
+        \end{align*}
+
+    \item \begin{enumerate}
+            \item We can add in $\mathbb{R}$, so the $+1$ doesn't restrict the domain. Taking the root of a number only gives a real solution for non-negative numbers, thus $7-x^2>0$ and $7>x^2$.
+
+                We find $D(f) = \{x\in\mathbb{R}\mid 7>x^2\} = \left(-\sqrt7,\sqrt7\right)$. We know then that the extreme values for $7-x^2$ are $0$ and $7$, and thus find $R(f) = \left(1,1+\sqrt7\right)$.
+
+            \item We have $f(x)=\frac{x-5}{(x-5)(x+2)}=(x+2)^{-1}$. We know that $\alpha^{-1}$ is defined for $\alpha\in\{x\in\mathbb{R}\mid x\neq0\}$, therefore $D(f)=\{x\in\mathbb{R}\mid x+2\neq0\} = \{x\in\mathbb{R}\mid x\neq-2\}$. Since the range of $y=x^{-1}$ is $(-\infty,\infty)\setminus\{0\}$, $R(f)=(-\infty,\infty)\setminus\{0\}$ as well.
+
+            \item This gives $f(x)=|x|^{-1}$. The modulus function does not restrict the domain. Since the domain of $y=x^{-1}$ is $\mathbb{R}\setminus\{0\}$, also $D(f)=\mathbb{R}\setminus\{0\}$. The range is then, similar to the above, $R(f)=(-\infty,\infty)\setminus\{0\}$.
+        \end{enumerate}
+
+    \item \begin{enumerate}
+            \item $$\lim_{x\to0}\frac{3(x-1)+3}{x} = \lim_{x\to0}\frac{3x}{x} = \lim_{x\to0}3 = 3.$$
+            \item $$\lim_{x\to2}\frac{x-2}{x^2+x-6} = \lim_{x\to2}\frac{x-2}{x^2+x-6} = \lim_{x\to2}(x+3)^{-1}.$$
+
+                $(x+3)^{-1}$ is continuous at $x=2$, therefore the limit equals $(2+3)^{-1}=\frac15$.
+            \item $$\lim_{x\to1}\frac{2x^2-4x+3}{x^2+x-2} = \lim_{x\to1}\frac{(x+4)(x-1)}{(x-1)(x+2)} = \lim_{x\to1}\frac{x+4}{x+2}.$$
+
+                $\frac{x+4}{x+2}$ is continuous at $x=1$, therefore the limit equals $\frac{1+4}{1+2}=\frac53$.
+        \end{enumerate}
+\end{enumerate}
+
+\end{document}
+