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diff --git a/assignment3.tex b/assignment3.tex new file mode 100644 index 0000000..ba1a7c4 --- /dev/null +++ b/assignment3.tex @@ -0,0 +1,111 @@ +\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} + |