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Sven Riwoldt
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solutions.tex
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solutions.tex
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\part{Solutions}
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\chapter*{Functions}
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\section*{Definition and examples of functions}
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% \noindent\begin{tcolorbox}[colback=red!5!white,colframe=red!75!black]
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\subsection*{Lösung 1}
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\label{FunctionsAnswer1}The domain is $\mathbb{R}$, because we can plug in any real number into a polynomial. The range is $\mathbb{R}$, which we see by noting that this is a cubic function, so as $x \rightarrow-\infty, f(x) \rightarrow-\infty$, and as $x \rightarrow \infty$, $f(x) \rightarrow \infty$
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%\end{tcolorbox}
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\subsection*{Lösung 2}
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% \noindent\fbox{%
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% \parbox{\textwidth}{
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\label{FunctionsAnswer2}
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For $\sin$ and cos: domain is $\mathbb{R}$; range is $[-1,1]$.
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For tan, the domain is $\left\{x \in \mathbb{R}: x \neq \frac{\pi}{2}+k \pi\right\}$; range is $\mathbb{R}$.
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% }}
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%
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\subsection*{Lösung 3}
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% \noindent\fbox{%
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% \parbox{\textwidth}{
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\label{FunctionsAnswer3}
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Domain is $\mathbb{R}$; range is $(0, \infty)$.
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% }}
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\subsection*{Lösung 4}
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% \noindent\begin{tcolorbox}[colback=red!5!white,colframe=red!75!black]
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\label{FunctionsAnswer4}Domain is $(0, \infty)$; range is $\mathbb{R}$. Notice how the domain and range of the exponential relate to the domain and range of the natural logarithm.
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% \end{tcolorbox}
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\subsection*{Lösung 5}
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% \noindent\begin{tcolorbox}[colback=red!5!white,colframe=red!75!black]
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\label{FunctionsAnswer5}$\sin ^{-1}$ is not a function, because one input has many outputs. For example, $\sin ^{-1}(0)=0, \pi, 2 \pi, \ldots$. By restricting the range of $\sin ^{-1}$ to $\left\lfloor-\frac{\pi}{2}, \frac{\pi}{2}\right\rfloor$, one gets the function arcsin.
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%\end{tcolorbox}
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%
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\section*{Operations on Functions}
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\subsection*{COMPOSITION}
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\subsubsection*{Lösung 1}
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\label{FunctionsAnswer6}We find that
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$$
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\begin{aligned}
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f \circ f(x) & =f(f(x)) \\
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& =f\left(\frac{1}{x+1}\right) \\
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& =\frac{1}{1 /(x+1)+1} \\
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& =\frac{x+1}{1+x+1} \\
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& =\frac{x+1}{x+2} .
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\end{aligned}
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$$
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\section*{Classes of Functions}
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% \subsection{POLYNOMIALS}
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% \subsection{RATIONAL FUNCTIONS}
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\subsection*{POWERS}
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\subsubsection*{Lösung 1}
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\label{FunctionsAnswer7}$x^0=1$
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\subsubsection*{Lösung 2}
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\label{FunctionsAnswer8}Recall a fractional power denotes root. For example, $x^{\frac{1}{2}}=\sqrt{x}$. The negative sign in the exponent means that we take the reciprocal. So, $x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}$.
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\subsubsection*{Lösung 3}
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\label{FunctionsAnswer9}One can rewrite this as $\left(x^{22}\right)^{1 / 7}$. That means we take $x$ to the 22 nd power and then take the 7 th root of the result.
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\subsection*{Additional Examples}
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\subsubsection*{Example 1}
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\label{FunctionsAnswer10}Note that the square root is only defined when its input is non-negative. Also, the denominator in a rational function cannot be 0 . So we find that this function is well-defined if and only if $x^2-3 x+2>0$. Factoring gives
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$$
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(x-2)(x-1)>0 \text {. }
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$$
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By plotting the points $x=1$ and $x=2$ (where the denominator equals 0 ) and testing points between them, one finds that $x^2-3 x+2>0$ when $x<1$ or $x>2$ :
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\begin{figure}
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\centering
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\includegraphics[width=0.7\linewidth]{PointChecking}
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%\caption{}
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\label{fig:pointchecking}
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\end{figure}
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So the domain of $f$ is $x<1$ or $2<x$. In interval notation, this is $(-\infty, 1) \cup(2, \infty)$.
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$$
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x^{\frac{22}{7}}=\sqrt[7]{x^{22}}
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$$
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\subsubsection*{Example 2}
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\label{FunctionsAnswer11}
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Since $\ln$ is only defined on the positive real numbers, we must have $x^3-6 x^2+8 x>0$. Factoring gives
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$$
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x\left(x^2-6 x+8\right)=x(x-2)(x-4)>0
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$$
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As in the above example, plotting the points where this equals 0 and then testing points, we find that the domain is $0<x<2$ and $4<x$. In interval notation, this is $(0,2) \cup(4, \infty)$.
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\chapter*{EXPONENTIALS}
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\section*{Euler's formula}
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\subsection*{Answer 1}
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\label{ExponentialsAnswer1}
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Setting $x=\pi$ in Euler's formula gives $e^{i \pi}=\cos \pi+i \sin \pi=-1$.
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%
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% \subsection{Euler's Formula}
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%
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% \subsection{Additional Examples}
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% \subsection{Lösung 1}
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% Note that the square root is only defined when its input is non-negative. Also, the denominator in a rational function cannot be 0 . So we find that this function is well-defined if and only if $x^2-3 x+2>0$. Factoring gives
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% $$
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% (x-2)(x-1)>0 \text {. }
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% $$
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%
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% By plotting the points $x=1$ and $x=2$ (where the denominator equals 0 ) and testing points between them, one finds that $x^2-3 x+2>0$ when $x<1$ or $x>2$ :
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%
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% \begin{figure}[h]
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% \centering
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% \includegraphics[width=0.5\linewidth]{Upenn004}
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% % \caption{}
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% \label{fig:upenn004}
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% \end{figure}
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%
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%
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% So the domain of $f$ is $x<1$ or $2<x$. In interval notation, this is $(-\infty, 1) \cup(2, \infty)$.
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%
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% \subsection{Lösung 2}
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%
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% Since $\ln$ is only defined on the positive real numbers, we must have $x^3-6 x^2+8 x>0$. Factoring gives
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% $$
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% x\left(x^2-6 x+8\right)=x(x-2)(x-4)>0
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% $$
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%
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% As in the above example, plotting the points where this equals 0 and then testing points, we find that the domain is $0<x<2$ and $4<x$. In interval notation, this is $(0,2) \cup(4, \infty)$.
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%
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% \chapter{The Exponential}
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% This module deals with a very important function: the exponential. The first question one might ask is: what is the exponential function $e^{\text {? }}$. We know certain values of the function such as $e^0=1$, but what about an
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