\part{Solutions}
\chapter*{Functions}
\section*{Definition and examples of functions}
%	\noindent\begin{tcolorbox}[colback=red!5!white,colframe=red!75!black]
	\subsection*{Lösung 1}
	\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 $x \rightarrow \infty$, $f(x) \rightarrow \infty$
	%\end{tcolorbox}
	\subsection*{Lösung 2}	
	%	\noindent\fbox{%
		%		\parbox{\textwidth}{
			\label{FunctionsAnswer2}
			For $\sin$ and cos: domain is $\mathbb{R}$; range is $[-1,1]$.
			For tan, the domain is $\left\{x \in \mathbb{R}: x \neq \frac{\pi}{2}+k \pi\right\}$; range is $\mathbb{R}$.
			%	}}
	%	
	\subsection*{Lösung 3}		
	%	\noindent\fbox{%
		%		\parbox{\textwidth}{
			\label{FunctionsAnswer3}
			Domain is $\mathbb{R}$; range is $(0, \infty)$.
			%	}}
	
	\subsection*{Lösung 4}		
	%		\noindent\begin{tcolorbox}[colback=red!5!white,colframe=red!75!black]
		\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.
		%	\end{tcolorbox}
	
	\subsection*{Lösung 5}		
	%	\noindent\begin{tcolorbox}[colback=red!5!white,colframe=red!75!black]
		
		\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.
		%\end{tcolorbox}
		
%		
\section*{Operations on Functions}
		\subsection*{COMPOSITION}	
		\subsubsection*{Lösung 1}			
		\label{FunctionsAnswer6}We find that
		$$
		\begin{aligned}
			f \circ f(x) & =f(f(x)) \\
			& =f\left(\frac{1}{x+1}\right) \\
			& =\frac{1}{1 /(x+1)+1} \\
			& =\frac{x+1}{1+x+1} \\
			& =\frac{x+1}{x+2} .
		\end{aligned}
		$$
		\section*{Classes of Functions}
%		\subsection{POLYNOMIALS}
%		\subsection{RATIONAL FUNCTIONS}
		\subsection*{POWERS}
  		\subsubsection*{Lösung 1}		
		\label{FunctionsAnswer7}$x^0=1$
		\subsubsection*{Lösung 2}		
		\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, $x^{-\frac{1}{2}}=\frac{1}{\sqrt{x}}$.
		\subsubsection*{Lösung 3}		
		\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.
		
		\subsection*{Additional Examples}
		\subsubsection*{Example 1}		
		\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
		$$
		(x-2)(x-1)>0 \text {. }
		$$
		
		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 $x>2$ :
		
		\begin{figure}
			\centering
			\includegraphics[width=0.7\linewidth]{PointChecking}
			%\caption{}
			\label{fig:pointchecking}
		\end{figure}
		
		
		So the domain of $f$ is $x<1$ or $2<x$. In interval notation, this is $(-\infty, 1) \cup(2, \infty)$.
		$$
		x^{\frac{22}{7}}=\sqrt[7]{x^{22}}
		$$
		
		\subsubsection*{Example 2}		
		\label{FunctionsAnswer11}
		Since $\ln$ is only defined on the positive real numbers, we must have $x^3-6 x^2+8 x>0$. Factoring gives
		$$
		x\left(x^2-6 x+8\right)=x(x-2)(x-4)>0
		$$
		
		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 $(0,2) \cup(4, \infty)$.
		

		\chapter*{EXPONENTIALS}
		\section*{Euler's formula}
		\subsection*{Answer 1}
		
		\label{ExponentialsAnswer1}
		Setting $x=\pi$ in Euler's formula gives $e^{i \pi}=\cos \pi+i \sin \pi=-1$.
		
		\subsection*{Answer 2}
		\label{ExponentialsAnswer2}
		Note that this is the long polynomial for $\cos x$, evaluated at $x=\pi$. So the value is $\cos \pi=-1$.
		
		\subsection*{Answer 3}
		\label{ExponentialsAnswer3}
		Computing the derivative term by term gives
		$$
		\begin{aligned}
			\frac{d}{d x} \sin (x) & =\frac{d}{d x}\left(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\ldots\right) \\
			& =1-3 \frac{x^2}{3!}+5 \frac{x^4}{5!}-\ldots \\
			& =1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots
		\end{aligned}
		$$
		which is the long polynomial for $\cos x$, as desired.

		\subsection*{Answer 4}
		\label{ExponentialsAnswer4}
		Beginning with $e^x \cdot e^y$, we find
		$$
		\begin{aligned}
			e^x \cdot e^y & =\left(1+x+\frac{x^2}{2!}+\cdots\right)\left(1+y+\frac{y^2}{2!}+\cdots\right) \\
			& =1+(x+y)+\left(\frac{x^2}{2!}+x y+\frac{y^2}{2!}\right)+\cdots \\
			& =1+(x+y)+\frac{x^2+2 x y+y^2}{2!}+\cdots \\
			& =1+(x+y)+\frac{(x+y)^2}{2!}+\cdots,
		\end{aligned}
		$$
		which is the long polynomial for $e^{x+y}$, as desired.
		
%		\subsection{Euler's Formula}
%		
%		\subsection{Additional Examples}
%		\subsection{Lösung 1}			
%		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
%		$$
%		(x-2)(x-1)>0 \text {. }
%		$$
%		
%		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 $x>2$ :
%		
%		\begin{figure}[h]
%			\centering
%			\includegraphics[width=0.5\linewidth]{Upenn004}
%	%		\caption{}
%			\label{fig:upenn004}
%		\end{figure}
%		
%		
%		So the domain of $f$ is $x<1$ or $2<x$. In interval notation, this is $(-\infty, 1) \cup(2, \infty)$.
%		
%		\subsection{Lösung 2}					
%		
%		Since $\ln$ is only defined on the positive real numbers, we must have $x^3-6 x^2+8 x>0$. Factoring gives
%		$$
%		x\left(x^2-6 x+8\right)=x(x-2)(x-4)>0
%		$$
%		
%		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 $(0,2) \cup(4, \infty)$.
%		
%		\chapter{The Exponential}
%		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