\documentclass[10pt,a4paper,addpoints]{exam}
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\begin{document}
\begin{coverpages}
\begin{center}{\huge MATH 1112 - EXAM 2 - FALL 2016}\\
\huge SOLUTION \\
\end{center}
\begin{flushleft}\vspace{0.5in}
Friday 14 October 2016\\
Instructor: Tom Cuchta
\end{flushleft}
\textbf{Instructions:}
\begin{itemize}
	\item  Show all work, clearly and in order, if you want to get full
	credit. If you claim something is true \textbf{you must show work backing up your claim}. I reserve the right to take off points if I cannot see how you arrived at your answer (even if your final answer is correct).
	\item Justify your answers algebraically whenever possible to ensure full credit. 	
	\item  Circle or otherwise indicate your final answers.
	\item  Please keep your written answers brief; be clear and to the point.
	\item  Good luck!
\end{itemize}
\vspace*{10pt}
\end{coverpages}

\begin{questions}
	\question[12] For the following function, state where it is increasing, decreasing, and constant.
	\begin{center}
		\begin{tikzpicture}[scale=0.75]
			\draw[very thick,<->] (0,-4)--(0,5);
			\draw[very thick,<->] (-4,0) -- (5,0);
			\draw[very thick] (-3,2) -- (-1,-2);
			\draw[very thick] (-1,-2) -- (1,-1.5);
			\draw[very thick] (1,-1.5) -- (2,3);
			\draw[very thick] (2,3) -- (4,3);
			
			\draw[fill=black] (-1,-2) circle (0.75ex) node[below] {$(-1,-2)$};
			\draw[fill=black] (1,-1.5) circle (0.75ex) node[below right] {$(1,-\frac{3}{2})$};
			\draw[fill=black] (2,3) circle (0.75ex) node[above] {$(2,3)$};
			\draw[fill=black] (4,3) circle (0.75ex) node[above] {$(4,3) $};
			\draw[fill=black] (-3,2) circle (0.75ex) node[above] {$(-3,2)$};
			\end{tikzpicture} \\
		\end{center}		
		Increasing: \underline{$(-1,2)$}\hfill Decreasing:\underline{$(-3,-1)$}\hfill Constant:\underline{$(2,4)$}\\
		
	\question[8] Let $$f(x)=\begin{cases}
						x+7, & -\infty<x< 5\\
						-2x+1, & 5 \leq x < \infty \\
	\end{cases} $$
		\begin{parts}
			\part[2] What is the value of $f(-10)$? \\
			\textit{Solution:} Since $-\infty < -10 < 5$,
			$$f(-1)=-10+7=-3.$$
			\part[2] What is the value of $f(2)$? \\
			\textit{Solution:} Since $-\infty < 2 < 5$,
			$$f(2)=2+7=9.$$
			\part[4] Graph $y=f(x)$. \\
			\textit{Solution:} \\
			\begin{tikzpicture}[scale=0.3]
				\draw[very thick, <->] (0,-10)--(0,20);
				\draw[very thick,<->] (-15,0)--(15,0);
				\draw[very thick,->] (5,12) -- (13,20);
				\draw[very thick,->] (5,-9) -- (-9,19);
				\draw[fill=white] (5,12) circle (1.7ex);
				\draw[fill=black] (5,-9) circle (1.7ex);
				\draw node at (7,-9) {$(5,-9)$};
				\draw node at (7,12) {$(7,12)$};
			\end{tikzpicture}
			
		\end{parts}
	
	\newpage
	
	\question[10] Let $f(x)=\sqrt{x}$ and $g(x)=3x-2$. 
		\begin{parts}
			\part[3] Compute $\left(f+g\right)(2)$ \\
			\textit{Solution:} Compute
			$$(f+g)(2)=f(2)+g(2)=\sqrt{2} + (3 \cdot 2-2) = \sqrt{2}+4.$$
			\part[3] Compute $\left(f-g\right)(x)$\\
			\textit{Solution:} Compute
			$$(f-g)(x)=f(x)-g(x)=\sqrt{x}-(3x-2)=\sqrt{x}-3x+2.$$
			\part[4] What is the domain of $\left( \dfrac{f}{g} \right)$? \\
			\textit{Solution:} Notice that 
			$$\left( \dfrac{f}{g} \right)(x) = \dfrac{f(x)}{g(x)}=\dfrac{\sqrt{x}}{3x-2}.$$
			The top forces $x \geq 0$ (which could be written as the interval $[0,\infty)$ and we must also remove any values of $x$ that makes $g(x)=0$, i.e. we must remove $x$ such that $3x-2=0$, i.e. we must remove $x$ such that $x=\dfrac{2}{3}$. Therefore the domain is 
			$$\left[ 0, \dfrac{2}{3} \right) \bigcup \left( \dfrac{2}{3}, \infty \right).$$
		\end{parts}
		
	\question[13] Suppose $f(x)=x^2+2$ and $g(x)=\sqrt{x-3}$. 
		\begin{parts}
			\part[3] Compute $\left(f\circ g\right)(5)$ \\
			\textit{Solution:} First note that 
			$$g(5)=\sqrt{5-3}=\sqrt{2}.$$
			Now compute
			$$(f \circ g)(5) = f(g(5)) = f(\sqrt{2})= (\sqrt{2})^2+2=2+2=4.$$
			\part[3] Compute $\left(f\circ g\right)(x)$ \\
			\textit{Solution:} Compute
			$$f(g(x))=f(\sqrt{x-3})=\left( \sqrt{x-3} \right)^2+2=x-3+2=x-1.$$				
			\part[3] Compute $\left(g\circ f\right)(1)$ \\
			\textit{Solution:} First note that
			$$f(1)=1^2+2=1+2=3.$$
			Now compute
			$$(g \circ f)(1)=g(f(1))=g(3) = \sqrt{3-3} = \sqrt{0}=0.$$
			\part[4] What is the domain of $\left(f\circ g\right)(x)$? \\
			\textit{Solution:} Note that
			$$(f \circ g)(x) = f(g(x)) = f(\sqrt{x-3}).$$
			Since $f$ is a polynomial, it will not restrict the domain. Since $g(x)=\sqrt{x-3}$ we must have $x-3 \geq 0$, i.e. $x \geq 3$. \\
			\textit{note: If the computation is continued, we see that $f(g(x))=x-1$, which is a polynomial. This does not mean the domain of $f \circ g$ is $\mathbb{R}$ -- we have to use the restriction given by $g$ since $g$ is part of the composition.}
		\end{parts}
		
	\newpage
	
	\question[9] Test the following for any symmetry. Circle yes or no for each option. \textit{Failure to show the work behind your decision will cause you to lose points.}
					$$3x^2-x+y^2=0$$ \\
					\textit{Solution:} To test for $x$-axis symmetry, replace $y$ with $-y$ and compare to the original:
					$$3x^2-x+(-y)^2=0.$$
					But $(-y)^2=(-y)(-y)=y^2$, so this is equivalent to
					$$3x^2-x+y^2=0,$$
					which matches the original. Hence we have $x$-axis symmetry. \\
					To test for $y$-axis symmetry, replace $x$ with $-x$ and compare to the original:
					$$3(-x)^2-(-x)+y^2=0.$$
					But $(-x)^2=x^2$ so we have
					$$3x^2+x+y^2=0,$$
					which does not match the original, so we do not have $y$-axis symmetry. \\
					To test for origin symmetry, replace $x$ with $(-x)$ and replace $y$ with $(-y)$ to get
					$$3(-x)^2-(-x)+(-y)^2=0,$$
					which is equivalent to
					$$3x^2+x+y^2=0,$$
					which does not match the original. Therefore it does not have origin symmetry.
		
	\question[6] Determine if the following function is even, odd, or neither: $$f(x) = 3x^2-3$$ 
	\textit{Failure to show the reasoning behind your decision will cause you to lose points.} \\
	\textit{Solution:} Compute
	$$f(-x)=3(-x)^2-3=x^2-3=f(x),$$
	therefore $f$ is even.
	
	\newpage 
	
	\question[16] Consider $y = -(x-2)^2$. Fill in the following information about transformations on the function. If a specific transformation is not used, state so.\\
	
		\begin{parts}
			\part[2] Base Function: $x^2$
			
			\part[2] Reflections: vertical reflection \\
			
			\part[2] Horizontal Stretch: none \\
			
			\part[2] Vertical Stretch: none \\
			
			\part[2] Horizontal Shift: $2$ to the right \\
			
			\part[2] Vertical Shift: none \\
			
			\part[4] Graph $y=-(x-2)^2$ with appropriate labeling. \\
			\textit{Solution:} \\
		\begin{tabular}{lll}
			\begin{minipage}{0.33\textwidth}
				\begin{tikzpicture}
					\draw[very thick] (0,0) -- (0,4);
					\draw[very thick] (-2,0) -- (2,0);
					\draw[fill=black] (0,0) circle (0.6ex);
					\draw[fill=black] (1,1) circle (0.6ex);
					\draw node at (1.5,0.5) {$(1,1)$};
					\draw node at (0.5,-0.5) {$(0,0)$};
					\draw[domain=-2:2,very thick] plot ({\x},{\x*\x});
					\draw node at (0,4.5) {$y=x^2$};
				\end{tikzpicture} 
			\end{minipage}
			&\begin{minipage}{0.33\textwidth}
				\begin{tikzpicture}
					\draw node at (-3,2) {\scalebox{2}{$\stackrel{\mathrm{h.shift}}{\longrightarrow}$}};
					\draw[very thick] (0,0) -- (0,4);
					\draw[very thick] (0,0) -- (4,0);
					\draw[fill=black] (2,0) circle (0.6ex);
					\draw[fill=black] (3,1) circle (0.6ex);
					\draw node at (3.5,0.5) {$(3,1)$};
					\draw node at (2.5,-0.5) {$(2,0)$};
					\draw[domain=0:4,very thick] plot ({\x},{(\x-2)*(\x-2)});
					\draw node at (2,4.5) {$y=(x-2)^2$};
				\end{tikzpicture} 			
			\end{minipage} \\
			&	\begin{tikzpicture}
					\draw node at (-3,2) {\scalebox{2}{$\stackrel{\mathrm{v.refl}}{\longrightarrow}$}};
					\draw[very thick] (0,2) -- (0,-4);
					\draw[very thick] (0,0) -- (4,0);
					\draw[fill=black] (2,0) circle (0.6ex);
					\draw[fill=black] (3,-1) circle (0.6ex);
					\draw node at (3.5,-0.5) {$(3,-1)$};
					\draw node at (2.5,0.5) {$(2,0)$};
					\draw[domain=0:4,very thick] plot ({\x},{-(\x-2)*(\x-2)});
					\draw node at (2,1.5) {$y=-(x-2)^2$};
				\end{tikzpicture} 					
	\end{tabular}
		\end{parts}	
	\newpage
	
	\question[16] Let $g(x) = 2\sqrt{x+3}-2$. Fill in the following information about transformations on the function. If a specific transformation is not used, state so.\\
	
	\begin{parts}
		\part[2] Base Function: $\sqrt{x}$ \\
		\part[2] Reflections: none \\
		\part[2] Horizontal Stretch: none \\
		\part[2] Vertical Stretch: none \\
		\part[2] Horizontal Shift: left $3$ \\
		\part[2] Vertical Shift: down $2$ \\
		
		\part[4] Graph $y=2\sqrt{x+3}-2$ with appropriate labeling.
	\end{parts}
		\begin{tabular}{lll}
			\begin{minipage}{0.33\textwidth}
				\begin{tikzpicture}
					\draw[very thick] (0,0) -- (0,2);
					\draw[very thick] (-2,0) -- (2,0);
					\draw[fill=black] (0,0) circle (0.6ex);
					\draw[fill=black] (1,1) circle (0.6ex);
					\draw node at (1.5,0.5) {$(1,1)$};
					\draw node at (0.5,-0.5) {$(0,0)$};
					\draw[domain=0:3,samples=100,very thick] plot ({\x},{\x^(0.5)});
					\draw node at (1,2.5) {$y=\sqrt{x}$};
				\end{tikzpicture} 
			\end{minipage}
			&\begin{minipage}{0.33\textwidth}
				\begin{tikzpicture}
				\draw node at (-5,1) {\scalebox{2}{$\stackrel{\mathrm{h.shift}}{\longrightarrow}$}};
					\draw[very thick] (0,0) -- (0,2);
					\draw[very thick] (-4,0) -- (1,0);
					\draw[fill=black] (-3,0) circle (0.6ex);
					\draw[fill=black] (-2,1) circle (0.6ex);
					\draw node at (-3,-0.5) {$(-3,0)$};
					\draw node at (-2,1.5) {$(-1,1)$};
					\draw[domain=-3:1,samples=100,very thick] plot ({\x},{(\x+3)^(0.5)});
					\draw node at (-1,2.5) {$y=\sqrt{x+3}$};
				\end{tikzpicture} 
			\end{minipage} \\
			&\begin{tikzpicture}
				\draw node at (-5,1) {\scalebox{2}{$\stackrel{\mathrm{v.stretch}}{\longrightarrow}$}};
					\draw[very thick] (0,0) -- (0,4);
					\draw[very thick] (-4,0) -- (1,0);
					\draw[fill=black] (-3,0) circle (0.6ex);
					\draw[fill=black] (-2,2) circle (0.6ex);
					\draw node at (-3,-0.5) {$(-3,0)$};
					\draw node at (-2.2,2.5) {$(-1,2)$};
					\draw[domain=-3:1,samples=100,very thick] plot ({\x},{2*(\x+3)^(0.5)});
					\draw node at (-1.5,3.5) {$y=2\sqrt{x+3}$};
				\end{tikzpicture} \\
			&\begin{tikzpicture}
				\draw node at (-5,1) {\scalebox{2}{$\stackrel{\mathrm{v.shift}}{\longrightarrow}$}};
					\draw[very thick] (0,-2) -- (0,2);
					\draw[very thick] (-4,0) -- (1,0);
					\draw[fill=black] (-3,-2) circle (0.6ex);
					\draw[fill=black] (-2,0) circle (0.6ex);
					\draw node at (-3,-2.5) {$(-3,-2)$};
					\draw node at (-2.2,0.5) {$(-1,0)$};
					\draw[domain=-3:1,samples=100,very thick] plot ({\x},{2*(\x+3)^(0.5)-2});
					\draw node at (-1.5,1.5) {$y=2\sqrt{x+3}-2$};
				\end{tikzpicture} 
		\end{tabular}	
	\newpage
	
	\question[10] Do the follow two problems.
	\begin{parts}
	\part[5] Hooke's law states that the distance $d$ that a spring will stretch is directly proportional to the mass $m$ of an object hanging from the spring. A $2$ $\mathrm{kg}$ mass stretches the spring $40$ $\mathrm{cm}$. How far will a $5$ $\mathrm{kg}$ mass stretch the spring? \\
	\textit{Solution:} The first sentence gives us the equation
	$$d = km.$$
	To find $k$ we plug in $d=40$ and $m=2$ to get
	$$40=2k,$$
	or
	$$20=k.$$
	This means the equation is actually
	$$d=20m.$$
	Now plug in $m=5$ to find the requested value for $d$:
	$$d=20(5)=100.$$
	\part[5] The current $I$ (in amperes) in a circuit is inversely proportional to its resistance $R$ (in ohms). When the current in a given circuit is $30$ amperes, the resistance is $8$ ohms. Find the resistance in the circuit when the current is $2$ amperes. \\
	\textit{Solution:} The first sentence gives us the equation
	$$I=\dfrac{k}{R}.$$
	To find $k$, we plug in $I=30$ and $R=8$ to get
	$$30 = \dfrac{k}{8},$$
	or
	$$k=240.$$
	Therefore the equation is
	$$I = \dfrac{240}{R}.$$
	Now plug in $I=2$ to find the requested value for $R$:
	$$2 = \dfrac{240}{R},$$
	so
	$$R = \dfrac{240}{2}=120.$$

		\end{parts}
\end{questions}

\end{document}