\documentclass[12pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{hyperref}
\usepackage{amsthm}
\begin{document}
\begin{center}
Problems PreCal 1508 Review, \today \\
PL: Alex Knaust, Lecturer: Yi-Yu Liao
\end{center}
\begin{enumerate}
\item Find the \emph{slope-intercept form} of the equation of the line passing through the two points $(1, -2), (2, -3)$.

\begin{proof}[Solution]
The first step is to determine the slope of the line (or use the point-point formula). The slope is given by
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - (-2)}{2 - 1} = \frac{-1}{1} = -1 \]
Thus we have $y = -x + b$ and we know that it go through the point $(1, -2)$. So in order to find $b$, we can simply plug in the point and solve
\[-2 = -(1) + b \Leftrightarrow -1 = b \]
So the \emph{slope-intercept form} of the equation is $y = -x - 1$ 
\end{proof}

\item Find the \emph{slope-intercept form} of the equation of a line passing through the point $(-9, 9)$ and perpendicular to line $y = 3x - 2$

\item Determine the implied domain for the following functions
\begin{enumerate}
\item $f(x) = x^3 + 2x + 9$
\begin{proof}[Solution]
The domain for all polynomials is known to be all real numbers, so the domain is $\mathbb{R}$ or alternatively $(-\infty, \infty)$.
\end{proof}
\item $h(x) = \frac{2x}{\sqrt{x - 3}}$
\item $q(x) = \frac{9}{x^2 - 9}$
\begin{proof}[Solution]
We can see that $q(x)$ is a rational function, a polynomial divided by another polynomial. And we know that we cannot divide by zero, thus we must exclude from the domain whatever values of $x$ make the denominator $0$. The denominator is $x^2 - 9$, and so we set $0 = x^2 - 9$ and solve.
\[ 0 = x^2 - 9 \Leftrightarrow 9 = x^2 \Leftrightarrow x = \pm 3 \]
So $-3, 3$ are not in the domain, and the domain can be written as $\mathbb{R} \setminus \{-3, 3\}$ or $(-\infty, -3)\cup(-3, 3)\cup(3,\infty)$
\end{proof}
\end{enumerate}

\item For the following functions find $f \circ g$, $f - g$, $\frac{f}{g}$
\begin{enumerate}
\item $f(x) = 3x - 9,\quad g(x) = x^2 + 7$
\item $f(x) = \sqrt{x - 1},\quad g(x) = x^4 + 2$
\begin{proof}[Solution]
	\begin{enumerate}
	\item \[(f \circ g)(x) = f(g(x)) = \sqrt{(x^4 + 2) - 1} = \sqrt{x^4 + 1} \neq x^2 + 1 \]
	\item \[(f - g)(x) = f(x) - g(x) = \sqrt{x - 1} - (x^4 + 2) = \sqrt{x - 1} - x^4 - 2\]
	\item \[\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x - 1}}{x^4 + 2}\]
	Note that since the denominator doesn't have any real zeros, that the domain of $\frac{f}{g}$ is all real numbers except those that would prevent you from taking the square root in the numerator. So the domain is all reals $x$, so that $x-1 \geq 0 \Longleftrightarrow x \geq 1$.
	\end{enumerate}
\end{proof}
\end{enumerate}

\item Find the vertex and the x-intercepts of the following quadratic functions
\begin{enumerate}
\item $f(x) = x^2 - 9x + 2$
\item $f(x) = 4x - 1 - 4x^2$
\begin{proof}[Solution]
We can rewrite $f(x)$ as $f(x) = ax^2 + bx + c$, so it is $f(x) = -4x^2 + 4x - 1$. The vertex of a quadratic function is given by $(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)$. Since $a=-4$ and $b=4$ we have...
\[-\frac{b}{2a} = -\frac{4}{2(-4)} = -\frac{1}{-2} = \frac{1}{2} \]
and
\[ f\left(\frac{1}{2}\right) = -4\left( \frac{1}{2} \right)^2 + 4\left( \frac{1}{2} \right) - 1 = -1 + 2 -1 = 0\]
The vertex is on the $x-$axis, so we can deduce that it is the only $x$ intercept, however you can check this with the discriminant, $b^2 - 4ac$
\[ b^2 - 4ac = 4^2 - 4(-4)(-1) = 16 - 16 = 0\]. So the quadratic formula is 
\[ x = \frac{-b \pm 0}{2a} \]
Which is the same as the x-coordinate of the vertex.
\end{proof}
\end{enumerate}

\item Find the inverse function (if it exists) of the following functions
\begin{enumerate}
\item $f(x) = \frac{2}{\sqrt{x}}$
\begin{proof}[Solution]
Graphing will reveal that the function passes the horizontal line test, since the function is \emph{one-to-one}. We can continue by trying to find the inverse function. First write $f(x)$ as $y = \frac{2}{\sqrt{x}}$ and then switch the two variables and solve for $y$...
\[x = \frac{2}{\sqrt{y}} \Longleftrightarrow \sqrt{y}x = 2 \Longleftrightarrow  \sqrt{y} = \frac{2}{x} \Longleftrightarrow y = \frac{4}{x^2} = f^{-1}(x) \]
We still must consider the domain of $f^{-1}(x)$. Which we know to be the range of $f(x)$. The range of $f(x)$ is all positive real numbers except $0$, $(0, \infty)$ (as graphing would reveal), thus the domain of the inverse function is all positive real numbers except $0$.
\end{proof}
\item $h(x) = (x - 2)^2 - 9$
\begin{proof}[Solution]
Squaring usually means there will be no inverse function, we can verify this by graphing or by noticing that for $x = -2$ and $x = 6$
\[ h(-2) = ((-2) - 2)^2 -9 = (-4)^2 - 9 = 16 - 9 = 7 \]
\[ h(6) = (6 - 2)^2 - 9 = 16 -9  = 7 \]
Thus the function cannot pass the horizontal line test.
\end{proof}
\end{enumerate}

\item Find all zeros of the functions (Real and Imaginary)
\begin{enumerate}
\item $f(x) = x^2 + 1$
\begin{proof}[Solution]
There are exactly $2$ roots, since the degree of $f(x)$ is 2, so we apply the quadratic formula!
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \Longleftrightarrow x = \frac{\pm \sqrt{0-4}}{2} \Longleftrightarrow x = \frac{\pm 2\sqrt{-1}}{2} \Longleftrightarrow x = \pm i \]
So the roots are $\pm i$ and we can write $f(x)$ \\
as $f(x) = (x - i)(x + i)$
\end{proof}
\item $f(x) = x^2 + x + 2$
\item $f(x) = 2x^4 - 6x^2 - 4x^2 + 20x - 24,\quad \text{given  } f(1 + i) = 0$
\end{enumerate}

\item Write $f(x)$ as $f(x) = q(x)d(x) + r(x)$ (Remainder Theorem)
\begin{enumerate}
\item $f(x) = x^3 + 2x + 9,\quad d(x) = x - 3$
\begin{proof}[Solution]
Use Long or synthetic division to get the quotient and remainder
\[f(x) = (x-3)(11 + 3x +x^2) + 42 \]
\end{proof}
\item $f(x) = x^5 + 3x^4 + 9x + 2, \quad d(x) = x + 2$
\end{enumerate}

\item Determine if $-3$ is a root of $f(x) = 2x^4 - 6x^3 - 38x^2 + 54x + 180$ (use synthetic division)
\begin{proof}[Solution]
It is. You can divide by $2$ first to make your work easier. $f(x) = 2x^4 - 6x^3 - 38x^2 + 54x + 180 = 2(x-5)(x-3)(x+2)(x+3)$
\end{proof}

\item Write the definitions of the following
\begin{enumerate}
\item The inverse function of $f$
\item A polynomial
\item Root of a function
\item Rational Function
\end{enumerate}
\end{enumerate}
\begin{center}
(Select) Solutions to the problems will be available at \url{http://rylai.dyndns.org/review\_1.pdf}\\
You can reach me via e-mail at \url{awknaust@miners.utep.edu}\\
I expect you all to do well on the exam! Good Luck!
\end{center}
\end{document}