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\begin{center}\textbf{Problem Set 5 \hspace{.7in} Foundations \hspace{.7in} Due 3/4/24} \end{center}

\noindent Four of these problems will be graded (my choice, not yours!), with each problem worth 5 points.  Clear and complete justification is required for full credit.  You are welcome to discuss these problems with anyone and everyone, but must write up your own final submission without reference to any sources other than the textbook and instructor.  Submit as a pdf on Moodle.

\begin{enumerate}

\item If $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ are even functions with $g(x) \not= 0, \ \forall x \in \mathbb{R}$, then $f/g$ is an even function.

\item If $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ are odd, then $f - g$ is odd.

\item If $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ are increasing, then $f \cdot g$ is increasing.

\item If $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ are decreasing, then can we know $f-g$ is increasing? decreasing? not increasing? not decreasing?

\item If $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R} \to \mathbb{R}$ have positive derivatives $\forall x \in \mathbb{R}$, then $f+g$ is increasing.

\item Let $f$ and $g$ be bounded functions, both with domain $D$.  Then $f+g$ is a bounded function.

\item Let $f$ and $g$ be bounded functions, both with domain $D$.  Then $f-g$ is a bounded function.

\item Let $f$ and $g$ be bounded functions, both with domain $D$.  Then $f \cdot g$ is a bounded function.

\item Let $f$ and $g$ be bounded functions, both with domain $D$.  Then $f/g$ is a bounded function.

\item Critique the following proof of the proposition ``The product of two even functions, both with domain $D$, is even'': \\
Well, let $f$ and $g$ be two even functions, both with domain $D$, and let $f(x) = 0$ and $g(x) = x^2$, which we know are both even.  Then consider their product $f \cdot g$. We have 
$$(f \cdot g)(-x) = f(-x) \cdot g(-x) = 0 \cdot (-x)^2 = 0 = 0 \cdot x^2 = (f \cdot g)(x).$$  So the product is even, as desired.

\end{enumerate}

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