% TEMPLATE FOR STANDARD EXAM
% (with some sample problem types)
% by laura


\documentclass[11pt]{article}

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\begin{document}

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\section*{Exam Practice Problem Solutions}

\begin{enumerate}
\item h
\item f
\item c
\item e
\item f

\item f
\item g
\item c
\item c
\item f

\item $xe^x+37e^x$
\item d
\item d
\item g
\item c

\item b
\item b
\item c
\item d
\item g

\item c
\item c
\item h
\item $\left[\frac{2}{x}-\tan x-\frac{2x}{x^2+4}-\frac{1}{x-1}-\frac{1}{x\ln x}\right]\left(\frac{x^2\cos x}{(x^2+4)(x-1)\ln x}\right)$
\item 3/8

\item g
\item g
\item g
\item a
\item f

\item b
\item h
\item h
\item $x\sin x-36\cos x$
\item $\frac{48}{25\sqrt{25^2-12^2}}$

\item $x=2$.  (Be careful about signs; $x=6$ is \emph{not} a critical point, so $x=2$ is the only critical point.)  Use the first derivative test to see that this is optimal: when $x<2$, say $x=-100$, the derivative is clearly negative, while when $x>2$, say $x=100$, the derivative is clearly positive. So $x=2$ is the unique local minimum.)
\item $-6$
\item $i$ is $f$, $ii$ is $f'$, $iii$ is $f''$ (note that the exam required an explanation)
\item $3/4$
\item $y-1/2=\frac{\sqrt{2}}{3}(x-\sqrt{2}/2)$

\item $x=60$, $y=80$.  The second derivative is easy to find, and positive when $x$ is positive.
\item $0$ is the only critical point.  There are no local minima or maxima (the function is not even defined at $f(x)$).
\item $c$
\item You need $g(x)$
\item
  \begin{enumerate}
  \item $\pm 1$
  \item $(-\infty, -1)$ and $(1,\infty)$
  \item $(0,\infty)$
  \end{enumerate}

\end{enumerate}

\end{document}
