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\centerline{MATH 135 -- Calculus 1}
\centerline{Review for Final}
\centerline{December 13, 2019}
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\noindent
{\it Practice Questions}
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\begin{enumerate}
\item What is the limit definition of the derivative $f'(x)$?
\item Use the definition to compute $f'(x)$ for $f(x) = \sqrt{x - 4}$.
\item What is the equation of the tangent line to $y = \sqrt{x - 4}$ at $x = 8$?
\item Suppose $f(x)$ is a function such that $f'(x)$ and $f''(x)$
exist and are continuous for all real $x$.  Assume that $f'(x) > 0$ on $(-3,-2)$ 
and $(2,\infty)$,  while $f'(x) < 0$ on $(-\infty,-3)$ and $(-2,2)$. 
Also assume $f'(-3) = f'(-2) = f'(2) = 0$.  
\begin{enumerate}
\item What does the First Derivative Test tell you about
$x = -3$, $x = -2$, and $x = 2$?
\item Using the Mean Value Theorem, explain why the equation $f''(x) = 0$ must have a 
solution in the interval $(-3,-2)$ and another 
solution in the interval $(-2,2)$.
\item Suppose you also know $f(-3) = 4$, $f(-2) = 6$, $f(2) = 0$, and
$$\lim_{x \to \pm\infty} f(x) = 12.$$
Sketch a possible graph $y = f(x)$ that satisfies all of these
conditions.
\end{enumerate}
\item Let $f(x) = \frac{20x}{x^2 - 4x + 3}$, for which 
\begin{eqnarray*}
f'(x) &=& \frac{-20x^2 + 60}{(x^2 - 4x + 3)^2}\\
f''(x) &=& \frac{40x^3 - 360x + 480}{(x^2 - 4x + 3)^3}
\end{eqnarray*}
\begin{enumerate}
\item Where does $y = f(x)$ have vertical asymptotes?  Does it have a horizontal asymptote?  If so, where?
\item What are 
$$\lim_{x \to 1^-} f(x) \text{ and } \lim_{x \to 1^+} f(x)?$$
What about
$$\lim_{x \to 3^-} f(x) \text{ and } \lim_{x \to 3^+} f(x)?$$
\item Does $f(x)$ have any critical points?  Where are they located?  What are the critical values?
\item What is the concavity of the graph $y = f(x)$ on the interval $(3,\infty)$?
\item Sketch the graph $y = f(x)$.
\end{enumerate}
\end{enumerate}
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