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\centerline{MATH 135 -- Calculus 1}
\centerline{Review for Final}
\centerline{December 13, 2019}
\vskip 10pt
\noindent
{\it Practice Questions}
\vskip 10pt
\begin{enumerate}
\item What is the limit definition of the derivative $f'(x)$?
\item[] {\it Solution:}  
$$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h},$$
provided that the limit exists.  
\item Use the definition to compute $f'(x)$ for $f(x) = \sqrt{x - 4}$.
\item[] {\it Solution:}
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{\sqrt{x + h - 4} - \sqrt{x - 4}}{h}\\
&=& \lim_{h\to 0} \frac{\sqrt{x + h - 4} - \sqrt{x - 4}}{h} \cdot 
\frac{\sqrt{x + h - 4} + \sqrt{x - 4}}{\sqrt{x + h - 4} + \sqrt{x - 4}}\\
&=& \lim_{h\to 0} \frac{(x + h - 4) - (x - 4)}{h(\sqrt{x + h - 4} + \sqrt{x - 4})}\\
&=& \lim_{h\to 0} \frac{h}{h(\sqrt{x + h - 4} + \sqrt{x - 4})}\\
&=& \lim_{h\to 0} \frac{1}{\sqrt{x + h - 4} + \sqrt{x - 4}}\\
&=& \frac{1}{2\sqrt{x - 4}}.
\end{eqnarray*}
(Note that this agrees with the formula you would get by applying the chain
rule to $f(x) = (x - 4)^{1/2}$.)
\item What is the equation of the tangent line to $y = \sqrt{x - 4}$ at $x = 8$?
\item[] The equation is 
$$y - 2 = \frac{1}{2\sqrt{4}}(x - 8)$$
or 
$$y = \frac{1}{4}x.$$
\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[] {\it Solution:}  Since $f'$ changes from negative to positive at $x = -3$, $f$ has a local minimum
there.  Similarily, $f$ has local maximum at $x = -2$, and another local minimum at $x = 2$
\item Using the Mean Value Theorem, explain why $f$ must have a 
point of inflection on the interval $(-3,-2)$ and another 
point of inflection on the interval $(-2,2)$.
\item[] {\it Solution:}  Recall that the Mean Value Theorem says that if $g$ is continuous on
$[a,b]$ and differentiable on $(a,b)$, then there exists a $c$ in $(a,b)$ where 
$$g'(c) = \frac{g(b) - g(a)}{b - a}.$$
If we apply this to $g = f'$, then since $g(-3) = g(-2) = 0$, there must be
some point $c$ in $(-3,-2)$ where $g'(c) = f''(c) = 0$.  The same reasoning applies in the 
interval $(-2,2)$ as well.
\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.
\item[] {\it Solution:}  Omitted.  There are infinitely many correct graphs that
satisfy 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[] {\it Solution} Since $x^2 - 4x + 3 = (x - 1)(x - 3)$ there are vertical asymptotes
at $x = 1$ and $x = 3$.  We must understand $\lim_{x\to\pm\infty} f(x)$ to determine horizontal asymptotes.
By L'Hopital's Rule,
$$\lim{x\to\pm\infty} \frac{x}{x^2 - 4x + 3} = \lim_{x\to\pm\infty} \frac{1}{2x - 4} = 0$$
There is a horizontal asymptote at the line $y = 0$.  
\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[] {\it Solution:}  By comparing the signs of $20x$ and $x^2 - 4x + 3$ on either side of $x = 1$ and $x =3$,
we can see
$$\lim_{x \to 1^-} f(x) = +\infty \text{ and } \lim_{x \to 1^+} f(x) = -\infty$$
Similary
$$\lim_{x \to 3^-} f(x) = -\infty \text{ and } \lim_{x \to 3^+} f(x) = +\infty?$$
\item Does $f(x)$ have any critical points?  Where are they located?  What are the critical values?
\item[] {\it Solution:}  There are critical points at the solutions of $f'(x) = 0$ -- namely 
where $-20x^2 + 60 = 0$, so $x = \pm \sqrt{3}$.  The critical values are
$f(\sqrt{3}) \doteq -37.3 \text{ and } f(-\sqrt{3}) \doteq -2.68$.  
\item What is the concavity of the graph $y = f(x)$ on the interval $(3,\infty)$?
\item[] {\it Solution:}  At all $x$ in that interval, for instance at $x = 4$, $f''(x) > 0$.
So the graph $y = f(x)$ is concave up.
\item Sketch the graph $y = f(x)$.
\item[] {\it Solution:}  See Figure 1 above.
\begin{figure}
\begin{center}
\includegraphics[width=.3\linewidth]{LastGraph.png}
\caption{$y = \frac{20x}{x^2 - 4x + 3}$ for $-5 \le x \le 10$}
\end{center}
\end{figure}
\end{enumerate}
\end{enumerate}
\end{document}

