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\begin{document}
\centerline{MATH 135 -- Calculus 1}
\centerline{Sample Questions for Exam 2 -- Answers and Solutions}
\centerline{October 10, 2016}

\begin{enumerate}
\item{} An object moves along a straight line path with position given by 
$x(t) = 4t^2 + t - 7$, ($t$ in seconds, $x$ in feet).   
\begin{enumerate}
\item[(a)]  What is the average velocity of the object over the interval $[0,5]$ of $t$-values?
\item[] {\it Solution:}  The average velocity is:
$$v_{ave} = \frac{x(5) - x(0)}{5 - 0} = \frac{98 - (-7)}{5} = 21\ {\rm ft/sec}$$
\item[(b)]  Fill in the following table with average velocities computed over the indicated intervals.  Using
this information, estimate the \emph{instantaneous velocity} at $t = 2$.  
\item[] {\it Solution:}  
$$\begin{array}{ccccccccc}
\hline
{\rm interval} &\vline& [2,2.5] &\vline& [2,2.05] &\vline& [2,2.005] &\vline& [2,2.0005]\\ \hline
{\rm ave. vel.}&\vline&  19.0   &\vline&   17.2   &\vline&   17.02   &\vline&  17.002   \\ \hline 
\end{array}$$
It looks as though the average velocity is tending to $17$ as the length of the interval
goes to $0$.
\item[(c)]  Construct a similar table for intervals \emph{ending} at $t = 2$ and 
repeat the calculations in the previous part.  If you estimate the instantaneous
velocity at $t = 2$ using this new information, does your result agree with what you 
did before (it should!)
\item[ ] {\it Solution:}
$$\begin{array}{ccccccccc}
\hline
{\rm interval} &\vline& [1.5,2] &\vline& [1.95,2] &\vline& [1.995,2] &\vline& [1.9995,2]\\ \hline
{\rm ave. vel.}&\vline&  15.0   &\vline&   16.8   &\vline&   16.98   &\vline&  16.998   \\ \hline 
\end{array}$$
It looks as though the average velocity is tending to $17$ as the length of the interval
goes to $0$ again.
\end{enumerate}

\item{} 
\begin{enumerate} 
\item[(a)]  What is the slope of the secant line to the graph $y = x^3 + 1$ through the points
with $x = 1$ and $x = 2$?
\item[] {\it Solution:}  The slope of the secant line is 
$$m_{sec} = \frac{(2^3+1) - (1^3+1)}{2 - 1} = \frac{9 - 2}{1} = 7.$$
\item[(b)]  What is the slope of the secant line to the graph $y = x^3 + 1$ through the points
with $x = 1$ and $x = 1 + h$ for a general $h$?
\item[] {\it Solution:}  The slope of the secant line is
$$m_{sec} = \frac{(1 + h)^3 + 1 - (1^3 + 1)}{1+h - 1} = \frac{3h + 3h^2 + h^3}{h}.$$ 
\item[(c)]  The slope of the tangent line to $y = x^3 + 1$ at $x = 1$ would be
obtained from what limit?
\item[] {\it Solution:}  The slope of the tangent is 
$$m_{tan} = \lim_{h\to 0} \frac{3h + 3h^2 + h^3}{h}.$$
\item[(d)]  Estimate the limit in the previous part numerically (as in the first question).
\item[] {\it Solution:}
$$\begin{array}{ccccccccc}
\hline
h &\vline& 1 &\vline&  .5 &\vline& .05 &\vline& .005 \\ \hline
(3h + 3h^2 + h^3)/h &\vline& 7 &\vline & 4.75 &\vline& 3.1525 &\vline& 3.015025 \\ \hline 
\end{array}$$
It seems the limit as $h \to 0$ is about $3$.  
\item[(e)]  Evaluate the limit exactly using our algebraic techniques.
\item[] {\it Solution:} 
$$\lim_{h\to 0} \frac{3h + 3h^2 + h^3}{h} = \lim_{h\to 0}\frac{h(3 + 3h + h^2)}{h} = 
\lim_{h\to 0} 3 + 3h + h^2 = 3$$
(cancelling the $h$'s top and bottom, then using the continuity of $3 + 3h + h^2$ 
as a function of $h$).  

\end{enumerate}

\item The graph of a function~$f$ is shown
  below with several points marked. Find all the marked points at
  which the following are true, and give explanations for your answers.
  \begin{center}
   \setlength{\unitlength}{1in}
\includegraphics[width=.4\linewidth]{M2I.png}
  \end{center}

\begin{enumerate}

  \item $f$ has an infinite discontinuity -- {\it Answer:}  $x = 3$.  

  \item $f$ has jump discontinuity -- {\it Answer:}  $x = -1$ and $x = 2$.  We have
$$\lim_{x \to -1^-} f(x) \doteq -0.3, \text{ and } \lim_{x \to -1^+} f(x) = 2$$
and 
$$\lim_{x \to 2^-} f(x) = 0, \text{ and } \lim_{x \to 2^+} f(x) = -1.$$


  \item $f$ has a removable discontinuity -- {\it Answer:} There are no removable 
discontinuities for the function given here.  A removable discontinuity would 
be an $x = b$ where $\lim_{x\to b} f(x) = L$ exists (``from both sides'') but $L \ne f(b)$.  

  \item $f$ is continuous -- {\it Answer:}  $f$ is continuous at all points shown except
$x = -1, 2, 3$.  

\end{enumerate}


\item Compute the indicated limits. Show all work for full credit.

  \begin{enumerate}

  \item[(a)]
     $\ds\lim_{x\to1}\frac{3x^2-5x-2}{x^2-4x+4}$
\vskip 10pt
\item[ ] {\it Solution:} This is not an indeterminate form, and the denominator is not
zero at $1$, so the answer can be found by continuity:  
$$\ds\lim_{x\to1}\frac{3x^2-5x-2}{x^2-4x+4} = \frac{-4}{1} = -4.$$
  \item[(b)] 
     $\ds\lim_{x\to 2}\frac{3x^2-5x-2}{x^2-4x+4}$
\item[ ] {\it Answer:}  This is a $0/0$ form.  To try to evaluate, we aim to factor the bottom
and cancel factors of $x - 2$:
$$\lim_{x\to 2}\frac{3x^2-5x-2}{x^2-4x+4} = \lim_{x\to 2} \frac{(x - 2)(3x + 1)}{(x - 2)(x - 2)}
=\lim_{x \to 2} \frac{3x + 1}{x - 2}.$$
This is not indeterminate any more, but the bottom is still $0$ at $x = 2$.  So this limit
does not exist.  (The function has an infinite discontinuity at $x = 2$.)
\vskip 10pt
  \item[(c)] 
     $\ds\lim_{x\to\infty}\frac{3x^2-5x-2}{x^2-4x+4}$
\vskip 10pt
\item[] {\it Solution:}  For this one, we multiply the top and bottom by $\frac{1}{x^2}$:
\begin{eqnarray*}
lim_{x\to\infty}\frac{3x^2-5x-2}{x^2-4x+4} &=& 
\lim_{x\to\infty} \frac{(3x^2 - 5x - 2)\frac{1}{x^2}}{(x^2-4x+4)\frac{1}{x^2}}\\
&=& \lim_{x\to\infty} \frac{3 - \frac{5}{x} - \frac{2}{x^2}}{1 - \frac{4}{x} + \frac{4}{x^2}}\\
&=& \frac{3 - 0 - 0}{1 - 0 + 0} = 3.
\end{eqnarray*}
This says that the graph of this function has a horizontal asymptote at the line $y = 3$.

   \item[(d)] 
     $\ds\lim_{x\to 2}\frac{\displaystyle{\frac{1}{x^2} - \frac{1}{4}}}{x - 2}$
\vskip 10pt
\item[] {\it Solution:} 
\begin{eqnarray*}
\lim_{x\to 2}\frac{\displaystyle{\frac{1}{x^2} - \frac{1}{4}}}{x - 2}
&=& \lim_{x\to 2}\frac{4 - x^2}{4x^2(x-2)}\\
&=& \lim_{x\to 2}\frac{(2-x)(2 + x)}{4x^2(x-2)}\\
&=& \lim_{x\to 2}\frac{-(2+x)}{4x^2} = \frac{-1}{4}.
\end{eqnarray*}
   \item[(e)]
     $\ds\lim_{t\to 0} \frac{\sin(6t)}{\sin(7t)}$
\vskip 10pt
\item[] {\it Solution:}  For this one, we need to use the formula 
$\lim_{u\to 0} \frac{\sin(u)}{u} = 1$ from Section 2.6.  We have
\begin{eqnarray*}
\lim_{t\to 0} \frac{\sin(6t)}{\sin(7t)} &=&
 \lim_{t\to 0} \frac{6\cdot \sin(6t)/(6t)}{7 \cdot\sin(7t)/(7t)}\\
&=& 6/7.
\end{eqnarray*}

   \item[(f)]
     $\ds\lim_{h\to 0} \frac{\sqrt{h + 9} - \sqrt{9}}{h}$
\item[]{\it Solution:}  Multiply top and bottom by the conjugate radical:
\begin{eqnarray*}
\lim_{h\to 0} \frac{\sqrt{h + 9} - \sqrt{9}}{h} &=&
\lim_{h\to 0} \frac{(\sqrt{h + 9} - \sqrt{9})(\sqrt{h + 9} + \sqrt{9})}{h(\sqrt{h + 9} + \sqrt{9})}\\
&=& \lim_{h\to 0} \frac{(h + 9) - 9}{h(\sqrt{h + 9} + \sqrt{9})}\\
&=& \lim_{h\to 0} \frac{h}{h(\sqrt{h + 9} + \sqrt{9})}\\
&=& \lim_{h\to 0} \frac{1}{(\sqrt{h + 9} + \sqrt{9})}\\
&=& \frac{1}{2\sqrt{9}} = \frac{1}{6}.
\end{eqnarray*}

  \end{enumerate}

\item{}  Suppose you know each of the following conditions.
What can you say about $\lim_{x\to c} f(x)$ for the indicated $c$?  Why?
  
  \begin{enumerate}

    \item[(a)]

         $x^2 + x \le f(x) \le x^3 + 3$ for all real $x$, at $c = 0$. 
\item[] {\it Solution:}  $f(x)$ is not ``squeezed'' in this case because $l(x) = x^2 + x$
has the value $l(0) = 0$, which is strictly less than $u(0) = 3$ for $u(x) = x^3 + 3$.
The limit of $f(x)$ as $x \to 0$ could be any number between 0 and 3, or it might not
exist at all.  

    \item[(b)]
        
         $-x^2 + 2x \le f(x) \le x^4 - 4x^3 + 6x^2 - 4x + 2$ for all real $x$, at $c = 1$
\item[]{\it Solution:}  Now we have $l(1) = 1$ and $u(1) = 1$.  So $f(x)$ is ``squeezed''
and the Squeeze Theorem says $\lim_{x \to 1} f(x) = 1$ also. 

    \item[(c)]

         $f(x) = x \sin\left(\frac{1}{x}\right)$ for all real $x \ne 0$, at $c = 0$.
\item[] {\it Solution:}  This one is trickier because we are not given $l(x)$ and $u(x)$.
However, note that for all $x \ne 0$, $-1 \le \sin\left(\frac{1}{x}\right) \le 1$.
Therefore, $-x \le f(x) \le x$ for all $x \ne 0$.  Hence the Squeeze Theorem does apply
and we can see $\lim_{x \to 0} f(x) = 0$.  (This $f(x)$ has a graph that looks like the 
picture from Exercise 56 B on page 71.)
\end{enumerate}

\item
\begin{enumerate}
\item[(a)] What is the definition of the derivative of a function $f$ at $x = a$ in 
its domain?
\item[]{\it Solution:}  $f'(a) = \displaystyle{\lim_{h\to 0} \frac{f(a+h) - f(a)}{h}}$, if the limit exists.
The limit $\displaystyle{\lim_{x\to a} \frac{f(x) - f(a)}{x - a}}$ also computes $f'(a)$. 
These can be seen to be equivalent if you let $x - a = h$ in the second form.
\item{} Using the definition (not the shortcut rules), find $f'(x)$ for $f(x) = 3x^3 - 2x^2 + 1$
at a general $x = a$.
\item[(b)]{\it Solution:}  
\begin{eqnarray*}
f'(a) &=& \lim_{h\to 0}\frac{(3(a+h)^3 - 2(a+h)^2 + 1 - (3a^3 - 2a^2 + 1)}{h}\\
&=& \lim_{h\to 0} \frac{3a^3 + 9a^2h + 9ah^2 + 3h^3 - 2a^2 - 4ah - 2h^2 + 1 - 3a^3 + 2a^2 - 1}{h} \\
&=& \lim_{h\to 0} \frac{(9a^2 - 4a + 9ah - 2h + 3h^2)h}{h}\\
&=& \lim_{h\to 0} 9a^2 - 4a + 9ah + 2h + 3h^2\\
&=& 9a^2 - 4a.
\end{eqnarray*}
\item[(c)]  Using the definition (not the shortcut rules), find $f'(x)$ for $f(x) = x^{1/2}$
at a general $x = a > 0$.
\item[]{\it Solution:}  
\begin{eqnarray*}
f'(a) &=& \lim_{h\to 0} \frac{\sqrt{a + h} - \sqrt{a}}{h}\\
&=& \lim_{h\to 0}\frac{\sqrt{a + h} - \sqrt{a}}{h}\cdot \frac{\sqrt{a + h} + \sqrt{a}}{\sqrt{a + h} + \sqrt{a}}\\
&=& \lim_{h\to 0}\frac{a + h - a}{h \cdot (\sqrt{a + h} + \sqrt{a})}\\
&=& \lim_{h\to 0}\frac{1}{\sqrt{a + h} + \sqrt{a}}\\
&=& \frac{1}{2\sqrt{a}}.
\end{eqnarray*}
\item{}  Using the definition (not the shortcut rules), find $f'(x)$ for $f(x) = \frac{1}{x^2}$
at a general $x = a \ne 0$.
\begin{eqnarray*}
f'(a) &=& \lim_{h\to 0} \frac{\frac{1}{(a+h)^2} - \frac{1}{a^2}}{h}\\
&=& \lim_{h\to 0} \frac{a^2 - (a + h)^2}{h a^2 (a + h)^2}\\
&=& \lim_{h\to 0} \frac{-2ah - h^2}{h a^2(a + h)^2}\\
&=& \lim_{h\to 0} \frac{-2a - h}{a^2(a + h)^2}\\
&=& \frac{-2a}{a^4} = \frac{-2}{a^3}.
\end{eqnarray*}
\end{enumerate}
\item
\begin{enumerate}
\item{} State and prove the product rule for derivatives.  
\item{} {\it Solution:}  The rule says that if $f$ and $g$ are both differentiable
at $x$, then $f(x)g(x)$ is also differentiable at $x$ and 
$$\frac{d}{dx}(f(x)g(x)) = f(x)g'(x) + g(x)f'(x)$$
The reason this is true is that
\begin{eqnarray*}
\frac{d}{dx}(f(x)g(x)) &=& \lim_{h\to 0} \frac{f(x+h)g(x+h) - f(x)g(x)}{h}\\
&=&\lim_{h\to 0} \frac{f(x+h)g(x+h) - f(x+h)g(x) + f(x+h)g(x) f(x)g(x)}{h}\\
&=&\lim_{h\to 0} \frac{f(x+h)g(x+h) - f(x+h)g(x)}{h} + \lim_{h\to 0} \frac{f(x+h)g(x) f(x)g(x)}{h}\\
&=&\lim_{h\to 0} f(x+h) \cdot \lim_{h\to 0}\frac{g(x+h) - g(x)}{h} 
+ g(x)\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}\\
&=& f(x)g'(x) + g(x)f'(x)
\end{eqnarray*}
Along the way here, we used the fact that if $f'(x)$ exists, then $\lim_{h\to 0} f(x+h) = f(x)$
(since otherwise the difference quotient for $f$ would not be an indeterminate
form as $h\to 0$ and the limit could not exist).   
\item{} Use the product rule to find $f'(x)$ for $f(x) = (4x^3 - 12x^2 + 1)e^x$.
\item{} {\it Solution:}  
$$f'(x) = (4x^3 - 12x^2 + 1)e^x + e^x(12x^2 - 24x) = (4x^3-24x + 1)e^x$$
\item{} Find $f'(x)$ for 
$$f(x) = \frac{x^2 - 4x + 1}{x^3 + 2}$$
\item{} {\it Solution:}  By the quotient rule:
$$f'(x) = \frac{(x^3 + 1)(2x - 4) - (x^2 - 4x + 1)(3x^2)}{(x^3 + 2)^2} 
= \frac{-x^4 - 8x^3-3x^2+4x+8}{(x^3+2)^2}$$
\item{} Find $f'(x)$ two ways:  One, using the quotient rule, one {\it without} using
the quotient rule.  Verify that the result is the same in both cases.  Which is easier?
$$f(x) = \frac{x^6 - 3x^3 + x}{x^{1/2}}$$
\item{} {\it Solution:}  Using the quotient rule:
$$f'(x) = \frac{x^{1/2}(6x^5 - 9x^2) - (x^6 - 3x^3 + x)\cdot \frac{1}{2}x^{-1/2}}{(x^{1/2})^2}$$
Another way to do this is to divide $x^{1/2}$ into each term on the top, and then differentiate:
$$f(x) = x^{11/2} - 3x^{5/2} + x^{1/2},$$
so then
$$f'(x) = \frac{11}{2}x^{9/2} - \frac{15}{2}x^{3/2} + \frac{1}{2} x^{-1/2}.$$
It is a ``nice exercise'' to show that these define the same function(!) 
\end{enumerate}

\end{enumerate}
\end{document}
