%% Math 133 Midterm 3, Fall 2015
\documentclass[12pt]{article}
\usepackage{latexsym,amsmath,amssymb,epsfig,float,graphicx,subfig}

%\usepackage{my_teach}

%\pagestyle{myheadings}
\markright{MATH 133, section 1, Midterm 3}

\setlength{\topmargin}{-0.5in}
\setlength{\textwidth}{6.5in}
\setlength{\oddsidemargin}{0in}
\setlength{\textheight}{8.5in}

\newcommand{\Blank}{$\underline{\phantom{Please write your name in this
        space}}$} 
\newcommand{\Strut}{$\vphantom{\Big|}$}

\begin{document}

\begin{center}

\textbf{College of the Holy Cross, Fall Semester, 2015} \\
\textbf{MATH 133, section 1, Solutions for Midterm 3} \\
\textbf{Thursday, November 12}
\end{center}



\begin{enumerate}

\item{} 
\begin{enumerate}
\item[A)] (5) State the limit definition of the
  derivative~$f'(x)$ as an equation starting with $f'(x) = $.
\item[] {\it Solution:}  
$$f'(x) = \lim_{h\to 0} \frac{f(x+h) - f(x)}{h},$$
if the limit exists.  (If so we say $f$ is differentiable at $x$; if not
we say $f$ is not differentiable at $x$.)
\item[B)] (10) Use the limit definition to
    compute $f'(x)$ for $f(x) = x^2 - 5x + 2$.
\item[] {\it Solution:}
\begin{eqnarray*}
f'(x) &=& \lim_{h\to 0} \frac{(x+h)^2 - 5(x + h) + 2 - x^2 + 5x - 2}{h}\\
&=& \lim_{h\to 0} \frac{x^2 + 2xh + h^2 - 5x - 5h + 2 - x^2 + 5x - 2}{h}\\
&=& \lim_{h\to 0} \frac{2xh + h^2 - 5h}{h}\\
&=& \lim_{h\to 0} \frac{(2x - 5 + h)h}{h}\\
&=& \lim_{h\to 0} (2x - 5 + h) = 2x - 5.\\
\end{eqnarray*}
(This agrees with the shortcut rules, of course!)
\item[C)] (5) Find the equation of the line tangent to
    the graph $y = x^2 - 5x + 2$ at $x=2$.
\item[] {\it Solution:}  The $y$-coordinate is $f(2) = -4$ and the slope is $f'(2) = -1$
so the line is $y + 4 = (-1)(x - 2)$, or $y = -x - 2$. 
\end{enumerate}
\vskip 10pt

\item{} Compute the indicated derivatives of the following functions. You may use any correct
method.  You do not need to simplify your answers unless specifically directed to do so.
But you must show work for full credit.
\begin{enumerate}
\item[(A)] (10)  $f(x) = \displaystyle{6x^{4/3} + \frac{3}{x^5} - 4e^x}$.  Compute both
$f'(x)$ and $f''(x)$.
\item[] {\it Solution:}  First, write the $\frac{3}{x^5}$ as $3x^{-5}$.  Then
$$f'(x) = 8x^{1/3} - 15 x^{-6} - 4e^x$$
and 
$$f''(x) = (f')'(x) = \frac{8}{3} x^{-2/3} + 90 x^{-7} - 4e^x.$$
\item[(B)]  (10) $g(x) = \displaystyle{(x^4 + \cos(x))(x^2 - \sin(x))}$.  Compute $g'(x)$.
\item[] {\it Solution:}  Using the product rule and the derivative rules for $\sin(x)$ and 
$\cos(x)$:
$$g'(x) = (x^4 + \cos(x))(2x - \cos(x)) + (x^2 - \sin(x))(4x^3 - \sin(x)).$$
The problem did not say to simplify, so we can leave it here(!)
\item[(C)]  (10) $h(x) = \displaystyle{\frac{x^3 - x}{x^2 + 1}}$.  Compute $h'(x)$ and 
simplify your answer. 
\item[] {\it Solution:}  By the quotient rule:
\begin{eqnarray*} 
h'(x) &=& \frac{(x^2 + 1)(3x^2 - 1) - (x^3 - x)(2x)}{(x^2 + 1)^2}\\
&=& \frac{3x^4 + 2x^2 - 1 - 2x^4 + 2x^2}{(x^2 + 1)^2}\\
&=& \frac{x^4 + 4x^2 - 1}{(x^2 + 1)^2}.
\end{eqnarray*} 
\item[(D)]  (10) $j(x) = \displaystyle{\frac{1}{(x^5 + 3x + 1)^3}}$.  Compute $j'(x)$.
\item[] {\it Solution}:  This is $(x^5 + 3x + 1)^{-3}$, so by the chain rule,
$$j'(x) = -3(x^5 + 3x + 1)^{-4}(5x^4 + 3) = \frac{-(15x^4 + 9)}{(x^5 + 3x + 1)^4}.$$ 
\item[(E)] (10) $k(x) = \displaystyle{(x^2 + 3)e^{x^3 - 5\sin(2x)}}$.  Compute $k'(x)$.
\item[] {\it Solution:}  By the product rule and the chain rule,
$$k'(x) = (x^2 + 3)e^{x^3 - 5\sin(2x)}(3x^2 - 10\cos(2x)) + e^{x^3 - 5\sin(2x)}(2x).$$
\end{enumerate}

\item{}  All parts of this question refer to the plots in Figure 1.  Assume the whole domain
of the functions is the interval $[-2,10]$ shown. 
\begin{figure}
\centering
  \subfloat[$y = A(x)$]{\includegraphics[width=70mm]{M33PA.png}}
  \subfloat[$y = B(x)$]{\includegraphics[width=70mm]{M33PB.png}}
\caption{Plots for Problem 3}
\end{figure}

\begin{enumerate}
\item[(A)] (3) Which has the larger slope:  The secant line to $y = A(x)$ through
$(-2,A(-2))$ and $(10,A(10))$ or the tangent line to $y = A(x)$ at $(4,A(4))$? \\ 
Answer: We see the secant line has a positive slope, but the tangent line at $x = 4$
has a negative slope.  So the secant line has a larger slope.
\vskip 5pt
\item[(B)] (3)  Is $A'(6)$ positive or negative?  
Answer:  $A'(6)$ is the slope of the tangent line to $y = A(x)$ at $x = 6$, which is 
negative.  
\vskip 5pt
\item[(C)] (3) At how many different points is $B'(x) = 0$?  Estimate the 
$x$-values from the graph.  Answer: There are three such points where the tangent
line to $y = B(x)$ is horizontal -- at approximately
$x = -1.1$, $x =.85$, and $x = 4.25$. 
\vskip 5pt
\item[(D)] (3) On the $x$-interval $(2,4)$, is $A''(x)$ positive or negative?
Answer: We can see that the slopes of the tangent lines are starting positive, 
decreasing to 0 and then becoming negative.  Since $A'' = (A')'$ and $A'$
is decreasing, $A''(x)$ is negative on this interval.
\vskip 5pt
\item[(E)] (3) On the $x$-interval $(1,3)$ is $B'(x)$ positive or negative?
Answer: The tangent lines have negative slopes, so $B'(x)$ is negative
\vskip 5pt
\item[(F)] (3)  One of the two functions $A(x)$ and $B(x)$ is the derivative of 
the other.  Which is which?  Answer: 
$B(x)$ is the derivative of $A(x)$.  On intervals where $A(x)$ is increasing (resp. 
decreasing), $B(x)$ is positive (resp. negative).
\vskip 5pt
\item[(G)] (2)  Which plot is $y = e^{-x}(x^3 + 2x^2)$?  
Answer: That's $y = A(x)$.  The easiest way to see that is that $A(x)$ takes only 
positive values for $x > -2$, but $B(x)$ has negative values for some of those
$x$'s.  That says $B(x)$ cannot be $e^{-x}(x^3 + 2x^2) = e^{-x}x^2(x + 2)$, since
that formula gives only positive values for $x > -2$.  
\end{enumerate}
\vskip 10pt

\item{} (20) The bluebook value of a 2010 model car at time $t$ years after
2010 is approximated by $P(t) = 20e^{(-.16) t}$ (in units of \$1000). 
What was the value in 2015, and what was the rate of change of the value then? (Give
your answer with units.) Was the value increasing or decreasing?
\vskip 10pt
\noindent
{\it Solution:}  Since $t$ is the number of years after 2010, the year 2015 is $t = 5$, and
the value in 2015 is $P(5) = 20 e^{(-.16)(5)} \doteq 8.987$ in thousands of dollars, or
$\$8987$.  (Note: cars depreciate fast!!)  

The rate of change of the value then is $P'(5)$.  We compute
$P'(t) = 20(-.16) e^{(-.16)t} = -3.2 e^{(-.16)t}$ by the chain rule.  So $P'(5) =
-3.2 e^{-.8} \doteq -1.438$ in thousands of dollars per year.  
This means the rate of change of the value is about $-1438$ dollars per year.   
\end{enumerate}

\end{document}


