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\begin{document}
\centerline{MATH 135 -- Calculus 1}
\centerline{Solutions/Answers for Exam 3 Practice Problems}
\centerline{November 20, 2019}
\vskip 10pt
\begin{enumerate}
\item[I.]  Find the indicated derivative(s) and simplify.
\vskip 10pt
\begin{enumerate}
\item[(A)]  $$y = \ln(x)\left(x^7 - \frac{4}{\sqrt{x}}\right)$$
\item[] {\it Solution:}  By the product rule and the derivative rules for $\ln(x)$ and powers:
$$y' = \ln(x)(7x^6 + 2 x^{-3/2}) + x^6 - \frac{4}{x^{3/2}}$$
\item[(B)]  $$y = \sin^{-1}(e^{2x} + 2)$$
\item[] {\it Solution:} By the chain rule and the inverse sine derivative rule:
$$y' = \frac{1}{\sqrt{1 - (e^{2x} + 2)^2}}\cdot 2e^{2x} = \frac{2e^{2x}}{\sqrt{1 - (e^{2x} + 2)^2}}$$
\item[(C)]  $$y = \frac{\ln(x + 1)}{3x^4 - 1}$$
\item[] {\it Solution:}  Using the quotient rule:
$$y' = \frac{(3x^4 - 1)\cdot \frac{1}{x+1} - \ln(x+1)(12x^3)}{(3x^4-1)^2}$$
\item[(D)]  $$y = \frac{\sin(x)}{1 + \cos(x)}$$
\item[] {\it Solution:} By the quotient rule:
$$y' = \frac{(1 + \cos(x))\cos(x) + \sin^2(x)}{(1 + \cos(x))^2} = \frac{1}{1 + \cos(x)}$$
\item[(E)]  $$y = \tan^{-1}(x^2+x)$$
\item[] {\it Solution:}  By the inverse tangent rule:
$$\frac{dy}{dx} = \frac{2x + 1}{1 + (x^2 + 2x)^2}$$
\item[(F)]  Using implicit differentiation: $$xy^2 - 3y^3 + 2x^4 - 4xy = 2$$
\item[] {\it Solution:}  We have 
$$2xy\frac{dy}{dx} + y^2 - 9y^2 \frac{dy}{dx} + 8x^3 - 4x\frac{dy}{dx} - 4y = 0$$
so
$$\frac{dy}{dx} = \frac{4y - 8x^3 - y^2}{2xy - 9y^2 - 4x}$$ 
\item[(G)]  Find the equation of the line tangent to the curve from (F) at $(x,y) = (1,0)$
\item[] {\it Solution:}  The slope is 2, so $y = 2x - 2$.
\item[(H)]  Find
$$\frac{d}{dx}\left(5x\sqrt{x}-\frac{2}{x^3}+ 11x - 4\right)$$
\item[] {\it Solution:}  The function can also be written as $5x^{3/2} - 2x^{-3} + 11x - 3$.
In this form, we only need the power rule to differentiate:
$$y' = \frac{15}{2} x^{1/2} + 6x^{-4} + 11.$$
\item[(I)]
$$\frac{d}{dt}\left(\frac{t^2e^{3t}}{t^4 + 1}\right)$$
\item[] {\it Solution:}  By the quotient rule, product rule, and chain rule the derivative is:
$$\frac{(t^4 + 1)\cdot (3t^2e^{3t} + 2te^{3t}) - (t^2e^{3t})\cdot (4t^3)}{(t^4 + 1)^2}
=\frac{e^{3t}(3t^6 - 2t^5+3t^2 + 2t)}{(t^4 + 1)^2}.$$
\item[(J)]
$$\frac{d^2}{dz^2} \frac{z^2 - 2z + 4}{z^2 + 1}$$
\item[] {\it Solution:}   Again using the quotient rule, the first derivative is:
$$\frac{(z^2 + 1)(2z - 2) - (z^2 - 2z + 4)(2z)}{(z^2 + 1)^2} = 
\frac{2z^2 - 6z - 2}{(z^2 + 1)^2}.$$
So then differentiating again, the second derivative is
\begin{eqnarray*}
&=&\frac{(z^2+1)^2(4z - 6) - (2z^2 - 6z -2)(2)(z^2 + 1)(2z)}{(z^2 + 1)^4}\\
&=& \frac{(z^2+1)(4z - 6) - (4z)(2z^2 - 6z - 2)}{(z^2 + 1)^3} \\
&=& \frac{-4z^3+18z^2+12z-6}{(z^2+1)^3}. 
\end{eqnarray*}
\item[(K)] 
$$\frac{d}{dx}\left( \sin(x)\left(x^7 - \frac{4}{\sqrt{x}}\right)\right)$$
\item[] {\it Solution:}   Rewrite the fuction as $\sin(x)(x^7 - 4x^{-1/2})$.  Then by the 
product rule the derivative is:
$$\sin(x)(7x^6 + 2x^{-3/2}) + (x^7 - 4x^{-1/2})\cos(x).$$
\item[(L)] Find $y'$ (note this is just another way of asking the same question!)
$$y = (e^{2x} + 2)^3$$
\item[]  By the chain rule, the derivative is: 
$$3(e^{2x} + 2)^2(2e^{2x}) = 6e^{2x}(e^{2x} + 2)^3.$$
\item[(M)]  Find $y'$ and $y''$ 
$$y = \frac{x + 1}{3x^4 - 1}$$
\item[] {\it Solution}: By the quotient rule,
$$y' = \frac{(3x^4 - 1)(1) - (x+1)(12x^3)}{(3x^4-1)^2} = \frac{-9x^4-12x^3-1}{(3x^4-1)^2}.$$
So then differentiating again with the quotient rule, we get
$$y'' = \frac{(12x^2)(9x^5 + 15x^4 +5x +3)}{(3x^4-1)^3}$$
(There is a common factor of $3x^4 - 1$ that can be cancelled between the numerator
and the denominator after you apply the quotient rule the second time.)
\item[(N)] Find $y'$
$$y = \frac{\sin(x)}{1 + \cos(x)} + x^2 \cos(x^3 + 3)$$
\item[]  {\it Solution:}  By the quotient, product, and chain rules:
\begin{eqnarray*}
y' &=& \frac{(1 + \cos(x))\cos(x) - \sin(x)(-\sin(x))}{(1 + \cos(x))^2} - x^2\sin(x^3+3)(3x^2) + 2x \cos(x^3 + 3)\\
&=& \frac{1 + \cos(x)}{(1 + \cos(x))^2} - 3x^4\sin(x^3 + 3) + 2x \cos(x^3 + 3)\\
&=& \frac{1}{1 + \cos(x)} - 3x^4 \sin(x^3+3)+ 2x \cos(x^3 + 3).
\end{eqnarray*}
\end{enumerate}
\item[II.] The total cost (in $ \$ $) of repaying a car loan at 
interest rate of $r \%$ per year is $C = f(r)$.  
\begin{enumerate}
\item[(A)] What is the meaning of the statement $f(7) = 20000$?
\item[] {\it Solution:}  At an interest rate of $7\%$ per year, the cost of repaying
the loan is $20000$ dollars.  
\item[(B)] What is the meaning of the statement $f'(7) = 3000$? What are the units of $f'(7)$? 
\item[] {\it Solution:}  At an interest rate of $7\%$ per year, the rate of change
of the cost of repaying the loan is $3000$ dollars per ($\%$ per year).  
\end{enumerate}
\item[III.]  The quantity of a reagent present in a chemical 
reaction is given by $Q(t) = t^3 - 3t^2 + t + 30$ grams at time $t$ seconds for all $t \ge 0$.
(Note:  For a question like this, I could also give you the plot of the function and ask
questions like those below.  In this case you need to start from the formula and compute
$Q'(t)$; if you were given the graph, you need to make the connection between slopes of 
tangent lines and signs of $Q'(t)$ visually.)
\begin{enumerate}
\item[(A)]  Over which intervals with $t \ge 0$ 
is the amount increasing? (i.e. $Q'(t) > 0$) decreasing (i.e. $Q'(t) < 0$)?
\item[] {\it Solution:}  $Q'(t) = 3t^2 - 6t + 1$.  $Q'(t) = 0$ when 
$$t = \frac{6 \pm \sqrt{36-12}}{6} =  1\pm \frac{\sqrt{6}}{3} \doteq 1.816, .184.$$
Since this is a quadratic function with a positive $t^2$ coefficient, $Q'(t) > 0$
for $t > 1.816$ and $t < .184$.  $Q'(t) < 0$ for $.184 < t < 1.816$ ($t$ in seconds).
\item[(B)]  Over which intervals is the rate of change of $Q$ increasing? decreasing?
\item[] {\it Solution:}   The rate of change of $Q$ is increasing when $(Q')' > 0$ and decreasing
when $(Q')' < 0$.  The second derivative of $Q$ is $Q''(t) = 6t - 6$.  So 
$Q''(t) > 0$ for $t > 1$ and $Q''(t) < 0$ for $t < 1$ ($t$ in seconds).
\end{enumerate}
\vskip 10pt
\item[IV.]  Compute the following limits using L'Hopital's Rule or other methods,
as appropriate.
\begin{enumerate}
\item[(A)] $\displaystyle \lim_{x\to 0^+} x^{1/3}\ln(x)$  
\item[] {\it Solution:}  This is a $0\cdot \infty$ form.  We can write it as 
\[
\lim_{x\to 0^+} \frac{\ln(x)}{x^{-1/3}}
\]
which makes it an $\infty/\infty$ form.  Applying L'Hopital's Rule, this is
\begin{eqnarray*}
\lim_{x\to 0^+} \frac{1/x}{(-1/3) x^{-4/3}} &=& 
\lim_{x\to 0^+} -3\cdot \frac{x^{4/3}}{x}\\
 &=& \lim_{x\to 0^+} -3 x^{1/3} = 0.
\end{eqnarray*}
\item[(B)] $\displaystyle \lim_{x\to \infty} \frac{x^3}{e^{2x}}$
\item[] {\it Solution:}  This is an $\infty/\infty$ form.  Applying L'Hopital's Rule three times:
\begin{eqnarray*}
\lim_{x\to\infty} \frac{x^3}{e^{2x}} &=& \lim_{x\to\infty} \frac{3x^2}{2e^{2x}}\quad \text{still $\infty/\infty$}\\
&=& \lim_{x\to\infty} \frac{6x}{4e^{2x}}\quad \text{still $\infty/\infty$}\\
&=& \lim_{x\to\infty} \frac{6}{8e^{2x}}\quad \text{not $\infty/\infty$}\\
&=& 0
\end{eqnarray*}
\item[(C)] $\displaystyle \lim_{x\to 1} \frac{x^2 - 5x + 4}{x^2 - 3x + 2}$
\item[] {\it Solution:}  This one is $0/0$.  We can either use L'Hopital:
\[
\lim_{x\to 1} \frac{x^2 - 5x + 4}{x^2 - 3x + 2} = \lim_{x\to 1} \frac{2x - 5}{2x - 3} = \frac{-3}{-1} = 3.
\]
Or we can also factor the top and bottom and cancel:
\[
\lim_{x\to 1} \frac{x^2 - 5x + 4}{x^2 - 3x + 2} = \lim_{x\to 1} \frac{(x-1)(x-4)}{(x-1)(x-2)} = \lim_{x\to 1}\frac{x-4}{x-2}= \frac{-3}{-1} = 3.
\]
\item[(D)] $\displaystyle \lim_{x\to \infty} \left(1 + \frac{4}{x}\right)^x$
\item[] {\it Solution:}  This is a $1^\infty$ form.  We take logarithms, then apply L'Hopital's Rule:
\begin{eqnarray*}
\lim_{x\to\infty} \ln\left(\left(1 + \frac{4}{x}\right)^x\right) &=& \lim_{x\to\infty} x\ln\left(1 + \frac{4}{x}\right) \quad \text{this is $\infty\cdot 0$}\\
&=& \lim_{x\to\infty} \frac{\ln\left(1 + \frac{4}{x}\right)}{\frac{1}{x}}\quad \text{(this is $0/0$)}\\
&=& \lim_{x\to\infty} \frac{\frac{1}{1 + \frac{4}{x}}\cdot \frac{-4}{x^2}}{\frac{-1}{x^2}}\\
&=& \lim_{x\to\infty} \frac{4}{1 + \frac{x}{4}}\\
&=& 4
\end{eqnarray*}
Since we took logarithms to get the function whose limit is 4, 
the original limit is then found by exponentiating:  
\[
\lim_{x\to \infty} \left(1 + \frac{4}{x}\right)^x = e^4.
\]
\end{enumerate}
\vskip 10pt
\item[V.]   All parts of this question refer to $f(x) = 4x^3 - x^4$.
\vskip 10pt
\begin{enumerate}
\item[(A)]  Find and classify all the critical points of $f$
using the First Derivative Test.
\item[] {\it Solution:}  $f'(x) = 12x^2 - 4x^3 = 4x^2(3 - x)$.  This is defined
for all $x$ and equal to zero at
$x = 0$ and $x = 3$.  Note that $4x^2 \ge 0$ for all $x$.  So the sign of $f'(x)$
comes from the $3 - x$ factor.  That is negative for $x  > 3$ and positive for $x < 3$.
Hence $f'$ changes sign from positive to negative at $x = 3$ and the First Derivative
Test says $f$ has a local local maximum at $x = 3$.  On the other hand, $f'(x)$ does
not change sign at $x = 0$, so that critical point is neither a local maximum nor
a local minimum.  
\item[(B)]  Over which intervals is the  graph $y = f(x)$ concave
up?  concave down?
\item[] {\it Solution:}  $f''(x) = 24x - 12x^2 = 12x(2 - x)$, which is zero at
$x = 0$ and $x = 2$.  Then $f''(x) > 0$ and the graph $y = f(x)$ is concave
up on $(0,2)$ and $f''(x) < 0$ and the graph $y = f(x)$ is concave down on $(-\infty,0)$
and $(2,\infty)$.
\item[(C)]  Sketch the graph $y = f(x)$.  
\item[] {\it Solution:}  See Figure 1 on the back of this page.
\begin{figure*}
\begin{center}
\includegraphics[width=80mm]{Ex4SamplePlotV.png}
\caption{Plot of $y = f(x)$ for Problem V}
\end{center}
\end{figure*}
\item[(D)]  Find the absolute maximum and minimum of $f(x)$
on the interval $[1,4]$.  
\item[] {\it Solution:}  Only the critical point $x = 3$ is in this interval.
$f(1) = 8$, $f(3) = 27$ and $f(4) = 0$.  So $f(3) = 27$ is the maximum value and
$f(4) = 0$ is the minimum value on the interval $[1,4]$.  
\end{enumerate}
\vskip 10pt
\item[VI.]  All three parts of this question refer to the function $f(x)$
\emph{whose derivative is plotted in Figure 1.  NOTE:  This is the graph $y = f'(x)$
not $y = f(x)$.}
\begin{figure}
\begin{center}
\includegraphics[width=100mm]{Ex4SamplePlot.png}
\caption{Plot of $y = f'(x)$ for Problem VI}
\end{center}
\end{figure}
\begin{enumerate}
\item[(A)]  Give approximate values for all the critical points of $f(x)$
in the interval shown, and say whether $f$ has a
local maximum, a local minimum, or neither at each.
\item[] {\it Solution:}  By inspection of the plot in Figure 1, we see
that $f'(x) = 0$ at approximately $x = -5.2, -2$, and $1.2$.  Since $f'$
changes sign from $+$ to $-$ at $x = -5.2$ and $x = 1.2$, those are
local maxima for $f$.  Since $f'$ changes sign from negative to positive at $x = -2$,
that is a local minimum for $f$.
\item[(B)]  Find approximate values for all the inflection points of $f(x)$.
\item[] {\it Solution:}  $y = f(x)$ has inflection points where $f'$ changes from 
increasing to decreasing.  That happens here at roughly $x = -3.9$ and $x = -0.8$.
\item[(C)]  Over which intervals is $y = f(x)$ concave up? concave down?
\item[] {\it Solution:}  Following on from (B), $y = f(x)$ will be concave down 
on intervals where $f'(x)$ is decreasing -- roughly $(-6,-3.9)$ and $(-0.8,2)$.
$y = f(x)$ will be concave up on intervals where $f'(x)$ is increasing -- roughly
$(-3.9,-0.8)$.  
\end{enumerate}
\item[VII.]   A town wants to build a pipeline from a water station on a small
island 2 miles off the shore of its water reservoir to the town. 
One possible route is shown dotted in red. The town is 6 miles along the shore
from the point nearest the island.  It costs \$3 million   
per mile to lay pipe under the water and \$2 million per mile
to lay pipe along the shoreline.
\begin{figure}
\begin{center}
\includegraphics[width=2.5truein,height=2.5truein]{FVI.png} 
\end{center} 
\caption{Figure for problem VII.}
\end{figure}

\begin{enumerate} 
\item[A)] Give the cost $C(x)$ of constructing the pipeline
as a function of $x$.
\item[] {\it Solution:}  By the Pythagorean theorem and the given 
information about cost per mile, we have 
$$C(x) = 3\sqrt{4 + x^2} + 2(6 - x)$$
\item{B)} Where along the 
shoreline should the pipeline hit land to minimize the costs
of construction? 
\item[] {\it Solution:} To find the minimum of $C(x)$, we
can restrict to $x$ in the closed interval $[0,6]$, since it clearly does no
good to take $x < 0$ or $x > 6$.  The function $C(x)$ has
a critical number for $x > 0$ at the positive solution of $C'(x) = 0$:
\begin{eqnarray*}
0 &=& \frac{3x}{\sqrt{4 + x^2}} - 2, \text{or}\\
3x &=& 2 \sqrt{4 + x^2} \\
9x^2 &=& 16 + 4x^2 \\
5 x^2 &=& 16 \\
x &=& \frac{4}{\sqrt{5}} \doteq 1.79.
\end{eqnarray*}
We have $C(0) = 18$, $C(6) = 3\sqrt{40} \doteq 19.0$, and
$C\left(\frac{4}{\sqrt{5}}\right) \doteq 16.47$.  So 
the minimum cost is attained at $x = \frac{4}{\sqrt{5}}\doteq 1.79$ miles.
\end{enumerate}
\end{enumerate}

\end{document}
