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\noindent
\textbf{Math 105 Exam I answers.}

\vspace{0.2in}

\begin{questions}

%%%%%%%%%%%%%%%%

\question Suppose that a baker can sell a cake for \$7, so the
revenue function is $R(q) = 7q$.  The baker estimates the cost of
making $q$ cakes to be $C(q) = 2q+600$.  What is the baker's profit
function?

\noindent
The profit function is $R(q)-C(q) = (7q)-(2q+600) =
\underline{5q-600}$.

%%%%%%%%%%%%%%%%

\question What is the $y$-intercept of the line $-3x + 4y = 2$?

\noindent 
Solve for $y$ to get $y=\frac{3}{4} x + \frac{1}{2}$.  Thus the
$y$-intercept is \underline{$\frac{1}{2}$}.

%%%%%%%%%%%%%%%%

\question What is the slope of the line through the points $(1,1)$ and
$(3,-5)$?

\noindent 
The slope is rise over run, or difference in $y$-values over
difference in $x$-values, which is $\frac{1-(-5)}{1-3} = \frac{6}{-2}
= -3$.  So the answer is \underline{$-3$}.

%%%%%%%%%%%%%%%%

\question What is the equation of the line which goes through the
point $(0,3)$ and which is parallel to the line $y=-4x+5$?

\noindent 
The slope of the line is $-4$, so use the point-slope form for the
equation of the line: $y-3=-4(x-0)$.  Rewrite to get
\underline{$y=-4x+3$}.

%%%%%%%%%%%%%%%%

\question Let $\displaystyle f(x) = \frac{x-7}{x+1}$.  Find $f(a+2)$.

\noindent 
Plug in $a+2$ for $x$ to get
$\displaystyle \frac{a+2-7}{a+2+1} = \underline{\frac{a-5}{a+3}}$.

%%%%%%%%%%%%%%%%

\question Let $f(x)=2x^{2} - 8x - 5$.  Complete the square in this
quadratic function.  (We are looking for an answer of the form ``$f(x)
= A (x+B)^{2} + C$''.)

Following the procedure discussed in class gives the answer:
\underline{$2(x-2)^{2} - 13$}.

%%%%%%%%%%%%%%%%

\question Find the domain of the function $f(x) = \sqrt{x+3}$.

\noindent 
Because of the square root, the quantity $x+3$ can't be negative: $x+3
\geq 0$.  Solving for $x$ gives the answer: the domain is
\underline{all $x$ with $x \geq -3$}.

%%%%%%%%%%%%%%%%

\question Compute $\displaystyle (1/8)^{2/3}$.
Express your answer as a fraction.

\noindent 
Raising something to the power $2/3$ is the same as taking the cube
root and then squaring the result.  So $(1/8)^{2/3} =
(\sqrt[3]{1/8})^{2} = (1/2)^{2} = 1/4$.  So the answer is
\underline{$1/4$}. 

%%%%%%%%%%%%%%%%

\question A local theater owner realizes that when she charges \$18
per ticket, she averages 200 people per night.  When she raises prices
to \$21, the attendance drops to 170.  Let $p$ denote the ticket
price, and let $q$ denote the demand (the attendance).  

\begin{parts}

\part Assuming that the demand is a linear function of the price,
write demand in terms of price.

\noindent 
The answer is the equation of the line going through the two points
$(p_{1}, q_{1}) = (18, 200)$ and $(p_{2}, q_{2}) = (21, 170)$.  The
slope of this line is $\frac{200-170}{18-21} = \frac{30}{-3} = -10$,
so the equation is $q-200 = -10(p-18)$, or \underline{$q=-10p+380$}.

\part If the theater holds 260 people, how much should the owner
charge in order to fill the theater to capacity?

\noindent
Since the theater holds 260 people, set $q=260$ and solve for $p$:
$260=-10p+380$.  Solving for $p$ gives a price of \underline{\$12}.

\end{parts}

%%%%%%%%%%%%%%%%

\question A businessman selling widgets determines that the price $p$
of widgets and the demand $q$ for widgets are related by the function
$p=-6q+1300$.  Determine the revenue function $R(q)$ in terms of the
demand $q$.

\noindent 
The key thing here is that $R=pq$.  Since we have a formula for $p$ in
terms of $q$, just plug it in: $R=(-6q+1300) q
=\underline{-6q^{2}+1300q}$.

%%%%%%%%%%%%%%%%

\question A baseball cap company has a profit function of $-100p^{2} +
1800p - 7200$, where $p$ is the price charged per cap.  How much
should they charge to maximize profit?

\noindent 
Complete the square to find out which price maximizes the profit
function: the result of completing the square is $-100(p-9)^{2} +
900$, so the profit is maximized when the price $p$ is
\underline{\$9}.

%%%%%%%%%%%%%%%%

\question Use the quadratic formula to solve the equation
$2x^{2} +x - 6 = 0$.

\noindent 
The quadratic formula says the solutions are
\[
x=\frac{-1 \pm \sqrt{1-4(2)(-6)}}{4} = \frac{-1 \pm \sqrt{49}}{4} =
\frac{-1 \pm 7}{4}.
\]
So there are two solutions: \underline{$x=-2$ and $x=3/2$}.

%%%%%%%%%%%%%%%%

\question Consider the function $f(x) = \frac{2(x+2)}{(x-3)(x-1)}$.

\begin{parts}

\part Find the vertical asymptotes of $f(x)$, or write ``none'' if
there aren't any.

\noindent 
The vertical asymptotes are the places where the denominator is zero:
\underline{$x=3$ and $x=1$}.

\part Find the horizontal asymptotes of $f(x)$, or write ``none'' if
there aren't any.

\noindent 
Since the degree of the numerator is less than the degree of the
denominator, then \underline{$y=0$} is the horizontal asymptote.

\end{parts}

%%%%%%%%%%%%%%%%

\question The line $x=2$ is a vertical asymptote of the function 
$f(x) = \frac{x+1}{x-2}$.

\begin{parts}

\part Does this curve climb or fall as it approaches the asymptote
from the left? 

\noindent 
The function can only change sign (from plus to minus or vice versa)
at the vertical asymptote $x=2$ or when the function is zero, at
$x=-1$.  Just to the left of the asymptote, then, I can figure out the
sign of the function by plugging in any number between $-1$ and $2$,
say $x=0$.  Since $f(0)=-1/2$, then the function is negative, so the
curve \underline{falls} as it approaches the asymptote.

\part Does this curve climb or fall as it approaches the asymptote
from the right? 

\noindent 
To the right of the asymptote, the function does not change sign, so I
can plug in any number bigger than $2$, say $x=3$.  Since $f(3)=4$,
then the function is positive, so it \underline{climbs} as it
approaches the asymptote from the right.

\end{parts}

%%%%%%%%%%%%%%%%

\question Let $f(x) = \frac{3x-4}{2x-1}$.  Find the horizontal
asymptotes of $f(x)$.

\noindent 
Since the degree of the numerator is the same as the degree of the
denominator, then the horizontal asymptote occurs at the ratio of the
leading terms: at \underline{$y=3/2$}.

%%%%%%%%%%%%%%%%

\question A certain sling-shot will fire a ball straight up in the air
at a speed of 160 feet/second.  The height of the ball after $t$
seconds is given by the formula $h(t) = -16 t^{2} + 160 t$.  What is
the maximum height that the ball reaches?

\noindent 
Complete the square to maximize this formula.  Completing the square
yields $-16(t-5)^{2} + 400$.  Therefore the maximum height is
\underline{400 feet}.

%%%%%%%%%%%%%%%%

\question Consider the function 
\[
f(x) = \begin{cases}
3x+1      & \text{if $x \leq 0$} \\
x^{2} + 1 & \text{if $0 < x \leq 1$} \\
2-x^{2}   & \text{if $x>1$}.
\end{cases}
\] 
Is the function continuous?  If not, give the $x$-values for its
discontinuities.

\noindent 
Plug $x=0$ into $3x+1$ and $x^{2}+1$: you get $1$ both times.  Since
you get the same answer, the function is continuous at $x=0$.  Plug
$x=1$ into $x^{2}+1$ and $2-x^{2}$: you get $2$ and $1$, so the
function is \underline{discontinuous at $x=1$}.

%%%%%%%%%%%%%%%%

\end{questions}

\end{document}

