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\centerline{MATH 133 -- Calculus with Fundamentals 1}
\centerline{Problem Set 1B Solutions, September 8, 2017}
\vskip 10pt
\noindent
1.1/71.  Given: The function $f(x)$ has domain $[4,8]$ and range $[2,6]$.
\begin{enumerate}
\item[(a)]  The function $f(x) + 3$ has domain $[4,8]$ (the same as $f(x)$) and range $[5,9]$, 
since $2 \le f(x) \le 6$ is equivalent to $5 \le f(x) + 3 \le 9$.
\item[(b)]  The function $f(x + 3)$ has domain $[1,5]$ since $4 \le x + 3 \le 8$ is the same
as $1 \le x \le 5$.  The range is the same as the range of $f(x)$ -- $[2,6]$.
\item[(c)]  The function $f(3x)$ has domain $[4/3,8/3]$ since $4 \le 3x \le 8$ is equivalent 
to $4/3 \le x \le 8/3$.  The range is $[2,6]$, the same as for $f(x)$.
\item[(d)]  The function $3f(x)$ has domain $[4,8]$, the same as $f(x)$.  The range is $[6,18]$,
since $2\le f(x) \le 6$ is equivalent to $6 \le 3f(x) \le 6$.
\end{enumerate}
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\noindent
77.  Another way to state the definition is 
\begin{align*}
f(x) &= \begin{cases} x &\ {\rm if}\ x \ge 2 - x\\
                       2-x &\ {\rm if}\ 2 - x > x.
         \end{cases}\\
&= \begin{cases} x &\ {\rm if}\ x \ge 1\\
                       2-x &\ {\rm if}\ x < 1.
         \end{cases}\\
\end{align*}  
The domain is the set of all real $x$ (or $(-\infty,\infty)$).  The range is $[1,+\infty)$.
This shows the graph has the form given in Figure 1 on the back of this page.
From this, we see that this is the graph $y = |x|$, shifted to the right one unit and up
one unit.  Another formula is 
$$y = |x - 1| + 1.$$
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\noindent
1.2/22.  The line contains the points $(0,b)$ and $(a,0)$.  This means the slope is 
$$m = \frac{b - 0}{0 - a} = -\frac{b}{a}.$$
The point $(0,b)$ is the $y$-intercept, so the equation of the line has the slope-intercept
form:
$$y = -\frac{b}{a} x + b.$$
From this, we can add $\frac{b}{a} x$ to both sides to obtain
$$\frac{b}{a} x + y = b,$$
then divide both sides by $b$ to obtain
$$\frac{x}{a} + \frac{y}{b} = 1.$$
This is the form we wanted.  
\begin{figure}
\begin{center}
\includegraphics[width=1.5in]{shiftedabs.png}
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\caption{The graph $y = f(x)$ in 1.1/77}
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1.2/48.  Saying $f(x) = Ax^2 + Bx + C$ has a \emph{double} root means that its factored form is
$A(x - r)^2$ for some $r$ (equivalently, if you use the quadratic formula to solve $A x^2 + Bx + C = 0$, 
then the $B^2 - 4AC$ under the radical sign is \emph{zero} and you only get one root).  
Graphically, that means the parabola $y = Ax^2 + Bx + C$ is ``just touching'' the 
$x$-axis at $x = r$.  (Graph $y = (x - 2)^2$ to see exactly what that looks like.)  It is 
statement (a) that is correct, since any parabola can be \emph{shifted up or down} by a unique
constant $c$ to make it just touch the $x$-axis at one point.  (b) cannot be true because shifting
the parabola left or right does not change the number of times it crosses the $x$-axis; it just
changes the \emph{locations of the points} (if there are any) where the parabola crosses the $x$-axis.  

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