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\centerline{Mathematics 136 -- Advanced Placement Calculus}
\centerline{Solutions to Ungraded Problems on Problem Set 1}
\centerline{September 15, 2009}
\vskip 10pt
\noindent
1.2/15. (a)  We want $T$ as a function of $N$.  The slope of 
the linear function is given by $\frac{80 - 70}{173 - 113} = \frac{10}{60} = \frac{1}{6}$.
By the point-slope form, $T - 70 = \frac{1}{6}(N - 113)$, so $T = \frac{1}{6}N + \frac{307}{6}$.

(b)  The slope is $\frac{1}{6}$. The meaning of this is that each increase of 6 chirps per
minute corresponds to a 1 degree increase in temperature.

(c)  If $N = 150$, then $T = \frac{1}{6}(150) + \frac{307}{6} \doteq 76$ degrees Fahrenheit.
\vskip 10pt
\noindent
1.3/7.  The graph $y = \sqrt{3x - x^2}$
is shifted 4 units to the right, reflected over the $x$-axis and
shifted 1 unit down:  
$$y = -\sqrt{3(x + 4) - (x + 4)^2} - 1 = -\sqrt{-x^2 - 5x - 4} - 1.$$
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\noindent
1.3/16.  The graph $y = \frac{1}{x-4}$ is obtained by shifting the graph
$y = \frac{1}{x}$ four units horizontally to the right.  (This also 
shifts the vertical asymptote to the line $x = 4$.)
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\noindent
1.3/24.  The graph $y = |\cos(\pi x)|$ is obtained from $y = \cos(x)$ 
by compressing horizontally (the period becomes 2 rather than $2 \pi$), 
and reflecting the parts of the graph $y = \cos(\pi x)$ that
lie below the $x$-axis over the $x$-axis.  See Figure 1 (on the back page).
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\noindent
1.5/21.  Since $(1,6)$ and $(3,24)$ must be on the graph $y = Ca^x$,
we get the two equations 
\begin{align*}
6 &= Ca^1 = Ca\\
24 &= Ca^3.
\end{align*}
Solving for $C$ in the first equation, $C = \frac{6}{a}$.  So substituting
into the second equation, we get $24 = \frac{6}{a}a^3 = 6 a^2$.  Since $a > 0$,
this shows $a = 2$, and hence $C = 3$.  The equation is $y = 3\cdot 2^x$.  
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\noindent
1.6/26.  First we solve the equation $y = \frac{e^x}{1 + 2e^x}$ for $x$ as a function
of $y$:
\begin{align*}
(1 + 2e^x)y &= e^x \text{ (after clearing denominators) }\\
y + 2y e^x &= e^x \\
y &= (1 - 2y)e^x\\
\frac{y}{1 - 2y} &= e^x\\
\ln\left(\frac{y}{1 - 2y}\right) &= x \text{ (after taking logarithms to solve for $x$) }.
\end{align*}
Then $f^{-1}(x) = \ln\left(\frac{y}{1 - 2y}\right)$ is the inverse function (swap $x$ and $y$).
\begin{figure}
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\caption{Plot for 1.3/24}
\end{figure}
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\noindent
1.6/51.  Solve for $x$ in the exponent by taking logarithms.  Using natural log,
\begin{align*}
\ln\left(2^{x - 5}\right) &= \ln(3)\\
(x - 5) \ln(2) &= \ln(3)\\
\ln(2)x &= \ln(3) + 5\ln(2)\\
x&= \frac{ln(3) + 5\ln(2)}{\ln(2)} = \frac{\ln(96)}{\ln(2)}, \text{ or } 5 + \frac{\ln(3)}{\ln(2)}.
\end{align*}
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\noindent
2.2/5 (a) $\lim_{t\to 0^-} g(t) = -1$.

(b) $\lim_{t \to 0^+} g(t) = -2$.

(c) $\lim_{t \to 0} g(t)$ does not exist since the one-sided limits are different.

(d) $\lim_{t \to 2^-} g(t) = 2$.

(e) $\lim_{t \to 2^+} g(t) = 0$.

(f) $\lim_{t \to 2} g(t)$ does not exist (same reason as in (c)).

(g) $g(2) = 1$

(h) $\lim_{t \to 4} g(t) = 3$.
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\noindent
2.2/15.  There are infinitely many correct graphs here.  They would 
all have jump discontinuities at $x = 3$, and removable discontinuities
at $x = -2$.  
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