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\title{Math 165: Honors Calculus I\\
Exam II Solutions}
\date{November 11, 1999}

\begin{document}

\maketitle

\bigskip

\begin{enumerate}
\item
  a) The area between the graphs of $f(x)$ and $g(x)$ on $[a,b]$ is defined
     to be $\int_a^b |f(x) - g(x)|\,dx$.

  b) Consider a unit circle centered at the origin~$O = (0,0)$ and let
     $A$ be the point $(1,0)$. Given a number $x$ between $0$ and $2\pi$,
     let $P_x$ be the point on the unit circle such that $AOP_x$ forms
     a circular sector with area $x/2$. Then $\cos(x)$  and $\sin(x)$
     are the coordinates of $P_x = (\cos(x),\sin(x))$. The domain
     of $\sin(x)$ and $\cos(x)$ is extended to all real numbers by
     periodicity: $\sin(x + 2\pi) = \sin(x)$ and $\cos(x+2\pi) = \cos(x)$.

  c) $\lim_{x\to p}f(x) = A$ means that for any $\epsilon>0$ there
     is a $\delta >0$ such that if $0 < |x-p|<\delta$ then $|f(x) - A|
     < \epsilon$.

  d) A function $f(x)$ is one-to-one if for any two distinct numbers
     $x_1 \ne x_2$ in its domain, the function values are distinct, $f(x_1)
     \ne f(x_2)$.

  e) The average value of an integrable function on $[a,b]$ is
     $\overline{f} = \frac{1}{b-a}\int_a^b f(x)\,dx$.

\problemspace
\item
  a) \textsc{Basic Limit Theorems}.
     Let $f$ and $g$ be function such that $\lim_{x\to p}f(x) = A$ and
     $\lim_{x\to p}g(x) = B$. Then
     \begin{enumerate}[i)]
       \item $\lim_{x\to p}[f(x)+g(x)] = A + B$,
       \item $\lim_{x\to p}[f(x)-g(x)] = A - B$,
       \item $\lim_{x\to p}[f(x)\cdot g(x)] = A \cdot B$,
       \item $\lim_{x\to p}[f(x)/g(x)] = A/B$, if $B \ne 0$.
     \end{enumerate}

  b) \textsc{The Squeezing Principle}. If $g(x) \le f(x) \le h(x)$
     for $x$ in a neighborhood of $p$ ($x\ne p$), and if $\lim_{x\to p}
     g(x) = A = \lim_{x\to p} h(x)$, then $\lim_{x\to p} f(x) = A$.

  c) \textsc{Bolzano's Theorem}. Let $f$ be continuous on $[a,b]$ and
     assume $f(a)$ and $f(b)$ have opposite signs. Then there is a
     $c\in[a,b]$ such that $f(c) = 0$.

  d) \textsc{Intermediate Value Theorem}.  Let $f$ be continuous on
     $[a,b]$ and let $x_1<x_2$ be points in $[a,b]$ such that $f(x_1)
     \ne f(x_2)$. Then for any value $k$ between $f(x_1)$ and $f(x_2)$
     there is at least one point $c\in[x_1,x_2]$ such that $f(p) = c$.

  e) \textsc{Extreme Value Theorem}.  Let $f$ be continuous on
     $[a,b]$. Then $f$ has a maximum and a minimum in $[a,b]$, that is,
     there are points $c$, $d\in[a,b]$ such that $f(c) \le f(x) \le f(d)$
     for all $x \in [a.b]$.

\problemspace

\item
  a) $\lim_{x\to 1} \frac{\sin(x^2-1)}{x-1}
     =\lim_{x\to 1} (x+1)\frac{\sin(x^2-1)}{x^2-1}
     =\lim_{x\to 1} (x+1)\lim_{x\to 1} \frac{\sin(x^2-1)}{x^2-1}
     =2\cdot1 = 2$

  b) $\lim_{x\to 0} \frac{\sqrt{x^4+1}-1}{x^4}
     =\lim_{x\to 0}\frac{(x^4+1)-1}{x^4(\sqrt{x^4+1}+1)}
     =\lim_{x\to 0}\frac{1}{\sqrt{x^4+1}+1} = \frac{1}{2}$

  c) $\lim_{x\to2}\sqrt{x^3 - \frac{1}{x^3}} = \sqrt{2^3 -\frac{1}{2^3}} =
     \sqrt{\frac{63}{8}}$

  d) Since $\left[\frac{1}{x}\right]\le \frac{1}{x}$, we get
     $0 \le x^2\Big(1 + \left[\frac{1}{x}\right]\Big)
     \le x^2\Big(1 + \frac{1}{x}\Big) = x^2 + x$ for $x > 0$. Since
     $\lim_{x\to 0^+} x^2+x = 0$, the
     \textsc{Squeezing Principle} implies that 
     $\lim_{x\to 0^+} x^2\Big(1 + \left[\frac{1}{x}\right]\Big) = 0$

\problemspace

\item Assume $\lim_{x\to p} f(x) = A$ and $\lim_{x\to p} g(x)
= B$. Using the definition of limit, prove that $\lim_{x\to p} \big(f(x)
- g(x)\big) = A-B$.

\emph{Proof.} Let $\epsilon>0$. By assumption, there exists
$\delta_1>0$ and $\delta_2>0$ such that if $0 < |x-p| < \delta_1$ then
$|f(x) - A|<\epsilon/2$, and if $0 < |x-p| < \delta_2$ then $|g(x) -
B|<\epsilon/2$.  Let $\delta = \min\{\delta_1, \delta_2\}$. If $0 <
|x-p| < \delta$ then $|(f(x) - g(x)) - (A-B)| \le |f(x)-A| + |g(x)-B| <
\epsilon/2 + \epsilon/2 = \epsilon$. Therefore, $\lim_{x\to p} \big(f(x)
- g(x)\big) = A-B$.

\problemspace
\item Let $f$ be continuous at $p$ and let $g$ be continuous at $q
= f(p)$. Prove that $g\circ f$ is continuous at~$p$.

\emph{Proof.} Let $\epsilon > 0$. By assumption, there is a $\delta_1>0$
such that if $0<|y-q|<\delta_1$ then $|g(y)-g(q)| < \epsilon$. Also
by assumption, there is a $\delta_2>0$ such that if $0<|x-p|<\delta_2$
then $|f(x)-f(p)|< \delta_1$. By what we've just written, this implies
that $|g(f(x)) - g(f(p))|<\epsilon$.  Therefore, $g(f(x))$ is continuous
at $x=p$.

\problemspace
\item Let $f$ be integrable and positive on~$[a,b]$. Show that the
function $F(x) = \int_a^x f(t)\,dt$ is increasing on $[a,b]$.

\emph{Proof.} Let $a\le x_1 < x_2 \le b$. Then $F(x_2) - F(x_1) =
\int_a^{x_2} f(t)\,dt - \int_a^{x_1} f(t)\,dt = \int_{x_1}^{x_2}
f(t)\,dt$.  Since $f(t) > 0$ for $t\in[a,b]$, $\int_{x_1}^{x_2}
f(t)\,dt \ge \int_{x_1}^{x_2} 0\,dt = 0$ by the \textsc{Comparison
Theorem}. Therefore, $F(x_2) - F(x_1) \ge 0$ and hence $F(x)$ is
increasing on $[a,b]$.

\problemspace
\item Suppose the inverse of a function is $f^{-1}(x) =
\sqrt{\frac{1}{\sqrt{x}}-1}$ for $0 < x \le 1$. Determine the original function,
$f(x)$, including its domain.


\emph{Solution.}
By definition, $y=f^{-1}(x)$ if and only if $f(y) = x$. To find $f(x)$
we solve the equation $y = \sqrt{\frac{1}{\sqrt{x}}-1}$ for $x$ to get
$x = \frac{1}{(y^2+1)^2} =f(y)$, or, $f(x) =\frac{1}{(x^2+1)^2}$. The range
of $f^{-1}$ on the interval $(0,1]$ is $[0,\infty)$ and this becomes
the domain of $f$.

\end{enumerate}

\end{document}
