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\title{Math 165: Honors Calculus I\\
Exam II Solutions}
\date{November 19, 1998}

\begin{document}

\maketitle

\bigskip

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

  b) A function $f$ is continuous at $p$ if $f$ is defined in a
  neighborhood of $p$ and $\lim_{x\to p}f(x) = f(p)$.

  c) The inverse of $f(x)$  is a function $g(x)$ whose domain is the
  range of $f$ and satisfies: $y = f(x)$ if and only if $g(y) = x$.

  d) $f'(x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$.

\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{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 $f(p) = c$.

  c) \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 $x \in [a.b]$.

  d) \textsc{Mean Value Theorem for Integrals}.
  Let $f$ be continuous on $[a,b]$. Then $f$ attains its average value at
  some point in $[a,b]$, that is, there is a $p\in[a,b]$ such that
  \[ f(p) = \frac{1}{b-a}\int_a^b f(x)\,dx \]

\problemspace

\item
  a) $\lim_{x\to 1} (x-1) \sin \frac{x}{x-1} = 0$. This follows from the
  \textsc{Squeezing Principle}: $-1 \le \sin \frac{x}{x-1} \le 1$ for
  all $x\ne1$, so $-|x-1| \le (x-1)\sin \frac{x}{x-1}  \le |x-1|$. Since the
  outside terms, $\pm|x-1|$, approach~0, the middle term also approaches~0.

  b)
  \begin{eqnarray*}
  \lim_{x\to 0} \frac{\cos(x)-1}{x^2}
      &=&\lim_{x\to 0} \frac{(\cos(x)-1)(\cos(x)+1)}{x^2(\cos(x)+1)}
       =\lim_{x\to 0} \frac{\cos^2(x)-1}{x^2(\cos(x)+1)}\\
      &=&\lim_{x\to 0} \frac{-\sin^2(x)}{x^2}\frac{1}{\cos(x)+1}
       = \lim_{x\to 0} -\Big[\frac{\sin(x)}{x}\Big]^2\frac{1}{\cos(x)+1}\\
      &=& -1^2 \cdot \frac{1}{2} = -\frac{1}{2}
  \end{eqnarray*}

  c)
  \begin{eqnarray*}
  \lim_{x\to 2^-} x^3 - [x^3]
    &=& \lim_{x\to 2^-} x^3 - \lim_{x\to 2^-} [x^3] \quad\mbox{(both
    left-sided limits exist)}\\
    &=& 8 - 7 = 1\quad\mbox{(if $1.95<x<2$ then $[x^3] = 7$)}
  \end{eqnarray*}

  d) Since the right-hand limit,
  $\lim_{x\to 0^+} x\Big|1 + \frac{1}{x}\Big|
  = \lim_{x\to 0^+} \frac{x}{|x|}|x + 1| = 1\cdot|0+1| = 1$,
  does not equal the left-hand limit,
  $\lim_{x\to 0^-} \frac{x}{|x|}|x + 1| = -1\cdot|0+1| = -1$, the limit
  $\lim_{x\to 0} x\Big|1 + \frac{1}{x}\Big|$ does not exist.

\problemspace

\item  a) Let $f(x) = x^{2/3}+x^{-1}$, $x>0$. Then, by the power rule,
  $f'(x) = (2/3)x^{2/3-1} + (-1)x^{-1-1} = (2/3)x^{-1/3} - x^{-2}$.

  b) Let $a$ and $b$ be constants, not both zero, and let $f(x) =
  \frac{ax+b}{a-bx}$. Then, by the quotient rule,
  $f'(x) = \frac{a(a-bx) - (-b)(ax+b)}{(a-bx)^2} = \frac{a^2+b^2}{(a-bx)^2}
  > 0$ for all $x \ne a/b$.


  c) Let $f(x) = (ax+b)\cos(x) + (cx+d)\sin(x)$. Determine values of the
  constants $a$, $b$, $c$, $d$, such that $f'(x) = x\sin(x)$.

  \emph{Solution.} By the power rule, $f'(x) = a \cos(x) - (ax+b)\sin(x)
  + c\sin(x) + (cx+d)\cos(x) = (cx+d+a)\cos(x) + (-ax-b+c)\sin(x)$. This
  equals $x\sin(x)$ when $-a=1$, $-b+c=0$, $c=0$, and $d+a=0$, i.e., when
  $a=-1$, $b=c=0$, and $d=1$. Thus $f(x) = -x\cos(x) + \sin(x)$.

\problemspace
\item Let $f$ be an integrable function on $[a,b]$ and let
$F(x) = \int_a^x f(t)\,dt$. Prove that $F$ is continuous at $c \in [a,b]$.

\emph{Proof.} It is enough to show that $\lim_{x\to c} F(x) - F(c) = 0$.
First note that
\[F(x)-F(c) = \int_a^xf(x)\,dx - \int_a^cf(x)\,dx = \int_c^xf(x)\,dx\]
Since $f$ is integrable on $[a,b]$, it is bounded on $[a,b]$ by
definition, so there is a constant $M>0$ such that $-M \le f(x) \le M$
for all $x\in[a,b]$. This implies that $-M|x-c| \le \int_c^x f(x)\,dx
\le M|x-c|$, so $|F(x) - F(c)| < M |x-c|$. Given any $\epsilon>0$, let
$\delta = \epsilon/M$. If $|x-c|<\delta$, then $|F(x)-F(c)| < M|x-c| <
M\delta = \epsilon$.

\item
Let $n\ge2$ be a positive integer. Prove that the polynomial
$f(x) = x^n - nx + 1$ has a root in $[0,1]$.

\emph{Proof.} If $n=2$, then $f(x) = x^2-2x+1$ and $f(1) = 0$. So we
may assume $n>2$. Since $f(0) = 1$ and $f(1) = 1 - n + 1 < 0$, Balzano's
theorem implies that there is a $c\in[0,1]$ such that $f(c) = 0$.

\problemspace
\item Find the largest interval $I$ containing $x = 1$ on which
the function $f(x) = \frac{1}{1+x^2}$ has an inverse. Give a formula for
the corresponding inverse function, $f^{-1}(x)$, as a function of $x$ and
determine its domain.

\emph{Solution.} First note that $f$ is strictly decreasing for $x\ge0$:
If $0\le x_1<x_2$ then $1+x_1^2<1+x_2^2$ so $f(x_2) = \frac{1}{1+x_2}
< \frac{1}{1+x_1^2} = f(x_1)$. Therefore, $f$ is 1-1 on $[0,\infty)$.
Now $f$ cannot be 1-1 on a larger interval, since any such interval would
contain a negative number, say $-a<0$, and $f(-a) = f(a)$ showing that $f$
is not 1-1 on that interval.

To find $f^{-1}(x)$ we solve the equation $y = \frac{1}{1+x^2}$ for $x$
to get $x = +\sqrt{\frac{1}{y}-1}$.  The range of $f$ on the interval
$[0,\infty)$ is $(0,1]$: it is clear that $0 < \frac{1}{1+x^2} \le 1$ and
any number $0<y\le 1$ is achieved by $f(x)$ when $x =\sqrt{\frac{1}{y}-1}$.
Thus the inverse function is $f^{-1}(x) = \sqrt{\frac{1}{x}-1}$ and its
domain is $(0,1]$.

\problemspace
\item
  a) Using the definition of continuity, prove that $f(x) = x^2$ is
  continuous at any real number~$p$.

  \emph{Proof.} We must show that given any $\epsilon>0$ there
  is a $\delta>0$ such that if $|x-p|<\delta$ then $|f(x)-f(p)|<
  \epsilon$. If $x$ is within 1 unit of $p$, i.e. if $|x-p|<1$, then
  $|x| < |p|+1$, so $|x+p| < |x| + |p| < 2|p|+1$. This allows us to
  estimate $|f(x)-f(p)|$ in terms of $|x-p|$: $|f(x)-f(p)| = |x^2-p^2|
  = |x+p||x-p| < (2|p|+1)|x-p|$. Let $\delta = \min\{1, \epsilon/(2|p|+1)\}$.
  Then $|x-p|<\delta$ implies both $|x-p|<1$ and $|x-p|<\epsilon/(2|p|+1)$
  and we get $|f(x)-f(p)| < (2|p|+1)|x-p| < (2|p|+1)\epsilon/(2|p|+1)
  = \epsilon$.

  b) Using the definition of the derivative, prove that the derivative
  of $f(x) = \sqrt{x}$ is $f'(x) = \frac{1}{2\sqrt{x}}$.

  \emph{Proof.}
  \begin{eqnarray*}
  f'(x) &=& \lim_{h\to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}
   = \lim_{h\to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h}+\sqrt{x})}
   {h(\sqrt{x+h}+\sqrt{x})}\\
  &=&\lim_{h\to 0} \frac{(x+h) - x}{h(\sqrt{x+h}+\sqrt{x})}
   =\lim_{h\to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}} =
   \frac{1}{2\sqrt{x}}\\
  \end{eqnarray*}

\end{enumerate}

\end{document}
