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\title{Math 165: Honors Calculus I\\
Final Exam Solutions}
\date{December 15, 1999}

\begin{document}

\maketitle

\bigskip

\begin{enumerate}


\item Give complete definitions\dots

See the text.

\problemspace

\item State the following theorems\dots

See the text.

\problemspace

\item Calculate the following.

\begin{enumerate}

\item $\int_0^{2} [x^2]\, dx$ where $[u]$ is the greatest integer
${}\le u$.

\emph{Solution.}
\begin{eqnarray*}
\int_0^{2} [x^2]\, dx &=& \int_{0}^{1}\!0\,dx + \int_{1}^{\sqrt{2}}\!1\,dx
+\int_{\sqrt{2}}^{\sqrt{3}}\!2\,dx + \int_{\sqrt{3}}^{2}\!3\,dx\\
&=& (\sqrt{2}-1)+2(\sqrt{3}-\sqrt{2})+3(2-\sqrt{3}) = 5-\sqrt{3}-\sqrt{2}
\end{eqnarray*}

\item Find the area between the graphs of $\sin(x)$ and $\cos(x)$
on $[0,2\pi]$.

\emph{Solution.}
\begin{eqnarray*}
A &=& \int_0^{2\pi}\!\!|\sin(x)-\cos(x)|\,dx\\
&=&\int_0^{\pi/4}\!\!\cos(x)-\sin(x)\,dx
+\int_{\pi/4}^{5\pi/4}\!\!\sin(x)-\cos(x)\,dx
+\int_{5\pi/4}^{2\pi}\!\!\cos(x)-\sin(x)\,dx\\
&=&(\sin(x)+\cos(x))\Big|_0^{\pi/4}
\!\!+(-\cos(x)-\sin(x))\Big|_{\pi/4}^{5\pi/4}
\!\!+(\sin(x)+\cos(x))\Big|_{5\pi/4}^{2\pi}\!\!=4\sqrt{2}
\end{eqnarray*}

\item The average value of $f(x)=x(x-1)$ on the interval $[0,2]$.

\emph{Solution.}
$\overline{f} = \frac{1}{2}\int_0^2x^2-x\,dx =
\frac{1}{2}(\frac{1}{3}x^3-\frac{1}{2}x^2)\Bigg|_0^2 =
\frac{1}{2}(\frac{1}{3}8-\frac{1}{2}4) = \frac{1}{3}$

\end{enumerate}

\problemspace

\item Compute the following limits or prove they do not exist.

\begin{enumerate}

\item $\lim_{x\to 1} \frac{x}{x-1}\sqrt{1-\frac{2}{x}+\frac{1}{x^2}}$

\emph{Solution.} Since
\[ \sqrt{1-\frac{2}{x}+\frac{1}{x^2}}
= \sqrt{\Big(1-\frac{1}{x}\Big)^2}
= \sqrt{\Big(\frac{x-1}{x}\Big)^2}
= \Big|\frac{x-1}{x}\Big|\]
we find 
\[\frac{x}{x-1}\sqrt{1-\frac{2}{x}+\frac{1}{x^2}}
=\frac{x}{x-1}\Big|\frac{x-1}{x}\Big| = \left\{\begin{array}{rl}
-1&0<x<1\\1&x>1\end{array}\right.\]
This shows that the right-hand limit is 1, and the left-hand limit is $-1$.
Therefore, the original limit does not exist.

\item $\lim_{x\to0} \sin(x)\cos(1/x)$

\emph{Solution.}
Since $0\le|\cos(1/x)| \le 1$, we get $0\le|\sin(x)\cos(1/x)| \le
|\sin(x)|$ and since $\lim_{x\to0}|\sin(x)|=0$, the \textsc{Squeezing
Principle} implies that $\lim_{x\to0} \sin(x)\cos(1/x) = 0$

\end{enumerate}

\problemspace

\item

\begin{enumerate}

\item Find the maximum and the minimum of $f(x) = \frac{x^2+1}{x^3+2}$
on $[0,2]$.

\emph{Solution.}
\[f'(x) = \frac{2x(x^3+2)-3x^2(x^2+1)}{(x^3+2)^2}
=-\frac{x(x-1)(x^2+x+4)}{(x^3+2)^2}\]
Thus, $f'(x)=0$ when $x = 0$ or $x=1$. Evaluating $f(x)$ at these critical
points and the endpoints of the interval reveals a minimum value of $f(0)
= f(2) = 1/2$ and a maximum of $f(1) = 2/3$.

\item  Find the line tangent to the ellipse $x^2 + 4y^2 = 4$ at the
point $(\sqrt{3},-1/2)$.

\emph{Solution.}
Taking the derivative implicitly, we find $2x + 8yy' = 0$ or $y' =
-x/(4y)$. Evaluating the derivative at $(x,y)=(\sqrt{3},-1/2)$
gives $y' = \sqrt{3}/2$. The equation of the tangent line is thus $y+1/2 =
(\sqrt{3}/2)(x-\sqrt{3})$ or $y=(\sqrt{3}/2)x-2$.

\end{enumerate}

\problemspace

\item Use the definition of the derivative to prove that
$\frac{d}{dx}\sin(x) = \cos(x)$.

\emph{Solution.}
We first observe that
\begin{eqnarray*}
\lim_{h\to0}\frac{\cos(h)-1}{h} 
&=&\lim_{h\to0}\frac{\cos^2(h)-1}{h(\cos(x)+1)}
=\lim_{h\to0}-h\frac{\sin^2(h)}{h^2}\frac{1}{(\cos(x)+1)}\\
&=&-0\cdot1^2\cdot\frac{1}{2} = 0
\end{eqnarray*}
Using the addition formula for $\sin(x+h)$ we find
\begin{eqnarray*}
\frac{d}{dx}\sin(x)
&=& \lim_{h\to0}\frac{\sin(x+h)-\sin(x)}{h}\\
&=& \lim_{h\to0}\frac{\sin(x)\cos(h)+\cos(x)\sin(h)-\sin(x)}{h}\\
&=& \lim_{h\to0}\sin(x)\frac{\cos(h)-1}{h}
  + \cos(x)\frac{\sin(h)}{h}\\
&=& \sin(x)\cdot0 + \cos(x)\cdot1 = \cos(x)
\end{eqnarray*}


\problemspace

\item Use the definition of the limit to prove that $\lim_{x\to3} x^2 = 9$.

\emph{Solution.}
Given any $\epsilon>0$ let $\delta = \min\{1,\epsilon/7\}$. If
$0<|x-3|<\delta$, then $|x-3|<1$ and $|x-3|<\epsilon/7$. In
particular, $2<x<4$, so $|x+3|<7$. Thus $|x^2-9| =
|x+3||x-3|<7|x-3|<7(\epsilon/7)=\epsilon$.

\problemspace

\item Use the \textsc{Mean Value Theorem} to prove that if $f'(x) > 0$ on
an interval $(a,b)$, then $f(x)$ is strictly increasing on that interval.

\emph{Solution.} Let $x_1$, and $x_2$ be any two points in the interval
$(a,b)$ with $x_1<x_2$. We must show that $f(x_1)<f(x_2)$. The
\textsc{Mean Value Theorem} implies that there is a $c\in(x_1,x_2)$
such that $f'(c)(x_2-x_1) = f(x_2)-f(x_1)$. Since the left hand side is
positive, so is the right hand side, and therefore $f(x_1)<f(x_2)$.

\problemspace

\item

\begin{enumerate}

\item Show that if $f(x)$ is a polynomial of degree $n$, then $f'(x)$ is a
polynomial of degree $n-1$.

\emph{Solution.} 
A polynomial a degree $n$ is a function of the form $f(x) = \sum_{k=0}^n
c_kx^k$ with $c_n\ne0$. The power rule and linearity of derivatives implies
that $f'(x) = \sum_{k=1}^nkc_kx^{k-1}$. The highest order term in this sum
is $nc_nx^{n-1}$ which means $f'(x)$ has degree $n-1$.

\item Use Rolle's Theorem, induction, and part a) to prove that if $f(x)$
is a polynomial of degree $n$ then $f(x)$ has at most $n$ distinct roots.

\emph{Solution.} 
Let us prove the assertion using induction on the degree~$n$ of $f(x)$.
The assertion is true for $n=1$: In this case, $f(x) = c_1 x + c_0$ has
exactly one root, $x = -c_0/c_1$. We now assume the assertion is true for
polynomials of degree $n-1\ge1$, and prove it is true for polynomials
of degree~$n$. By part a), if $f(x)$ has degree $n$, then $f'(x)$ has
degree $n-1$. By the induction hypothesis, we know that $f'(x)$ can have
at most $n-1$ distinct roots. Rolle's theorem implies that between any
two roots of $f(x)$ there is a root of $f'(x)$. So, if $f(x)$ had more
than $n$ distinct roots, say $r_1,\dots,r_m$ with $m>n$, then $f'(x)$
would have more than $n-1$ roots, i.e., at least one root in each interval
$(r_i,r_{i+1})$, $i=1,\dots,m-1$. This contradiction implies that $f(x)$
can have at most $n$ distinct roots.

\end{enumerate}

\problemspace

\item Let $f(x) = \frac{2}{5}x^5 - 3x^3 + 7x$.

\begin{enumerate}

\item Determine the intervals on which $f$ is increasing and decreasing.

\emph{Solution.} The roots of $f'(x) = 2x^4-9x^2 + 7$ are $x=\pm1$
and $x=\pm\sqrt{7/2}$. Moreover, $f'(x)>0$ and $f(x)$ is increasing
for $x<-\sqrt{7/2}$, $-1<x<1$, and $x>\sqrt{7/2}$; $f'(x)<0$ and $f(x)$
is decreasing for $-\sqrt{7/2}<x<-1$, and $1<x<\sqrt{7/2}$.

\item Determine the relative maxima and minima of $f$.

\emph{Solution.} The \textsc{First Derivative Test} implies that $f(x)$
has relative maxima at $x=-\sqrt{7/2}$ and $x=1$ ($f(-\sqrt{7/2})=
-(7/5)\sqrt{7/2}$, $f(1) = 22/5$), and relative minima at $x=-1$, and
$x=\sqrt{7/2}$ ($f(-1) = -22/5$ $f(\sqrt{7/2})= (7/5)\sqrt{7/2}$).

\item Determine the intervals on which $f$ is convex and concave.

\emph{Solution.} The roots of $f''(x) = 8x^3 - 18x$ are $x=\pm3/2$ and
$x=0$. $f''(x)>0$ and $f(x)$ is convex for $-3/2<x<0$ and $x>3/2$.
$f''(x)<0$ and $f(x)$ is concave for $x<-3/2$ and  $0<x<3/2$.

\item Sketch the graph of $f$.

\includegraphics{graph.eps}

\end{enumerate}

\problemspace

\end{enumerate}

\end{document}
