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%\markright{\bf Math 131--FALL 2002 \ \ \ \  EXAM 1 \hfill page \hspace{1ex}}




\begin{document}

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\begin{center}
{\bf 
SOLUTION OF EXAM 1 MATH 131 Fall 2002}
\end{center}



\begin{enumerate}
%***********************************************************************
% Limits and limit laws (sec 2.2 and 2.3)
%***********************************************************************
\item
a) (10) Evaluate the limit
\[
\lim_{x\rightarrow 2}\frac{2x^2-5x+2}{x^3-4x}
\]
by an algebraic procedure. Show all your algebraic steps.

\[
\lim_{x\rightarrow 2}\frac{2x^2-5x+2}{x^3-4x}=
\lim_{x\rightarrow 2}\frac{(2x-1)(x-2)}{x(x-2)(x+2)}=
\lim_{x\rightarrow 2}\frac{2x-1}{x(x+2)}=\frac{2(2)-1}{2(2+2)}=\frac{3}{8}
\]



b)
(5) Support your result in part a) with a calculator.
Describe your procedure and show as much data as possible.


\begin{tabular}{l|l}
x & y\\
\hline
1.9 & .377868\\
1.99 & .375310\\
1.9999 & .375003\\
1.999999 & .375000\\
 & \\
2.1 & .371661\\
2.01 & .374685\\
2.0001 & .374997\\
2.000001 & .374999\\
\end{tabular}

As the values of $x$ approach 2 either from the right or from the
left (but not equal to 2) the values of $y$ approach $.375=3/8$.

%***********************************************************************
% Continuity and Differentiability (sec 2.5 and 2.8)
%***********************************************************************
\item
%\pagebreak
Given the function
\[
f(x) \ = \ \left\{
\begin{array}{ccc}
x^2+1 & \mbox{for} & x\leq 1
\\
ax+b & \mbox{for} & x > 1,
\end{array}
\right.
\]
where $a$, $b$ are constants.

a) (5)
Find
${\displaystyle \lim_{x\rightarrow 1^{-}}f(x)}$ and
${\displaystyle \lim_{x\rightarrow 1^{+}}f(x)}$.


\[
\lim_{x\rightarrow 1^{-}}f(x)=\lim_{x\rightarrow
1^{-}}x^2+1=(1)^2+1=2,
\]

\[
\lim_{x\rightarrow 1^{+}}f(x)=\lim_{x\rightarrow
1^{+}}ax+b=a(1)+b=a+b.
\]


b) (5)
What conditions do $a$ and $b$ have to satisfy, so that $f(x)$ is
continuous at $x=1$.

\medskip

For $f(x)$ to be continuous at $x=1$ we need the limit
${\displaystyle \lim_{x\rightarrow 1}f(x)}$ exist and equal
$f(1)$. Since $f(1)=2$ and ${\displaystyle \lim_{x\rightarrow
1}f(x)}$ exists if and only if ${\displaystyle \lim_{x\rightarrow
1^{-}}f(x)}={\displaystyle \lim_{x\rightarrow 1^{+}}f(x)}$ we see
that the condition on $a$ and $b$ is $a+b=2$.

\medskip

c) (5)
Find $a$ and $b$, such that $f(x)$ is everywhere differentiable.

\medskip

First $f(x)$ must be continuous everywhere, thus $b=2-a$ from part
b). Notice that $f(x)$ is differentiable at each $x\neq 1$. The
derivative at $x=1$ is the limit ${\displaystyle
\lim_{x\rightarrow 1}\frac{f(x)-f(1)}{x-1}}$. It exists if and
only if the left hand and the right hand derivative limits are
equal. We have

\[
\lim_{x\rightarrow 1^{-}}\frac{f(x)-f(1)}{x-1}=\lim_{x\rightarrow
1^{-}}\frac{x^2+1-2}{x-1}=\lim_{x\rightarrow
1^{-}}\frac{(x-1)(x+1)}{x-1}=(1)+1=2,
\]

\[
\lim_{x\rightarrow 1^{+}}\frac{f(x)-f(1)}{x-1}=\lim_{x\rightarrow
1^{+}}\frac{ax+b-2}{x-1}=\lim_{x\rightarrow
1^{+}}\frac{ax+(2-a)-2}{x-1}=\lim_{x\rightarrow
1^{+}}\frac{a(x-1)}{x-1}=a,
\]
Therefore $a=2$ and $b=0$.

%***********************************************************************
% Horizontal and vertical asymptotes (sec 2.6)
%***********************************************************************
\item
%\pagebreak
a) (10) Find all horizontal and vertical asymptotes for the function
\[
f(x) \ = \ \frac{3x^2+1}{(2x+3)(x-1)}
\]
Explain your reasons.

\medskip

The vertical asymptotes are the lines $x=a$ such that the
denominator is zero at $x=a$, but the numerator is not. We have
$(2x+3)(x-1)=0$, i.e. $x=-3/2$ or $x=1$. Since
$3(-3/2)^2+1=31/4\neq 0$ and $3(1)^2+1=4\neq 0$ the vertical
asymptotes are $x=-3/2$ and $x=1$.

The horizontal asymptotes are the lines $y=b$ where $b$ is the
limit of $f(x)$ as $x$ approaches positive or negative infinity.
We have

\[
\lim_{x\rightarrow
+\infty}\frac{3x^2+1}{(2x+3)(x-1)}=\lim_{x\rightarrow
+\infty}\frac{3+1/x^2}{(2+3/x)(1-1/x)}=\frac{3+0}{(2+0)(1-0)}=\frac{3}{2}.
\]

\[
\lim_{x\rightarrow
-\infty}\frac{3x^2+1}{(2x+3)(x-1)}=\lim_{x\rightarrow
-\infty}\frac{3+1/x^2}{(2+3/x)(1-1/x)}=\frac{3+0}{(2+0)(1-0)}=\frac{3}{2}.
\]

Therefore the only horizontal asymptote is $y=3/2$.

\medskip

b) (5) Graph the function in part a) indicating all horizontal and vertical
asymptotes.

%***********************************************************************
%  Derivative (sec 2.8)
%***********************************************************************
\item
%\pagebreak
(10) Determine the derivative $f'(x)$ of the given function
{\bf using the definition of the derivative as a limit}
\[
f(x) \ = \ \frac{x+1}{x-1}
\]


One can first write
$f(x)=\frac{x+1}{x-1}=\frac{x-1+2}{x-1}=1+\frac{2}{x-1}$. Then

\[
f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to
0}\frac{1+\frac{2}{x+h-1}-1-\frac{2}{x-1}}{h}
\]
\[
=\lim_{h\to 0}\frac{2((x-1)-(x+h-1))}{(x+h-1)(x-1)h}=\lim_{h\to
0}\frac{-2}{(x+h-1)(x-1)}=-\frac{2}{(x-1)^2}.
\]

Another way is
\[
f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to
a}\frac{\frac{x+1}{x-1}-\frac{a+1}{a-1}}{x-a}
\]
\[
=\lim_{x\to
a}\frac{(x+1)(a-1)-(a+1)(x-1)}{(x-1)(a-1)(x-a)}=\lim_{x\to
a}\frac{-2(x-a)}{(x-1)(a-1)(x-a)}=-\frac{2}{(a-1)^2}.
\]



%***********************************************************************
%  The derivative as a function (sec 2.9)
%***********************************************************************
\item
%\pagebreak
A ball is dropped from a tower of height $H$ feet.
Its height after $t$ seconds,
in feet above the ground, is given by the formula
\[
h(t) \ = \ H - 16t^2.
\]
a) (5) Find the velocity of the ball after $t$ seconds.

\medskip

The velocity function is the derivative of the height function
with respect to the time $t$. Thus $v(t)=h'(t)=-32t$.

\medskip

b) (5) It is observed, that the ball hits the ground with
a speed of $128$ feet per second. How long was the ball in the air?
Hint: The speed is the absolute value of the velocity.

\medskip

At the moment when the ball hits the ground the velocity is $-128$
ft/s (the velocity vector points downward). We get $-32t=-128$,
i.e. $t=4$, which means that the ball was 4 seconds in the air.

\medskip

c) (5)
Use the information in part b) to determine the height of the tower.

\medskip

When the ball hits the ground the height of the ball is zero. On
the other hand, it is the height of the ball after 4 seconds, thus
we get $h(4)=0$, i.e. $H-16(4)^2=0$ and $H=256$.


%c) (??) What is the average velocity of the ball during the
%first three seconds?

%***********************************************************************
%  Derivatives of exponential functions (sec 3.1)
%***********************************************************************
\item
%\pagebreak
a) (8) Suppose $a$ is a number, such that
\[
\lim_{h\rightarrow 0}\frac{a^h-1}{h} \ = \ 17.
\]
Find an equation for the tangent line to the graph of $y=a^x$ at the point
$(0,1)$.

\medskip

The slope of the tangent line to the graph of $y=a^x$ at the point $x=0$
is equal to the derivative of $a^x$ at $x=0$. By definition the derivative
of $a^x$ at $x=0$ is the limit ${\displaystyle \lim_{h\rightarrow 0}\frac{a^h-1}{h}}$.
We are given that this limit is equal to 17, so the slope of the tangent
is 17. Since the tangent passes through the
point $(0,1)$ the equation of the tangent is $y-1=17(x-0)$, i.e. $y=17x+1$.


\medskip

b) (7) Find a formula for the derivative $f'(x)$ of the following function
\[
f(x) \ = \ x^{1984} + \frac{1}{x\sqrt{x}} - 2002 e^x.
\]

\[
(x^{1984})'=1984x^{1983},\quad (2002e^x)'=2002e^x,
\]

\[
\left(\frac{1}{x\sqrt{x}}\right)'=(x^{-1}x^{-1/2})'=(x^{-3/2})'=-3/2x^{-5/2}=-\frac{3}{2x^2\sqrt{x}},
\]

Therefore,
\[
f'(x)=1984x^{1983}-\frac{3}{2x^2\sqrt{x}}-2002e^x.
\]


%***********************************************************************
% Quotient rule (sec 3.2)
%***********************************************************************
\item
%\pagebreak
a) (10)
Find a formula for the derivative $f'(x)$ of the function
\[
f(x) \ = \ \frac{x}{x-5}.
\]

By the quotient rule

\[
f'(x)=\frac{(x)'(x-5)-x(x-5)'}{(x-5)^2}=\frac{(x-5)-x}{(x-5)^2}=-\frac{5}{(x-5)^2}
\]


b) (5)
Find the equation of the tangent line to the graph of $y=f(x)$
at the point $(4,-4)$.
%that are parallel to the line $y=5-5x$.

\medskip

The slope of the tangent line to the graph of $y=f(x)$ at $(4,-4)$
is $f'(4)$. By part a) $$f'(4)=-\frac{5}{(4-5)^2}=-5.$$ Since the
line passes through $(4,-4)$ we have $$y+4=-5(x-4),\quad\rm{ i.e. }\ \ y=-5x+16.$$


%c) (5)
%Find another point $(x_0,y_0)$ on the graph of $y=f(x)$,
%different from $(4,-4)$,
%such that the tangent lines at the two points are parallel.


%c) (??)
%Graph the curve $y=f(x)$ and the two tangent lines in part b)
%in the window $0\leq x \leq 10$ and $-10\leq y \leq 10$.













\end{enumerate}



\end{document}
