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Exam 1 Oct. 6, 2005 \hfil SHOW ALL WORK \hfil
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Math 25 Calculus I \hfil Either circle your answers or place on answer line. \hfil

[15] 1.) Calculate the following limit:  
$lim_{x \rightarrow +\infty} {\sqrt{4x^2 + 9x + 8} \over 5x + 4}$

\vfill
\centerline{Answer 1.) $\underline{\hskip 4in}$}

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[15] 2.) Find the derivative of $f(x) = \sqrt{x}$ by using the definition
of derivative.


\vfill
\centerline{Answer 2.) $\underline{\hskip 4in}$}
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Find the following derivatives

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[15]~  3.) ${d \over dx}[{e^x(x^2 - x + 3) \over cos(2x)}]$

\vfil
\centerline{Answer 3.) $\underline{\hskip 4in}$}
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[15]~  4.) ${d \over dx}[2sin(e^{x^3} + 4)]$

\vfil
\centerline{Answer 4.) $\underline{\hskip 4in}$}

\eject


5.)  Answer the following questions based on the graph of $f$ given below.

\centerline{\includegraphics[width=36ex]{graph25a}}


%%\centerline{\epsfbox{graph25a.eps}}

[2] 5a.)  domain of $f = \underline{\hskip 0.7in}$
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[2] 5b.)  range of $f = \underline{\hskip 0.7in}$
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[1] 5c.)  Is $f$ one-to-one? $\underline{\hskip 0.7in}$
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[2] 5d.)  Does $f^{-1}$ exist? $\underline{\hskip 0.7in}$

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[1] 5e.)  $f(1) = \underline{\hskip 0.7in}$
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[2] 5g.)  $f\hskip 1pt'(-1) = \underline{\hskip 0.7in}$
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[2] 5f.)  Solve $f(x) = 1: \underline{\hskip 0.7in}$


\vskip 15pt

[2] 5i.)  $lim_{x \rightarrow +\infty} f(x) = \underline{\hskip 0.7in}$
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[2] 5j.)  $lim_{x \rightarrow -\infty} f(x) = \underline{\hskip 0.7in}$
\vskip 15pt

[2] 5k.)  $lim_{x \rightarrow 3^+} f(x) = \underline{\hskip 0.7in}$
\hfil
[2] 5l.)  $lim_{x \rightarrow 3^-} f(x) = \underline{\hskip 0.7in}$
\vskip 15pt

[2] 5m.)  State all points where $f$ is not continuous: $\underline{\hskip 1.7in}$

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[2] 5n.)  State all points where $f$ is not differentiable: $\underline{\hskip 1.7in}$

%%\centerline{\epsfbox{graph25.eps,width=6truein}}

\includegraphics[width=83ex]{graph25}

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6.)  Given the graph of $y = f(x)$ below, draw the following graphs:


[4] 6a)  y =







\hfil





\end
Review Problems

Ch 3 p.270-274

Concept check:  1, 3

TF Quiz:  1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 13

Exercises:  1 - 14, 17, 19, 21-24, 26, 30, 32, 35, 37, 39, 41, 42, 44, 55-57, 59, 61, 63, 67, 68, 84, [85-87 (but ignore acceleration)], 88, 100, 101, 102


Ch 2 p. 176 - 179

Concept check:  2, 4, 5, 7, 8, 9, 10, 11, 12, 13, 14

TF Quiz:  All

Exercises:  [1, 2], 3-12, [13], 14, [15-21], 25, 26, [31, 32], [35], 37, [39], 40, 41ab, [44-46], [47ab], 49

Ch 1:

Concept check:  1, 3, 4, 6, 7, 8, 9, 11, 12, 13

Exercises:  1, 2, 5-8, 10, 19, 24, 25, 26

Tuesday Quiz:  Proof + 3.6

  Prove 

$(f + g)' = f' + g' $

$(cf)' = c(f')$ 

$f$ differentiable implies $f$ continuous:

Show $lim_{x \rightarrow a} f(x) = f(a)$ 

Show $lim_{x \rightarrow a} [f(x) - f(a)] = 0$ 

$f$ differentiable at $a$ implies $f'(a)$ exists and $lim_{x \rightarrow a} {f(x) - f(a) \over x - a} = f'(a)$

Hence $lim_{x \rightarrow a} [f(x) - f(a)] = lim_{x \rightarrow a} {f(x) - f(a) \over x - a} (x-a) = lim_{x \rightarrow a} f'(a) (x - a)   
f'(a) lim_{x \rightarrow a} (x - a) = f'(a) (0) = 0$.

 $lim_{x \rightarrow a} f(x) = lim_{x \rightarrow a} [f(x) - f(a) + f(a)] = lim_{x \rightarrow a} [f(x) - f(a)] + lim_{x \rightarrow a} f(a) =   
0 + f(a)$

\end