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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:  
$lim_{x \rightarrow +\infty} {\sqrt{4x^2 + 9x + 8} \over 5x + 4}$
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$lim_{x \rightarrow +\infty} {\sqrt{4x^2 + 9x + 8} \over 5x + 4} \cdot
{{1 \over \sqrt{x^2}} \over {1 \over \sqrt{x^2}}}$
$= lim_{x \rightarrow +\infty} {\sqrt{4x^2 + 9x + 8} \over 5x + 4} \cdot
{{1 \over \sqrt{x^2}} \over {1 \over |x|}}$

$= lim_{x \rightarrow +\infty} {\sqrt{4x^2 + 9x + 8} \over 5x + 4} \cdot
{{1 \over \sqrt{x^2}} \over {1 \over x}}$, since $x \rightarrow +\infty$, we can assume $x$ is positive.  

\rightline{What if $x \rightarrow -\infty$?}



$ = lim_{x \rightarrow +\infty} {\sqrt{4 + {9 \over x} + {8 \over x^2}} \over 
5 + {4 \over x}}$
$ = {\sqrt{4} \over 5}$
$ = {2 \over 5}$

Alternate method:  Factor out highest power in denominator:

$lim_{x \rightarrow +\infty} {\sqrt{4x^2 + 9x + 8} \over 5x + 4}$
$= lim_{x \rightarrow +\infty} {\sqrt{x^2(4 + {9 \over x} + {8 \over x^2})} \over x(5 + {4 \over x})}$
$= lim_{x \rightarrow +\infty} {\sqrt{x^2}\sqrt{4 + {9 \over x} + {8 \over x^2}} \over x(5 + {4 \over x})}$

$= lim_{x \rightarrow +\infty} {|x|\sqrt{4 + {9 \over x} + {8 \over x^2}} \over x(5 + {4 \over x})}$
$= lim_{x \rightarrow +\infty} {x\sqrt{4 + {9 \over x} + {8 \over x^2}} \over x(5 + {4 \over x})}$, since $x \rightarrow +\infty$, we can assume $x$ is positive.  

\rightline{What if $x \rightarrow -\infty$?}


$ = lim_{x \rightarrow +\infty} {\sqrt{4 + {9 \over x} + {8 \over x^2}} \over 
5 + {4 \over x}}$
$ = {\sqrt{4} \over 5}$
$ = {2 \over 5}$

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\centerline{Answer 1.) $\underline{~~ = {2 \over 5}~~}$}

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


$f'(x) = lim_{h \rightarrow 0}{f(x + h) - f(x) \over h}
= lim_{h \rightarrow 0}{\sqrt{x + h} - \sqrt{x} \over h}$

$= lim_{h \rightarrow 0}{\sqrt{x + h} - \sqrt{x} \over h}
\cdot {\sqrt{x + h} + \sqrt{x} \over \sqrt{x + h} + \sqrt{x}}$

$= lim_{h \rightarrow 0}{x + h - x \over h(\sqrt{x + h} + \sqrt{x})}$

$= lim_{h \rightarrow 0}{h \over h(\sqrt{x + h} + \sqrt{x})}$

$= lim_{h \rightarrow 0}{1 \over (\sqrt{x + h} + \sqrt{x})}$
$= {1 \over (\sqrt{x} + \sqrt{x})}$
$= {1 \over 2\sqrt{x}} = {1 \over 2}x^{-1 \over 2}$  IF $x > 0$.

Extra credit if you noticed that $x$ must be $> 0$.


\vfill
\centerline{Answer 2.) $\underline{~~{1 \over 2}x^{-1 \over 2}, ~x > 0}~~$}
\eject

Find the following derivatives

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


${[e^x(x^2 - x + 3)]'cos(2x) - e^x(x^2 - x + 3)[cos(2x)]' \over cos^2(2x)}$

$= {[(e^x)'(x^2 - x + 3) + e^x(x^2 - x + 3)' ]cos(2x) - e^x(x^2 - x + 3)[-sin(2x)](2x)' \over cos^2(2x)}$

$= {[(e^x)(x^2 - x + 3) + e^x(2x - 1) ]cos(2x) - e^x(x^2 - x + 3)[-sin(2x)](2) \over cos^2(2x)}$

$= {(e^x)(x^2 + x + 2)cos(2x) + 2e^x(x^2 - x + 3)sin(2x) \over cos^2(2x)}$

Note it is better if you don't show all the intermediate steps.
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\centerline{Answer 3.) $\underline{~~{(e^x)(x^2 + x + 2)cos(2x) + 2e^x(x^2 - x + 3)sin(2x) \over cos^2(2x)}~~}$}
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[15]~  4.) ${d \over dx}[2sin(e^{x^3} + 4)]$


$2[sin(e^{x^3} + 4)]' = 2cos(e^{x^3} + 4)[e^{x^3} + 4]' $

$= 2[cos(e^{x^3} + 4)][e^{x^3}(x^3)' + 0]$

$= 2[cos(e^{x^3} + 4)][e^{x^3}(3x^2)]$


$= 6x^2e^{x^3}cos(e^{x^3} + 4)$


Note it is better if you don't show all the intermediate steps.
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\centerline{Answer 4.) $\underline{~~6x^2e^{x^3}cos(e^{x^3} + 4)~~}$}

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

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


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[2] 5a.)  domain of $f = \underline{~~{R}~~}$
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[2] 5b.)  range of $f = \underline{~~(-\infty, 5]~~}$
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[1] 5c.)  Is $f$ one-to-one? $\underline{~~No~~}$
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[2] 5d.)  Does $f^{-1}$ exist? $\underline{~~No~~}$

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[1] 5e.)  $f(1) = \underline{~~2~~}$
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[2] 5g.)  $f\hskip 1pt'(-1) = \underline{~~0~~}$
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[2] 5f.)  Solve $f(x) = 1: \underline{~~-4, 4~~}$


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[2] 5i.)  $lim_{x \rightarrow +\infty} f(x) = \underline{~~4~~}$
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[2] 5j.)  $lim_{x \rightarrow -\infty} f(x) = \underline{~~-\infty~~}$
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[2] 5k.)  $lim_{x \rightarrow 3^+} f(x) = \underline{~~-\infty~~}$
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[2] 5l.)  $lim_{x \rightarrow 3^-} f(x) = \underline{~~5~~}$
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[2] 5m.)  State all points where $f$ is not continuous: $\underline{~~
x = 3 (or (3, 5))~~}$

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[2] 5n.)  State all points where $f$ is not differentiable: $\underline{
 ~~x = 3, -3, 1, (or (3, 5), (-3, 2), (1, 2)) ~~}$

Note there is a corner at $x = -3$.  The transition where $f$ is not constant for $x < -3$ to where $f$ is constant between -3 and 1 is not a smooth transition.  However, if you thought it was a smooth transition due to my lack of drawing skills, you will not be docked if you missed this point.

 

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

\includegraphics[width=83ex]{graph25AA}

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


[4] 6a)  y =







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

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

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