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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 77 "Lab # 4:  Analysis of Fun
ctions using first and second derivative information" }}{PARA 256 "" 
0 "" {TEXT -1 278 "In this lab we analyze functions with regards to in
tervals where the function is  increasing or decreasing, concave up or
 down, and the location of inflection points and relative extrema.  So
me of the Maple commands that aid in this task are demonstrated with a
 sample problem. " }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 60 "Sample Prob
lem: Plotting Derivatives and Solving for Zeros. " }}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 29 "Consider the function        " }{XPPEDIT 18 0 "f(x
) = sqrt(5)*x+3*x^(2/3);" "6#/-%\"fG6#%\"xG,&*&-%%sqrtG6#\"\"&\"\"\"F'
F.F.*&\"\"$F.)F'*&\"\"#F.F0!\"\"F.F." }{TEXT -1 6 "      " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }{TEXT 263 4 "(1) " }{TEXT -1 43 "Plot the f, f',
 and f'' on the same axes.  " }}{PARA 0 "" 0 "" {TEXT 259 4 "(2) " }
{TEXT -1 46 "Find all relative minimum and maximum points. " }}{PARA 
0 "" 0 "" {TEXT 260 3 "(3)" }{TEXT -1 33 " Find all points of inflecti
on   " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "First try to plot the f,
 f' and f'' on the same axes. " }}{PARA 0 "" 0 "" {TEXT -1 101 "In ord
er to plot x^(2/3) we must use the surd function in order to get an ou
tput when x is negative. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
38 "f := x -> sqrt(5) * x + 3*surd(x^2,3);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"fGf*6#%\"xG6\"6$%)operatorG%&arrowGF(,&*&-%%sqrtG6#
\"\"&\"\"\"9$F2F2*&\"\"$F2-%%surdG6$*$)F3\"\"#F2F5F2F2F(F(F(" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "df := x -> D(f)(x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#dfGf*6#%\"xG6\"6$%)operatorG%&arrow
GF(--%\"DG6#%\"fG6#9$F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
12 "eval(df(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&*$\"\"&#\"\"\"\"
\"#F'*(F(F'-%%surdG6$*$)%\"xGF(F'\"\"$F'F/!\"\"F'" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 85 "If you want to see what this looks like you can ev
aluate this with the eval command. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "ddf := x -> D(df)(x);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%$ddfGf*6#%\"xG6\"6$%)operatorG%&arrowGF(--%\"DG6#%#dfG6#9$F(F(
F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eval(ddf(x));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&#\"\"#\"\"$\"\"\"*&-%%surdG6$*$)%
\"xGF&F(F'F(F/!\"#F(!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 110 "Now
 plot all three on the same axes. There are some disontinuities in the
 derivatives so we use discont=true. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 75 "plot([f(x),df(x),ddf(x)],x=-3..3,y=-10..10,discont=tr
ue,thickness=[3,2,1]);" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 
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$\"+)3k#)=#F_cn$!+v\"=mM#F[cn7$$\"+Nqj[AF_cn$!+J:*HE#F[cn7$$\"+fTu6BF_
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cn7$$\"+&RY-]#F_cn$!+J'eX'>F[cn7$$\"+Kd'Qc#F_cn$!+&HJ)**=F[cn7$$\"+p#e
pi#F_cn$!+DwBR=F[cn7$$\"+fP$\\o#F_cn$!+(\\wky\"F[cn7$$\"+7-Q^FF_cn$!+7
U=H<F[cn7$$\"+62\"3\"GF_cn$!+,(31o\"F[cn7$$\"+?g<uGF_cn$!+N#*QJ;F[cn7$
$\"+0Z#[$HF_cn$!+kTf'e\"F[cn7$Fa\\lFibn-Ff\\l6&Fh\\lFi\\lFi\\lF\\]l-F^
]l6#\"\"\"-%+AXESLABELSG6%Q\"x6\"Q\"yFchp-%%FONTG6#%(DEFAULTG-%%VIEWG6
$;F(Fa\\l;$F[cnF*$\"#5F*" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 
45.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 153 "The thick curve is f(x), the medium curve is f '(x), a
nd the light curve is f ''(x).  Since the derivative is negative for x
 < 0 and positive for x > 0, " }{TEXT 285 36 "there is a relative mini
mum at (0,0)" }{TEXT -1 181 ".  This fact is evident from the graph. T
here is also a relative maximum near x = -.7 found by putting the arro
w where f'(x) = 0. We can have Maple find this value of x exactly by  \+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "solve(df(x)=0,x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*(\"\")\"\"\"\"#D!\"\"\"\"&#F&\"\"#F
(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "So there is a relative maxim
um at x = " }{XPPEDIT 18 0 "-8/25*5^(1/2);" "6#,$*(\"\")\"\"\"\"#D!\"
\"\"\"&#F&\"\"#F(" }{TEXT -1 39 ".  The relative maximum has a value o
f " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(%);" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6##\"\"%\"\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 286 
37 "The relative maximum maximum point is" }{TEXT -1 2 " (" }{XPPEDIT 
18 0 "-8/25*5^(1/2);" "6#,$*(\"\")\"\"\"\"#D!\"\"\"\"&#F&\"\"#F(" }
{TEXT -1 8 ",  4/5) " }{TEXT 287 31 "or approximately (-0.72,  0.80)" 
}{TEXT -1 32 " rounded to two decimal places. " }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 173 "By examining the curve describing the second deriva
tive we see that it is undefined at x=0 but negative everywhere else. \+
So the curve does not change concavity and there is " }{TEXT 288 19 "n
o inflection point" }{TEXT -1 2 ". " }}}}{SECT 1 {PARA 3 "" 0 "" 
{TEXT -1 10 "Assignment" }}{PARA 0 "" 0 "" {TEXT -1 219 "You may do th
e analysis for these problems by hand or with Maple.  The above exampl
e illustrates the techniques you may use to identify the relative maxi
mum and minimum points as well as inflection points. Make sure to " }
{TEXT 291 24 "fit these onto one page." }{TEXT -1 5 "     " }}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 277 11 "Graph #1   " }{TEXT 279 
285 "Generate a graph of the following function and its first two deri
vatives over the interval x = -3..3.  Label each curve accordingly.  A
lso label all relative maximum and minimum points as well as inflectio
n points.  You may do this by hand. Make sure the x and y values are r
ounded to " }{TEXT 292 19 "two decimal places." }{TEXT 293 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 289 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }{TEXT 280 1 " " }{TEXT -1 2 "  " }{XPPEDIT 18 0 "f(x) =
 (x-1)^3*(x+3.2)+50;" "6#/-%\"fG6#%\"xG,&*&),&F'\"\"\"F,!\"\"\"\"$F,,&
F'F,-%&FloatG6$\"#KF-F,F,F,\"#]F," }{TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }{TEXT 281 11 "Problem #2 " }{TEXT 290 297 "Generate a
 graph of the following function and its first two derivatives over th
e interval x = -3..3.  Label each curve accordingly.  Also label all r
elative maximum and minimum points as well as inflection points.  You \+
may do this by hand. Make sure your points are rounded to two decimal \+
places." }}{PARA 0 "" 0 "" {TEXT 282 1 " " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }{TEXT 284 1 " " }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 "  \+
" }{XPPEDIT 18 0 "f(x) = (x^2-2*x)/(x^2-2*x+2)+4;" "6#/-%\"fG6#%\"xG,&
*&,&*$)F'\"\"#\"\"\"F.*&F-F.F'F.!\"\"F.,(*$F,F.F.*&F-F.F'F.F0F-F.F0F.
\"\"%F." }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "1" 0 }
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