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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "Lesson 2 Maple Activities \+
for August 31:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 31 "Let's consider the function  f(" }{TEXT 257 1 "x" }{TEXT 
-1 4 ") = " }{XPPEDIT 18 0 "2*x^2-3*x+4;" "6#,(*&\"\"#\"\"\"*$%\"xG\"
\"#F&F&*&\"\"$F&F(F&!\"\"\"\"%F&" }{TEXT -1 96 " using Maple.  The fol
lowing command is how we tell Maple that we want  f  is be a function \+
of  " }{TEXT 260 1 "x" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "f := x -> 2*x^2 - 3*x +
 4;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 28 "Now, let's evaluate  f  at  " }{TEXT 259 1 "x" }{TEXT -1 78 ", \+
1, - 2, and 0.  That is, let's find what the function  f  corresponds \+
with  " }{TEXT 258 1 "x" }{TEXT -1 17 ", 1, - 2, and 0.." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 5 "f(1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f(-
2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(0);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 66 "Now, let'
s see what the graph of the function  f  looks like from " }{TEXT 261 
1 "x" }{TEXT -1 10 " = - 4 to " }{TEXT 262 1 "x" }{TEXT -1 62 " = 5.  \+
We will give a title to this graph and store it to \"p\":" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot(f(x), x=-4..5, title=\"y = f(x
)\"); p := %:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 59 "Recall: In order to graph the function  f  described b
y  f(" }{TEXT 263 1 "x" }{TEXT -1 4 ") = " }{XPPEDIT 18 0 "2*x^2-3*x+4
;" "6#,(*&\"\"#\"\"\"*$%\"xG\"\"#F&F&*&\"\"$F&F(F&!\"\"\"\"%F&" }
{TEXT -1 13 ", you set  f(" }{TEXT 264 1 "x" }{TEXT -1 40 ") equal to \+
y and graph the equation y = " }{XPPEDIT 18 0 "2*x^2-3*x+4;" "6#,(*&\"
\"#\"\"\"*$%\"xG\"\"#F&F&*&\"\"$F&F(F&!\"\"\"\"%F&" }{TEXT -1 27 ".  S
ince the variable of   " }{TEXT 265 1 "y" }{TEXT -1 34 "  in this equa
tion represents the " }{TEXT 266 1 "y" }{TEXT -1 56 "-coordinate of po
ints on the graph of the equation and  " }{TEXT 267 1 "y" }{TEXT -1 5 
" = f(" }{TEXT 268 1 "x" }{TEXT -1 40 "), then for any point on the gr
aph of   " }{TEXT 269 1 "y" }{TEXT -1 5 " = f(" }{TEXT 270 1 "x" }
{TEXT -1 7 "), the " }{TEXT 271 1 "y" }{TEXT -1 62 "-coordinate of the
 point is the function  f  evaluated at the " }{TEXT 272 1 "x" }{TEXT 
-1 149 "-coordinate of the point.  Thus, consider the following three \+
points which are on the graph of the function  f.  We will store these
 points to \"pts\"." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "plot
([ [-2,f(-2)], [0,f(0)], [3,f(3)] ], style=point, symbol=box,color=blu
e); pts := %:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 19 "Now, let's use the " }{TEXT 256 11 "with(plots)" }
{TEXT -1 77 " command to show that these three points are on the graph
 of the function  f." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "wit
h(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "display(p,pts)
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
32 "Now, let's use Maple to find  f(" }{TEXT 273 1 "x" }{TEXT -1 3 " +
 " }{TEXT 274 1 "h" }{TEXT -1 2 "):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "f(x+h); expand(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "In calculus, not only will we h
ave a need to find f(" }{TEXT 275 1 "x" }{TEXT -1 3 " + " }{TEXT 276 
1 "h" }{TEXT -1 39 "), but we will also have a need to find" }}{PARA 
0 "" 0 "" {TEXT -1 4 "( f(" }{TEXT 277 1 "x" }{TEXT -1 3 " + " }{TEXT 
278 1 "h" }{TEXT -1 6 ") - f(" }{TEXT 279 1 "x" }{TEXT -1 6 ") ) / " }
{TEXT 280 1 "h" }{TEXT -1 46 ".  Let's see now we might do this using \+
Maple:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "f(x+h) - f(x); ex
pand(%); q := %;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "q/h; si
mplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 43 "Now, let's consider the following function:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "g := x -> sqrt(7 - 2*x);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 52 "Le
t's evaluate  g  at  - 9, - 5, - 1, - 21/2, and 6:" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 36 "g(-9); g(-5); g(-1); g(-21/2); g(6);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 180 "W
hat is the value of  g  at  6?  This function  g  motivates the defini
tion of the domain of a function.  Before we talk about the domain of \+
a function, let consider the following:" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{PARA 0 "" 0 "" {TEXT -1 80 "Notice what kind of result you get i
f you ask Maple to find g(-5.) and g(-10.5):" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 17 "g(-5.); g(-10.5);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 "Note:  If you give Maple \+
a decimal input, it will give you a decimal output." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}}{MARK "1 0 0" 26 }{VIEWOPTS 1 1 0 1 1 1803 }