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257 48 "Dr. Sarah's Maple Demo on 1-1 and Onto Functions" }}}{EXCHG 
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{TEXT -1 88 "Before executing the following sections, try and think of
 an example of such a function." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 
47 "A function g:R-->R that is not onto and not 1-1" }}{PARA 0 "" 0 "
" {TEXT -1 67 "cos(x) is a function g:R->R so that f is not onto and f
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"" {MPLTEXT 1 0 21 "g:=unapply(cos(x),x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"gG%$cosG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
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0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 95 "We know from calculus that cos \+
x is a function.  But, it is not so easy to rigorously prove - t" }
{TEXT 309 51 "ry it without assuming what you are trying to prove" }
{TEXT -1 61 "!  Let's assume that we have proved that cos x is a funct
ion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 310 30 "To prove that c
os x is not 1-1" }{TEXT -1 60 " we must produce x1 not equal to x2 so \+
that cos(x1)=cos(x2)." }}{PARA 0 "" 0 "" {TEXT -1 73 "Take x1=Pi and x
2=3Pi.  Notice that x1 and x2 are real and unequal. Also," }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "cos(Pi);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "cos
(3*Pi);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#!\"\"" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 47 "and so cos(x1)=cos(x2).  Hence cosx is not 1-1." }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 311 31 "To prove that cos x is no
t onto" }{TEXT -1 67 " we must produce y so that cos(x) is not equal t
o y for all x in R." }}{PARA 0 "" 0 "" {TEXT -1 66 "Take y=2.  Assume \+
for contradiction that cos(x)=2 for some x in R." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 25 "evalf(solve(cos(x)=2,x));" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#,$%\"IG$\"+(*y&pJ\"!\"*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 105 "Then x=1.316957897 I.  We have arrived at a contradictio
n to the fact that x is in R, since x is complex." }}{PARA 0 "" 0 "" 
{TEXT -1 73 "Hence, cos(x) is not equal to 2 for all x in R, and so co
s x is not onto." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 46 "A function \+
o:R-->R that is not onto but is 1-1" }}{PARA 0 "" 0 "" {TEXT -1 61 "e^
(x) is a function o:R-->R so that f is not onto but is 1-1." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "o:
=unapply(exp(x),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"oG%$expG" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(o(x),x=-5..5);" }}
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0 "" {MPLTEXT 1 0 13 "diff(o(x),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#-%$expG6#%\"xG" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "We know from c
alculus that e^x is a function.  But, it is not so easy to rigorously \+
prove - t" }{TEXT 303 51 "ry it without assuming what you are trying t
o prove" }{TEXT -1 59 "!  Let's assume that we have proved that e^x is
 a function." }}{PARA 0 "" 0 "" {TEXT -1 281 "Recall that the derivati
ve of e^x is e^x itself.  We know that (e^x) is always positive for al
l x by definition of a positive number(2.71828...) to a power x.  Sinc
e the derivative is always positive, we know that the function o(x) is
 increasing.  Thus if a<b then e^(a) < e^(b).  " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 302 24 "Proof tha
t o(x) is  1-1 " }{TEXT -1 46 "assuming that we know that o(x) is incr
easing." }}{PARA 0 "" 0 "" {TEXT -1 50 "So, assume for contradiction t
hat o(x) is not 1-1." }}{PARA 0 "" 0 "" {TEXT -1 101 "Then, we can fin
d x1 and x2 that are not equal so that o(x1)=o(x2).  But, if x1 and x2
 are not equal," }}{PARA 0 "" 0 "" {TEXT -1 100 "then one must be smal
ler than the other - without loss of generality, assume that x1<x2.  S
ince o(x)" }}{PARA 0 "" 0 "" {TEXT -1 94 "is increasing, we know that \+
o(x1)<o(x2).  We have arrived at a contradiction to the fact that " }}
{PARA 0 "" 0 "" {TEXT -1 36 "o(x) is not 1-1.  Hence o(x) is 1-1." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
304 27 "Proof that o(x) is not onto" }{TEXT -1 1 "." }}{PARA 0 "" 0 "
" {TEXT -1 100 "To show that o(x) is not onto, we'll produce a y in R \+
so that o(x) is not equal to y for all x in R." }}{PARA 0 "" 0 "" 
{TEXT -1 66 "Take y=-1.  Assume for contradiction that o(x)=-1 for som
e x in R." }}{PARA 0 "" 0 "" {TEXT -1 28 "Then e^x=-1 for some x in R.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve(exp(x)=-1,x);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#*&%\"IG\"\"\"%#PiGF%" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 104 "But then x=Pi*I, which is complex, and so we h
ave arrived at a contradiction to the fact that x is in R." }}{PARA 0 
"" 0 "" {TEXT -1 70 "Hence o(x) is not equal to -1 for all x in R, and
 so o(x) is not onto." }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 43 "A func
tion f:R-->R that is onto but not 1-1" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "We'll prove that 2x^3-x is a f
unction from f:R-->R so that f is onto but not one-to-one." }}{PARA 0 
"" 0 "" {TEXT -1 29 "Let's first look at the plot." }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 20 "f:=unapply(x^3-x,x);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"fGR6#%\"xG6\"6$%)operatorG%&arrowGF(,&*$)9$\"\"$\"
\"\"\"\"\"F/!\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "
plot(f(x),x=-2..2);" }}{PARA 13 "" 1 "" {INLPLOT "6%-%'CURVESG6$7S7$$!
\"#\"\"!$!\"'F*7$$!1LLL$Q6G\">!#:$!1(Q82bte3&F07$$!1nm;M!\\p$=F0$!1MW)
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QFen7$$!1,++v#\\N)\\Fen$\"1WqJpn%eu$Fen7$$!1pmmmCC(>%Fen$\"1+OT&z?yX$F
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m')fdL8F0$\"1j;Nq()3Q5F07$$\"1nmm,FT=9F0$\"1lURUTGN9F07$$\"1LL$e#pa-:F
0$\"1ZN!3.u'*)=F07$$\"1+++Sv&)z:F0$\"10FwWwQjBF07$$\"1LLLGUYo;F0$\"16*
*zV7;wHF07$$\"1nmm1^rZ<F0$\"18mxK[p!f$F07$$\"1++]sI@K=F0$\"1\">U@!\\`=
VF07$$\"1++]2%)38>F0$\"1l^RMxj)3&F07$$\"\"#F*$\"\"'F*-%'COLOURG6&%$RGB
G$\"#5!\"\"F*F*-%+AXESLABELSG6$Q\"x6\"%!G-%%VIEWG6$;F(Fhz%(DEFAULTG" 
2 247 247 247 2 0 1 0 2 9 0 4 2 1 45 45 10030 0 10056 10074 0 0 0 
20530 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 112 "It looks like f is a function since it passes \+
the vertical lines tests:  Each verticle line hits the graph of f," }}
{PARA 0 "" 0 "" {TEXT -1 193 "and a verticle lines doesn't hit the gra
ph more than once.  But this is not a proof since we don't have the en
tire graph of f to refer to at once (we can't graph from -infinity to \+
infinity!).  " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 282 47 "We can prove th
at f(x) is a function from R-->R" }{TEXT -1 85 ", because f is a polyn
omial in x, defined on all of R (ie we use the e.c. from PS 4)." }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 159 "It looks like f is not 1-1 since \+
it fails the one-to-one horizontal line test - we can draw a horizonta
l line that hits the graph of f more than once (ie y=0)." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 281 26 "To prove that f is not 1-1" }{TEXT -1 
63 ", we must produce two x values that correspond to the same y.  " }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "fsolve(f(x) = 0);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6%$!+++++5!\"*\"\"!$\"\"\"F&" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 90 "So, take x1=0 and x2=1.  Notice that x1 i
s not equal to x2, but we'll see that f(x1)=f(x2)" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 5 "f(0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"
!" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(1);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "Hence,
 we have produced x1 not equal to x2 so that f(x1)=f(x2).  Thus, f: R-
->R is not one-to-one." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "It look
s like f is onto since it looks like each horizontal line hits the gra
ph at least once. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 283 23 "To prove t
hat f is onto" }{TEXT -1 83 ", we let y=a be an element of R be arbitr
ary.  We must produce an x so that f(x)=a." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 18 "solve(f(x) = a,x);" }}{PARA 12 "" 1 "" {XPPMATH 20 
"6%,&*$),&%\"aG\"$3\"*$-%%sqrtG6#,&!#7\"\"\"*$)F'\"\"#\"\"\"\"#\")F3\"
#7#F/\"\"$F3#F/\"\"'*&F3F3*$)F&#\"\"\"F7F3!\"\"F2,(F$#!\"\"F5F:FB*(%\"
IGF/-F+6#F7F3,&F$F8F:!\"#F/#F/F2,(F$FAF:FBFC#FBF2" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 19 "d:=unapply(%[1],a);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"dGR6#%\"aG6\"6$%)operatorG%&arrowGF(,&*$),&9$\"$3\"
*$-%%sqrtG6#,&!#7\"\"\"*$)F0\"\"#\"\"\"\"#\")F<\"#7#F8\"\"$F<#F8\"\"'*
&F<F<*$)F/#\"\"\"F@F<!\"\"F;F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 112 "I have chosen d to be the first solution that Maple gave me.  \+
I didn't choose the other solutions since they had" }}{PARA 0 "" 0 "" 
{TEXT -1 49 "I in them (complex instead of real).  Notice that" }
{TEXT 284 44 " if  |a|  bigger than or equal to sqrt(12)/9" }{TEXT -1 
79 ", then sqrt(-12+81a^2) is real, and so we get a real solution x so
 that f(a)=x." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "So, given any y=
a>sqrt(12)/9, we can find an x so that f(x)=a by plugging into d:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "evalf(d(20));" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"+q'Qr$G!\"*" }}}{EXCHG {PARA 11 "" 1 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "What about " }
{TEXT 285 22 "if  |a| <sqrt(12)/9.  " }{TEXT -1 74 "Then the sqrt(-12+
81a^2) is complex.  But, we can refer back to the graph:" }}{PARA 0 "
" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 12 "" 1 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "plot(f(x),x=-1..1,y=-sqrt(12
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10030 0 10056 10074 0 0 0 20530 0 12020 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 
0 0 0 0 0 }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 277 "Since we have the en
tire graph for these y values, we can use a graphing argument to show \+
that f is onto these remaining y values.  For any y in this region, dr
aw a horizontal line at y.  Notice that this hits the graph of f, so c
hoose x as (one of) the corresponding x values." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 11 "" 1 "" {TEXT -1 0 "" }}}{EXCHG 
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