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{SECT 0 {PARA 3 "" 0 "" {TEXT -1 35 "The Fundamental Theorem of Calcul
us" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{SECT 1 
{PARA 4 "" 0 "" {TEXT -1 31 "The Fundamental Theorem, Part I" }}{PARA 
0 "" 0 "" {TEXT -1 82 "The first part of the Fundamental Theorem state
s that for any continuous function " }{XPPEDIT 18 0 "f" "6#%\"fG" }
{TEXT -1 5 " on [" }{XPPEDIT 18 0 "a,b" "6$%\"aG%\"bG" }{TEXT -1 15 "]
 the function " }{XPPEDIT 18 0 "F(x)=int(f(t), t=a..x)" "6#/-%\"FG6#%
\"xG-%$intG6$-%\"fG6#%\"tG/F.;%\"aGF'" }{TEXT -1 23 " is differentiabl
e on [" }{XPPEDIT 18 0 "a,b" "6$%\"aG%\"bG" }{TEXT -1 5 "] and" }}
{PARA 0 "" 0 "" {TEXT -1 9 "         " }{XPPEDIT 18 0 "d/dx" "6#*&%\"d
G\"\"\"%#dxG!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "int(f(t), t=a..x) =
 f(x)" "6#/-%$intG6$-%\"fG6#%\"tG/F*;%\"aG%\"xG-F(6#F." }{TEXT -1 1 ".
" }}{PARA 0 "" 0 "" {TEXT -1 29 "In other words, if we have a " }
{TEXT 256 169 "definite integral with variable upper limit, then the r
ate of change with respect to the upper limit is precisely the value o
f the integrand evaluated at the upper limit" }{TEXT -1 1 "." }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "The following procedure produces a
n animation which illustrates part I of the Fundamental Theorem." }}
{PARA 0 "" 0 "" {TEXT -1 17 "As input it takes" }}{PARA 16 "" 0 "" 
{TEXT -1 11 "a function " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 1 ",
" }}{PARA 16 "" 0 "" {TEXT -1 26 "an interval, specified by " }
{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "b" "6#
%\"bG" }{TEXT -1 1 "," }}{PARA 16 "" 0 "" {TEXT -1 12 "a parameter " }
{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 57 " determining the number of f
rames used for the animation," }}{PARA 16 "" 0 "" {TEXT -1 12 "a param
eter " }{XPPEDIT 18 0 "withF;" "6#%&withFG" }{TEXT -1 5 ". If " }
{XPPEDIT 18 0 "withF = 1;" "6#/%&withFG\"\"\"" }{TEXT -1 19 " then the
 function " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 45 " is also includ
ed in the plot (otherwise set " }{XPPEDIT 18 0 "withF=0" "6#/%&withFG
\"\"!" }{TEXT -1 3 "). " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 48 "The information printed during the animation is " }}
{PARA 16 "" 0 "" {TEXT -1 39 "the current area under the graph, i.e. \+
" }{XPPEDIT 18 0 "F(x)=int(f(t),t=a..x)" "6#/-%\"FG6#%\"xG-%$intG6$-%
\"fG6#%\"tG/F.;%\"aGF'" }{TEXT -1 27 ", for the current value of " }
{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 1 "," }}{PARA 16 "" 0 "" {TEXT 
-1 47 "the area under the graph in the previous frame," }}{PARA 16 "" 
0 "" {TEXT -1 91 "the approximate rate of change of the area, i.e. (cu
rrent area - previous area)/(change in " }{XPPEDIT 18 0 "x" "6#%\"xG" 
}{TEXT -1 6 "), or " }{XPPEDIT 18 0 "(F(x[i])-F(x[i-1]))/(Delta*x[i]);
" "6#*&,&-%\"FG6#&%\"xG6#%\"iG\"\"\"-F&6#&F)6#,&F+F,\"\"\"!\"\"F3F,*&%
&DeltaGF,&F)6#F+F,F3" }{TEXT -1 1 "," }}{PARA 16 "" 0 "" {TEXT -1 34 "
the exact rate of change given by " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%
\"xG" }{TEXT -1 1 "," }}{PARA 16 "" 0 "" {TEXT -1 34 "the exact rate o
f change given by " }{XPPEDIT 18 0 "d/dx" "6#*&%\"dG\"\"\"%#dxG!\"\"" 
}{TEXT -1 1 " " }{XPPEDIT 18 0 "F(x)" "6#-%\"FG6#%\"xG" }{TEXT -1 1 ".
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 2095 "FT_graph := proc(f, a, b, n, \+
withF)\n   local dF, F, i, j, deltaF, M, m, cp, ip, ap, tp1, tp2, tp3,
 tp4, tp5, textmargin, T, H;\n   F := unapply(int(f(t), t=a..x), x);\n
   dF := D(F);\n   M := max(0, evalf(maximize(f(x), x, a..b)));\n   m \+
:= min(0, evalf(minimize(f(x), x, a..b)));\n   textmargin := (M-m)*.2;
\n   T := textmargin/5;\n   H := M + textmargin;\n   m := m - (M-m)*.0
5;\n   x.0 := a:\n   for i from 1 to n do\n      x.i := a+(i-1)/(n-1)*
(b-a);\n      F.i := F(x.i);\n      if i > 1 then\n         deltaF := \+
(F.i-F.(i-1))/(x.i-x.(i-1));\n      else\n         deltaF := 0;\n     \+
 fi;\n      s.i:=[[x.i,0], [x.(i-1),0], seq([x.(i-1)+j*(x.i-x.(i-1))/5
,f(x.(i-1)+j*(x.i-x.(i-1))/5)], j=0..5)]:\n      cp := plot(f(t), t=a.
.x.i, color=red, labels=[t,'f']):\n      if withF = 1 then\n         i
p := plot(F(x), x=a..x.i, color=blue):\n      fi;\n      ap:=plots[pol
ygonplot](\{seq(s.j, j=1..i)\}, color=cyan, view=[a..b,m..H], style=pa
tchnogrid):\n      tp1:=plots[textplot]([(a+b)/2,H,cat(\"current area:
 \",convert(evalf(F.i,5),string))], color=black):\n      if i > 1 then
 \n         tp2:=plots[textplot]([(a+b)/2,M+3.75*T,cat(\"previous area
: \",convert(evalf(F.(i-1),5),string))], color=black):\n         tp3:=
plots[textplot]([(a+b)/2,M+2.5*T,cat(\"approx. rate of change: \",conv
ert(evalf(deltaF,5),string))], color=black):\n         tp4:=plots[text
plot]([(a+b)/2,M+1.25*T,cat(\"f(x): \",convert(evalf(f(x.i),5),string)
)], color=black):\n         tp5:=plots[textplot]([(a+b)/2,M,cat(\"D(F)
(x): \",convert(evalf(dF(x.i),5),string))], color=black):\n      else
\n         tp2:=plots[textplot]([(a+b)/2,M+3.75*T,\"previous area: \"]
, color=black):\n         tp3:=plots[textplot]([(a+b)/2,M+2.5*T,\"appr
ox. rate of change: \"], color=black):\n         tp4:=plots[textplot](
[(a+b)/2,M+1.25*T,\"f(x): \"], color=black):\n         tp5:=plots[text
plot]([(a+b)/2,M,\"D(F)(x): \"], color=black):\n      fi; \n      if w
ithF = 1 then\n         p.i := plots[display](cp, ip, ap, tp1, tp2, tp
3, tp4, tp5):\n      else\n         p.i := plots[display](cp, ap, tp1,
 tp2, tp3, tp4, tp5):\n      fi;\n   od:\n   plots[display](seq(p.i, i
=1..n), insequence=true);\nend:" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
25 "Here is a sample call to " }{TEXT 256 8 "FT_graph" }{TEXT -1 18 " \+
for the function " }{XPPEDIT 18 0 "F(x) = int(sqrt(abs(t))-3*t/2+1,t =
 -1 .. x);" "6#/-%\"FG6#%\"xG-%$intG6$,(-%%sqrtG6#-%$absG6#%\"tG\"\"\"
*(\"\"$F3F2F3\"\"#!\"\"F7\"\"\"F3/F2;,$\"\"\"F7F'" }{TEXT -1 5 " and \+
" }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 17 " in the interval " }
{XPPEDIT 18 0 "[-1,3]" "6#7$,$\"\"\"!\"\"\"\"$" }{TEXT -1 2 ". " }}
{PARA 0 "" 0 "" {TEXT -1 47 "Run the animation and observe what is goi
ng on." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "f:=x->sqrt(abs(x))-3*x/2+
1;\na:=-1:\nb:=3:\nn:=20:\nFT_graph(f,a,b,n,0);" }}}}{SECT 1 {PARA 4 "
" 0 "" {TEXT -1 32 "The Fundamental Theorem, Part II" }}{PARA 0 "" 0 "
" {TEXT -1 85 "The second part of the Fundamental Theorem states that \+
for every continuous function " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT 
-1 5 " on [" }{XPPEDIT 18 0 "a,b" "6$%\"aG%\"bG" }{TEXT -1 1 "]" }}
{PARA 0 "" 0 "" {TEXT -1 16 "                " }{XPPEDIT 18 0 "int(f(x
),x=a..b) = F(b)-F(a)" "6#/-%$intG6$-%\"fG6#%\"xG/F*;%\"aG%\"bG,&-%\"F
G6#F.\"\"\"-F16#F-!\"\"" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 
6 "where " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 27 " is some antider
ivative of " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 17 "Thus, this gives " }{TEXT 256 36 "a way to evaluat
e definite integrals" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 25 "We use the same function " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT 
-1 40 " as above and compute an antiderivative " }{XPPEDIT 18 0 "F" "6
#%\"FG" }{TEXT -1 1 "." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "f:=x->sqr
t(abs(x))-3*x/2+1;\nF:=unapply(int(f(x), x),x);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 28 "Let's evaluate the integral " }{XPPEDIT 18 0 "int(f(
x),x=-1..3)" "6#-%$intG6$-%\"fG6#%\"xG/F);,$\"\"\"!\"\"\"\"$" }{TEXT 
-1 18 " by using Maple's " }{TEXT 256 3 "int" }{TEXT -1 102 " command \+
as a black box, and by using the second part of the Fundamental Theore
m, i.e., by evaluating " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 40 " a
t the limits of the definite integral." }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 65 "Int(f(x), x=-1..3) = int(f(x), x=-1..3);\n'F(3)-F(-1)'=F(3)-F(
-1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Finally, we plot " }
{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "F" "6#
%\"FG" }{TEXT -1 28 " together over the interval " }{XPPEDIT 18 0 "[-1
,3]" "6#7$,$\"\"\"!\"\"\"\"$" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 10 "Note that " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 35 " here is
 not quite the same as the " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 
68 " in example illustrating the first part of the Fundamental Theorem
. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "plot([f(x), F(x)], x=-1..3, c
olor=[red,green]);" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 13 "Assignmen
t 11" }}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "Ex.1:" }}{PARA 0 "" 0 "" 
{TEXT -1 35 "How can the difference between the " }{XPPEDIT 18 0 "F" "
6#%\"FG" }{TEXT -1 89 "s in the illustration of the the two parts of t
he Fundamental Theorem above be explained?" }}}{SECT 1 {PARA 4 "" 0 "
" {TEXT -1 5 "Ex.2:" }}{PARA 0 "" 0 "" {TEXT -1 18 "Use the procedure \+
" }{TEXT 256 8 "FT_graph" }{TEXT -1 77 " to study the statement of the
 Fundamental Theorem, part I, for the function " }{XPPEDIT 18 0 "f(x)=
2*x^2-x^3" "6#/-%\"fG6#%\"xG,&*&\"\"#\"\"\"*$F'\"\"#F+F+*$F'\"\"$!\"\"
" }{TEXT -1 17 " on the interval " }{XPPEDIT 18 0 "[-1,3]" "6#7$,$\"\"
\"!\"\"\"\"$" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 104 "Run the
 animation step by step (using the ->| button) and describe in your ow
n words what is happening. " }}{PARA 0 "" 0 "" {TEXT -1 21 "Include th
e graph of " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 40 " in the animat
ion (and your discussion)." }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "Ex.
3:" }}{PARA 0 "" 0 "" {TEXT -1 4 "Let " }{XPPEDIT 18 0 "f(x) = x^3-4*x
^2+3*x" "6#/-%\"fG6#%\"xG,(*$F'\"\"$\"\"\"*&\"\"%F+*$F'\"\"#F+!\"\"*&
\"\"$F+F'F+F+" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "[a,b]=[0,4]" "6#/7
$%\"aG%\"bG7$\"\"!\"\"%" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 
9 "Also let " }{XPPEDIT 18 0 "F(x) = int(f(t),t = a .. x);" "6#/-%\"FG
6#%\"xG-%$intG6$-%\"fG6#%\"tG/F.;%\"aGF'" }{TEXT -1 42 ", as in part I
 of the Fundamental Theorem." }}{PARA 0 "" 0 "" {TEXT -1 7 "a) Use " }
{TEXT 256 8 "FT_graph" }{TEXT -1 9 " to plot " }{XPPEDIT 18 0 "f" "6#%
\"fG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 16 " \+
together over [" }{XPPEDIT 18 0 "a,b" "6$%\"aG%\"bG" }{TEXT -1 2 "]." 
}}{PARA 0 "" 0 "" {TEXT -1 9 "b) Solve " }{XPPEDIT 18 0 "D(F)(x) = 0;
" "6#/--%\"DG6#%\"FG6#%\"xG\"\"!" }{TEXT -1 50 ". What can you see to \+
be true about the graphs of " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 
5 " and " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 17 " at points where \+
" }{XPPEDIT 18 0 "D(F)(x)=0" "6#/--%\"DG6#%\"FG6#%\"xG\"\"!" }{TEXT 
-1 2 "? " }}{PARA 0 "" 0 "" {TEXT -1 39 "c) Over what intervals is the
 function " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 24 " increasing/dec
reasing? " }}{PARA 0 "" 0 "" {TEXT -1 19 "What is true about " }
{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 22 " over those intervals?" }}
{PARA 0 "" 0 "" {TEXT -1 28 "d) Calculate the derivative " }{XPPEDIT 
18 0 "D(f)" "6#-%\"DG6#%\"fG" }{TEXT -1 27 " and plot it together with
 " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" 
{TEXT -1 9 "e) Solve " }{XPPEDIT 18 0 "D(f)(x)=0" "6#/--%\"DG6#%\"fG6#
%\"xG\"\"!" }{TEXT -1 49 ". What can you see to be true about the grap
h of " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 17 " at points where " }
{XPPEDIT 18 0 "D(f)(x)=0" "6#/--%\"DG6#%\"fG6#%\"xG\"\"!" }{TEXT -1 1 
"?" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 5 "Ex.4:" }}{PARA 0 "" 0 "" 
{TEXT -1 4 "The " }{TEXT 256 16 "Fresnel function" }{TEXT -1 15 " is d
efined as " }{XPPEDIT 18 0 "S(x) = int(sin(Pi*t^2/2), t=0..x)" "6#/-%
\"SG6#%\"xG-%$intG6$-%$sinG6#*(%#PiG\"\"\"*$%\"tG\"\"#F0\"\"#!\"\"/F2;
\"\"!F'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 21 "a) At what val
ues of " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 46 " does this functio
n have local maximum values?" }}{PARA 0 "" 0 "" {TEXT -1 81 "b) Find t
he coordinates of the first inflection point to the right of the origi
n." }}{PARA 0 "" 0 "" {TEXT -1 57 "c) Does the Fresnel function have a
 horizontal asymptote?" }}{PARA 0 "" 0 "" {TEXT -1 8 "d) Plot " }
{XPPEDIT 18 0 "sin(Pi*x^2/2)" "6#-%$sinG6#*(%#PiG\"\"\"*$%\"xG\"\"#F(
\"\"#!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "S" "6#%\"SG" }{TEXT 
-1 13 " together on " }{XPPEDIT 18 0 "[0, 10];" "6#7$\"\"!\"#5" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 63 "e) Use the graph of the \+
Fresnel function to solve the equation " }{XPPEDIT 18 0 "S(x)=0.2" "6#
/-%\"SG6#%\"xG$\"\"#!\"\"" }{TEXT -1 30 " correct to one decimal place
." }}{PARA 0 "" 0 "" {TEXT -1 15 "f) Use Maple's " }{TEXT 256 6 "fsolv
e" }{TEXT -1 37 " command to verify your answer in e)." }}}{SECT 1 
{PARA 4 "" 0 "" {TEXT -1 5 "Ex.5:" }}{PARA 0 "" 0 "" {TEXT -1 75 "A hi
gh-tech company purchases a new computer system whose initial value is
 " }{XPPEDIT 18 0 "V" "6#%\"VG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" 
{TEXT -1 39 "The system will depreciate at the rate " }{XPPEDIT 18 0 "
f=f(t)" "6#/%\"fG-F$6#%\"tG" }{TEXT -1 51 " and will accumulate mainte
nance costs at the rate " }{XPPEDIT 18 0 "g=g(t)" "6#/%\"gG-F$6#%\"tG
" }{TEXT -1 8 ", where " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 33 " i
s the time measured in months. " }}{PARA 0 "" 0 "" {TEXT -1 70 "The co
mpany wants to determine the optimal time to replace the system." }}
{PARA 0 "" 0 "" {TEXT -1 65 "a) Why will the company choose to replace
 the system at the time " }{XPPEDIT 18 0 "T" "6#%\"TG" }{TEXT -1 23 " \+
for which the function" }}{PARA 0 "" 0 "" {TEXT -1 8 "        " }
{XPPEDIT 18 0 "C(t) = 1/t" "6#/-%\"CG6#%\"tG*&\"\"\"\"\"\"F'!\"\"" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "int(f(s)+g(s), s=0..t)" "6#-%$intG6$,&-
%\"fG6#%\"sG\"\"\"-%\"gG6#F*F+/F*;\"\"!%\"tG" }{TEXT -1 1 " " }}{PARA 
0 "" 0 "" {TEXT -1 31 "attains its minimum? What does " }{XPPEDIT 18 
0 "C" "6#%\"CG" }{TEXT -1 11 " represent?" }}{PARA 0 "" 0 "" {TEXT -1 
16 "b) Suppose that " }{XPPEDIT 18 0 "f(t) = V/15 - V/450*t" "6#/-%\"f
G6#%\"tG,&*&%\"VG\"\"\"\"#:!\"\"F+*(F*F+\"$]%F-F'F+F-" }{TEXT -1 5 " f
or " }{XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 23 " between 0 and 30, and
 " }{XPPEDIT 18 0 "f(t)=0" "6#/-%\"fG6#%\"tG\"\"!" }{TEXT -1 5 " for \+
" }{XPPEDIT 18 0 "t >30" "6#2\"#I%\"tG" }{TEXT -1 2 ". " }}{PARA 0 "" 
0 "" {TEXT -1 29 "Determine the length of time " }{XPPEDIT 18 0 "T" "6
#%\"TG" }{TEXT -1 28 " for the total depreciation " }{XPPEDIT 18 0 "D(
t)=int(f(s),s=0..t)" "6#/-%\"DG6#%\"tG-%$intG6$-%\"fG6#%\"sG/F.;\"\"!F
'" }{TEXT -1 27 " to equal the inital value " }{XPPEDIT 18 0 "V" "6#%
\"VG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "Hint: We d
efine " }{XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 105 " as a piecewise fu
nction. The resulting integral can be converted back to a piecewise fu
nction using the " }{HYPERLNK 17 "convert,piecewise" 2 "convert,piecew
ise" "" }{TEXT -1 26 " command (click for help)." }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 70 "f := t->piecewise(t<=30, V/15-V*t/450,\n             \+
   t>30, 0);\nf(t);" }}}{PARA 0 "" 0 "" {TEXT -1 24 "c) Further, assum
e that " }{XPPEDIT 18 0 "g(t)=V*t^2/12900" "6#/-%\"gG6#%\"tG*(%\"VG\"
\"\"*$F'\"\"#F*\"&+H\"!\"\"" }{TEXT -1 5 " for " }{XPPEDIT 18 0 "t>0" 
"6#2\"\"!%\"tG" }{TEXT -1 47 ".\nIt can be shown that the critical num
bers of " }{XPPEDIT 18 0 "C" "6#%\"CG" }{TEXT -1 12 " occur when " }
{XPPEDIT 18 0 "C(t)=f(t)+g(t)" "6#/-%\"CG6#%\"tG,&-%\"fG6#F'\"\"\"-%\"
gG6#F'F," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 34 "Determine the
 absolute minimum of " }{XPPEDIT 18 0 "C" "6#%\"CG" }{TEXT -1 4 " on \+
" }{XPPEDIT 18 0 "[0,T]" "6#7$\"\"!%\"TG" }{TEXT -1 1 "." }}{PARA 0 "
" 0 "" {TEXT -1 9 "d) Graph " }{XPPEDIT 18 0 "C" "6#%\"CG" }{TEXT -1 
5 " and " }{XPPEDIT 18 0 "f+g" "6#,&%\"fG\"\"\"%\"gGF%" }{TEXT -1 27 "
 together in one plot over " }{XPPEDIT 18 0 "[0,T]" "6#7$\"\"!%\"TG" }
{TEXT -1 59 " in order to verify the claim about the critical points o
f " }{XPPEDIT 18 0 "C" "6#%\"CG" }{TEXT -1 13 " made above.." }}{PARA 
0 "" 0 "" {TEXT -1 82 "e) Create another plot showing the total deprec
iation, total maintenance cost and " }{XPPEDIT 18 0 "C" "6#%\"CG" }
{TEXT -1 15 " together over " }{XPPEDIT 18 0 "[0,T]" "6#7$\"\"!%\"TG" 
}{TEXT -1 10 ". Comment." }}}}}{MARK "2" 0 }{VIEWOPTS 1 1 0 1 1 1803 }