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Cell[CellGroupData[{
Cell["Calculus: The Language of Change", "Title"],
Cell[TextData[{
StyleBox["Exercise set 5.5.1-CD\nContinuity and the Derivative",
FontVariations->{"Underline"->True}],
"\n\t",
StyleBox[" ", "Subsubtitle"]
}], "Subtitle",
PageWidth->640,
CellMargins->{{Inherited, 0}, {Inherited, Inherited}}],
Cell[TextData[StyleBox["by Brandon W. Buster", "Subsubtitle"]], "Subtitle",
FontSize->18],
Cell["1. ", "Subtitle"],
Cell[CellGroupData[{
Cell["\<\
Part 1.\
\>", "Section"],
Cell[TextData[
"Suppose that f[x] is smooth on an interval around a so that the \
\"microscope\" increment equation is valid. Suppose that x\[TildeTilde]a so \
that, x=a + \[Delta]x for \[Delta]x\[TildeTilde]0. Show that f[x]=f[a + \
\[Delta]x]\[TildeTilde]f[a]; in other words, show that smooth real functions \
are continuous at real points.\n"], "Text"],
Cell[CellGroupData[{
Cell["\<\
Solution to Part 1.
\
\>", "Subsection"],
Cell[TextData[
"We know that \[Delta]x = a - x, so we use the approximation\n\tf[a]=f[x + \
\[Delta]x]= f[x]+ f'[x]\[Delta]x + \[Epsilon]\[Delta]x.\nwhere [f'[x] + \
\[Epsilon]]\[Delta]x is a medium value times a small value which gives us a \
small value. Therefore, f[a]\[TildeTilde]f[x].\n"], "Text"]
}, Open ]]
}, Open ]],
Cell[CellGroupData[{
Cell["\<\
Part 2.\
\>", "Section"],
Cell[TextData[{
"Consider the real function f[x] = ",
Cell[BoxData[
\(TraditionalForm\`1\/x\)]],
", which is undefined at x = 0. We could extend the definition by simply \
assigning f[0] = 0. Show that this function is not continuous at x = 0, but \
is continuous at every other real x.\n"
}], "Text"],
Cell[CellGroupData[{
Cell["\<\
Solution to Part 2.
\
\>", "Subsection"],
Cell[TextData[
"Let a = 0 and x = a + \[Delta]x for \[Delta]x \[TildeTilde] 0. If f[x] is \
continuous at 0, then f[0] \[TildeTilde] f[x]. We are given that f[0] = 0. \
If \[Delta]x \[TildeTilde] 0 then f[x] is very large and cannot be \"close\" \
to zero. So f[0] is not close to f[x] and f[x] is not continuous at zero. \
Evaluate the cell below to see this discontinuity in the graph.\n"], "Text"],
Cell[BoxData[{
\(\(Clear[j, x]; \)\),
\(\(j[x_] := 1\/x; \)\),
\(j[x]\),
\(\(Plot[j[x], {x, \(-10\), 10}, PlotStyle -> {Thickness[0.01], Blue}];
\)\)}], "Input"]
}, Open ]]
}, Open ]],
Cell[CellGroupData[{
Cell["\<\
Part 3.\
\>", "Section"],
Cell["\<\
Give an intuitive graphical description of the definition of continuity in \
terms of powerful microscopes and explain why it follows that smooth \
functions must be continuous.
\
\>", "Text"],
Cell[CellGroupData[{
Cell["\<\
Solution to Part 3.
\
\>", "Subsection"],
Cell[TextData[
"Graphically, a function is continuous if f[x] is close to f[a] when x is \
close to a, for every x \[TildeTilde] a, f[x] is defined and f[x] \
\[TildeTilde] f[a]. So for any magnification we should be able to see both \
f[x] and f[a]. From Chapter 3 we know that a highly magnified view of a \
smooth graph appears as a straight line. Since we can see f[x] for any x in \
our window, it follows that smooth functions must be continuous.\n"], "Text"]
}, Open ]]
}, Open ]],
Cell[CellGroupData[{
Cell["\<\
Part 4.\
\>", "Section"],
Cell[TextData[{
"The function f[x] = ",
Cell[BoxData[
\(TraditionalForm\`\@x\)]],
" is defined for x \[GreaterEqual] 0; there is nothing wrong with f[0]. \
However, our increment computation for ",
Cell[BoxData[
\(TraditionalForm\`\@x\)]],
" above was not valid at x = 0 because a microscopic view of the graph \
focused at x = 0 looks like a vertical ray (or half-line). Explain why this \
is so, but show that f[x] is still continuous \"from the right;\" that is, if \
0\[Precedes] x \[TildeTilde] 0, then ",
Cell[BoxData[
\(TraditionalForm\`\@x\)]],
" \[TildeTilde] 0 ,but ",
Cell[BoxData[
FormBox[
FractionBox[
FormBox[\(\@x\),
"TraditionalForm"], "x"], TraditionalForm]]],
"is very large.\n"
}], "Text"],
Cell[CellGroupData[{
Cell["\<\
Solution to Part 4.
\
\>", "Subsection"],
Cell["\<\
Evaluate the cell below to see what happens to our magnified view of f[x] \
near x = 0 as we increase the magnification.
\
\>", "Text"],
Cell[BoxData[
\(\(f[x_] := Sqrt[x]; \nx = 0.001; \nSetOptions[Plot, Ticks\ -> \ None];
\nPrint["\", x]; \n
Do[delta\ = \ 1/\((1.5^n)\); \n\ \ \t
\(Plot[f[x + dx] - f[x], {dx, \(- .001\), delta}, \n\ \ \t\t
PlotRange\ -> \ {\(-0.001\), delta},
PlotStyle -> {Thickness[0.01], Blue}, AxesOrigin -> {0, 0}]; \)\n
\t, {n, 0, 16}]\n\)\)], "Input"],
Cell[TextData[{
"\nWe see that the view becomes more and more vertical as we zoom in. From \
Example 5.6 we know that\n\tf'[x] = ",
Cell[BoxData[
FractionBox["1",
RowBox[{"2",
FormBox[\(\@x\),
"TraditionalForm"]}]]]],
"\nSo for positive x \[TildeTilde] 0, f'[x] becomes large. Additionally, ",
Cell[BoxData[
FormBox[
FractionBox[
FormBox[\(\@x\),
"TraditionalForm"], "x"], TraditionalForm]]],
" = ",
Cell[BoxData[
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FormBox[\(\@x\),
"TraditionalForm"]]]],
" becomes very large as x becomes very small.\n\n"
}], "Text"],
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}, Open ]]
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