126 lines
2.9 KiB
Text
126 lines
2.9 KiB
Text
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#import "@youwen/zen:0.1.0": *
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#import "@preview/cetz:0.3.2"
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#show: zen.with(
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title: "Homework 8",
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author: "Youwen Wu",
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)
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#set heading(numbering: none)
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#show heading.where(level: 2): it => [#it.body.]
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#show heading.where(level: 3): it => [#it.body.]
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#set par(first-line-indent: 0pt)
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#show figure: it => {
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pad(
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y: 10pt,
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it,
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)
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}
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Problems:
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- 4.2: \#1g, 2b, 3b, 6, 7, 8, 14bd
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- 4.3: \#1bdeh, 2bdeh, 3ab, 5, 6, 7, 8
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#outline()
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= 4.2
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== 1
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=== g
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$f(x) = 1 / (1-x)$
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$f^(-1) (x) = 1 - 1 / x$
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== 2
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=== b
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$f(x) = x^2 + 2x$, $g(x) = 2x + 1$.
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$
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f compose g &= (2x+1)^2 + 2(2x+1) = 4x^2 + 8x + 3 \
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g compose f &= 2(x^2 + 2x) + 1 = 2x^2 + 4x + 1
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$
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== 3
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=== b
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$"Dom"(f compose g) = RR$, $"Dom"(g compose f = RR$.
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$"Rng"(f compose g) = [3,infinity]$, $"Rng"(g compose f) = [1, infinity]$.
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== 6
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#theorem[
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If $f : A -> B$, then $I_B compose f = f$.
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]
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#proof[
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$"Dom"(I_B compose f) = "Dom"(f) = A$. Now suppose $x in A$, then $(I_B
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compose f)(x) = I_B (f(x)) = f(x)$. Therefore $I_B compose f = f$.
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]
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== 7
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#theorem[
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Let $f : A -> B$ with $"Rng"(f) = C$. If $f^(-1)$ is a function, $f compose f^(-1) = I_C$.
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]
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#lemma[
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$"Dom"(f^(-1)) = "Rng"(f)$.
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#proof[
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Suppose $(x,y) in f$. Then $(y,x) in f^(-1)$. Therefore for any $x$ in $"Dom"(f)$, $x$ is also in $"Rng"(f^(-1))$, so $"Dom"(f) subset.eq "Rng"(f^(-1))$.
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Suppose $(x,y) in f^(-1)$. Then $(y,x) in f$. So for any $y in "Rng"(f^(-1))$, $y$ is also in $"Dom"(f)$. So $"Rng"(f^(-1)) subset.eq "Dom"(f)$. Therefore $"Dom"(f) = "Rng"(f^(-1))$.
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]
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]<dom-inverse-rng>
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#proof[
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Suppose that $f : A -> B$ and $f^(-1)$ is a function. By @dom-inverse-rng, $"Dom"(f^(-1)) = "Rng"(f) = "Dom"(I_C) = C$.
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Suppose that $y in C$. Because $(y,f^(-1) (y)) in f^(-1)$, $(f^(-1) (y),y) in f$. Then $f compose f^(-1) = f(f^(-1) (y)) = y = I_C$.
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]
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== 8
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The graph of $lr(f|)_A$ is simply the collection of points ${(1,1),(2,-2),(3,-5),(4,-8)}$
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The graph of $lr(f|)_[-1,3]$ is simply the line given by $f(x) = 4 - 3x$
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restricted between the $x$-coordinates $[-1,3]$.
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The graph of $lr(f|)_(2,4]$ is the line given by $f(x) = 4 - 3x$ between $x =
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2$ to $x = 4$, with a hole at $x = 2$.
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The graph of $lr(f|)_{6}$ is simply the point at $(6,-14)$.
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== 14
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=== b
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$
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&h : [-1, infinity) -> RR, &&h(x) = x^2 + 1 \
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&g : (-infinity, -1] -> RR, &&g(x) = x + 3
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$
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$h union g$ is a function only if $h(x) = g(x)$ where their domain overlaps at
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-1. We check $h(-1) = g(-1) = 2$ is indeed true. Everywhere else $h union g$ is
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either $g(x)$ or $h(x)$ which are both functions. Therefore it is a function.
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=== d
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$
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&h : (-infinity,2] -> RR, &&h(x) = cos x \
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&g : [2, infinity) -> RR, &&g(x) = x^2
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$
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We check that $h$ and $g$ agree at their overlapping domain. $h(2) = cos 2$,
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but $g(2) = 4$. Without further computation note that $cos(x)$ gives values
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only in $[-1,1]$ so $h(2) != g(2)$. $h union g$ would contain both $(2,4)$ and
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$(2,cos 2)$ so it's not a function.
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