#import "@youwen/zen:0.1.0": *
#import "@preview/cetz:0.3.2"

#show: zen.with(
  title: "Homework 8",
  author: "Youwen Wu",
)

#set heading(numbering: none)
#show heading.where(level: 2): it => [#it.body.]
#show heading.where(level: 3): it => [#it.body.]
#set par(first-line-indent: 0pt)
#show figure: it => {
  pad(
    y: 10pt,
    it,
  )
}

Problems:

- 4.2: \#1g, 2b, 3b, 6, 7, 8, 14bd

- 4.3: \#1bdeh, 2bdeh, 3ab, 5, 6, 7, 8

#outline()

= 4.2

== 1

=== g

$f(x) = 1 / (1-x)$

$f^(-1) (x) = 1 - 1 / x$

== 2

=== b

$f(x) = x^2 + 2x$, $g(x) = 2x + 1$.

$
  f compose g &= (2x+1)^2 + 2(2x+1) = 4x^2 + 8x + 3 \
  g compose f &= 2(x^2 + 2x) + 1 = 2x^2 + 4x + 1
$

== 3

=== b

$"Dom"(f compose g) = RR$, $"Dom"(g compose f = RR$.

$"Rng"(f compose g) = [3,infinity]$, $"Rng"(g compose f) = [1, infinity]$.

== 6

#theorem[
  If $f : A -> B$, then $I_B compose f = f$.
]

#proof[
  $"Dom"(I_B compose f) = "Dom"(f) = A$. Now suppose $x in A$, then $(I_B
compose f)(x) = I_B (f(x)) = f(x)$. Therefore $I_B compose f = f$.
]

== 7

#theorem[
  Let $f : A -> B$ with $"Rng"(f) = C$. If $f^(-1)$ is a function, $f compose f^(-1) = I_C$.
]

#lemma[
  $"Dom"(f^(-1)) = "Rng"(f)$.

  #proof[
    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 $"Dom"(f) subset.eq "Rng"(f^(-1))$.

    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 $"Dom"(f) = "Rng"(f^(-1))$.
  ]
]<dom-inverse-rng>

#proof[
  Suppose that $f : A -> B$ and $f^(-1)$ is a function. By @dom-inverse-rng, $"Dom"(f^(-1)) = "Rng"(f) = "Dom"(I_C) = C$.

  Suppose that $y in C$. Because $(y,f^(-1) (y)) in f^(-1)$, $(f^(-1) (y),y) in f$. Then $f compose f^(-1) = f(f^(-1) (y)) = y = I_C$.
]

== 8

The graph of $lr(f|)_A$ is simply the collection of points ${(1,1),(2,-2),(3,-5),(4,-8)}$

The graph of $lr(f|)_[-1,3]$ is simply the line given by $f(x) = 4 - 3x$
restricted between the $x$-coordinates $[-1,3]$.

The graph of $lr(f|)_(2,4]$ is the line given by $f(x) = 4 - 3x$ between $x =
2$ to $x = 4$, with a hole at $x = 2$.

The graph of $lr(f|)_{6}$ is simply the point at $(6,-14)$.

== 14

=== b

$
  &h : [-1, infinity) -> RR, &&h(x) = x^2 + 1 \
  &g : (-infinity, -1] -> RR, &&g(x) = x + 3
$

$h union g$ is a function only if $h(x) = g(x)$ where their domain overlaps at
-1. We check $h(-1) = g(-1) = 2$ is indeed true. Everywhere else $h union g$ is
either $g(x)$ or $h(x)$ which are both functions. Therefore it is a function.

=== d

$
  &h : (-infinity,2] -> RR, &&h(x) = cos x \
  &g : [2, infinity) -> RR, &&g(x) = x^2
$

We check that $h$ and $g$ agree at their overlapping domain. $h(2) = cos 2$,
but $g(2) = 4$. Without further computation note that $cos(x)$ gives values
only in $[-1,1]$ so $h(2) != g(2)$. $h union g$ would contain both $(2,4)$ and
$(2,cos 2)$ so it's not a function.