$\color{blue}\textrm{A. Pendahuluan}$
$\begin{aligned}&\\\textrm{Fungsi}\: &\textrm{atau pemetaan dari A ke B adalah}\\ &\textrm{suatu relasi khusus yang memasangkan}\\ &\textrm{setiap} \: \: x\in A\: \: \textrm{ke tepat satu}\: \: y\in B.\\ & \end{aligned}$
$\color{red}\begin{array}{|l|l|}\hline \textrm{Notasi}&\color{black}f:x \rightarrow y\: \: \: \color{red}\textrm{atau}\: \: \: \color{black}f:x \rightarrow f(x) \\\hline \textrm{Dibaca}&\textrm{fungsi}\: \: \color{black}f\: \: \color{red}\textrm{memetakan}\: \: \color{black}x\in A\: \: \color{red}\textrm{ke}\: \: \color{black}y\in B\\\hline \: \: \: \: \color{black}A&\textrm{Domain atau daerah asal fungsi atau}\quad \color{black}D_{f}\\\hline \: \: \: \: \, \color{black}x&\textrm{prapeta(sebelum dipetakan)}\\\hline \: \: \: \: \color{black}B&\textrm{Kodomain atau daerah kawan fungsi atau}\quad \color{black}K_{f}\\\hline \: \: \: \: \, \color{black}y&\textrm{peta(bayangan dari prapeta) adalah Range}\\ &\textrm{atau}\quad \color{black}R_{f}\\\hline \end{array}$
Sebagai ilustrasi perhatikanlah gambar berikut!
Sebagai misal, diberikan
$\begin{aligned}&\LARGE\boxed{f:x\rightarrow f(x)=3x+2}\\ &\textit{dibaca}:\\ &\color{red}\textrm{sebuah fungsi}\: \: f\: \: \textrm{memetakan}\: \: x\: \: ke\: \: 3x+2 \end{aligned}$
$\color{blue}\textrm{B. Sifat-Sifat Fungsi}$
$\color{purple}\begin{array}{|c|c|l|}\hline \textrm{Injektif}&\textrm{Surjektif}&\qquad\qquad\textrm{Bijektif}\\ (\textrm{satu-satu})&(\textrm{pada})&(\textrm{korespondensi satu-satu})\\\hline \begin{aligned}&\textrm{Jika setiap anggota}\\ &\textrm{himpunan A memiliki}\\ &\textrm{bayangan berbeda di}\\ &\textrm{himpunan B} \end{aligned}&\begin{aligned}&\textrm{Jika setiap anggota}\\ &\textrm{himpunan di B}\\ &\textrm{mempunyai prapeta}\\ &\textrm{di himpunan A} \end{aligned}&\begin{aligned}&\textrm{Jika fungsi yang injektif}\\ &\textrm{sekaligus juga surjektif}\\ &\\ & \end{aligned}\\\hline \end{array}$
$\color{blue}\textrm{C. Operasi Aljabar Fungsi}$
$\color{purple}\begin{array}{|l|l|}\hline \qquad\color{black}\textrm{Aljabar Fungsi}&\quad\color{black}\textrm{Daerah Asal}\\\hline (f+g)(x)=f(x)+g(x)&D_{_{(f+g)}}=D_{_{f}}\cap D_{_{g}}\\\hline (f-g)(x)=f(x)-g(x)&D_{_{(f-g)}}=D_{_{f}}\cap D_{_{g}}\\\hline (f.g)(x)=f(x).g(x)&D_{_{(f.g)}}=D_{_{f}}\cap D_{_{g}}\\\hline \left ( \displaystyle \frac{f}{g} \right )(x)=\displaystyle \frac{f(x)}{g(x)}&D_{_{\left ( \frac{f}{g} \right )}}=D_{_{f}}\cap D_{_{g}}\: , \\ &\textrm{dengan}\: \: g(x)\neq 0 \\\hline \end{array}$
$\color{blue}\textrm{D. Macam-Macam Fungsi}$
$\color{purple}\begin{array}{|l|l|}\hline \textrm{Fungsi Konstan}&\textrm{Berupa konstanta}\\ &f(x)=c\\\hline \textrm{Fungsi Identitas}&\textrm{Nilainya dirinya sendiri}\\ &f(x)=x\\\hline \textrm{Fungsi linear}&\textrm{Fungsi berupa garis lurus}\\ &f(x)=ax+b\\\hline \textrm{Fungsi Kuadrat}&\textrm{Fungsi Kuadrat/parabola}\\ &f(x)=ax^{2}+bx+c,\quad a\neq 0\\\hline \textrm{Fungsi Rasional}&\textrm{Fungsi Pecahan}\\ &f(x)=\displaystyle \frac{p(x)}{q(x)}\\\hline \textrm{Fungsi Khusus 1}&\textrm{Fungsi Modulus(nilai mutlak)}\\ &f(x)=\left | x \right |\\\hline \textrm{Fungsi Khusus 2} &\textrm{Fungsi tangga}\\ &f(x)=\left \lfloor x \right \rfloor\\\hline \textrm{Fungsi Khusus 3}&\textrm{Fungsi genap dan ganjil}\\ &\begin{cases} \textrm{Fungsi ganjil} & f(-x)=-f(x) \\ \textrm{Fungsi genap} & f(-x)=f(x) \end{cases}\\\hline \end{array}$
$\LARGE\colorbox{yellow}{CONTOH SOAL}$
$\begin{array}{ll}\\ 1.&\textrm{Diketahui 2 humpuan sebagai berikut}:\\ &\begin{cases} \textrm{P} & =\left \{ -2,-1,0,1,2 \right \} \\ \textrm{Q} & =\left \{ 0,1,2,5,7 \right \} \end{cases}\\ &\textrm{Di antara relasi dari P ke Q berikut manakah }\\ &\textrm{yang merupakan fungsi}\\ &\textrm{a}.\quad \textrm{A}=\left \{ (-2,0),(-1,0),(0,0),(1,0),(2,0) \right \}\\ &\textrm{b}.\quad \textrm{B}=\left \{ (-2,1),(-1,2),(0,5),(1,7),(-2,2) \right \}\\ &\textrm{c}.\quad \textrm{C}=\left \{ (-2,0),(-1,1),(0,2),(1,5),(2,7) \right \}\\\\ &\color{blue}\textrm{Jawab}:\\ &\textrm{Semuanya Fungsi kecuali}\textbf{ poin b)} \end{array}$
$\begin{array}{ll}\\ 2.&\textrm{Relasi berikut yang merupakan fungsi adalah}.... \end{array}$
$.\qquad \begin{array}{|c|l|l|}\hline \textrm{Poin}&\textrm{Jenis}&\textrm{Keterangan}\\\hline \textrm{a}&\textrm{Fungsi}&\begin{aligned}&\textrm{Sesuai definisi}\\ \end{aligned}\\ &&\begin{aligned}&\textrm{yaitu}:\\ &\textrm{Setiap prepeta}\\ &\textrm{(anggota himpunan A) }\\ &\textrm{memiliki peta di }\\ &\textrm{himpunan B tepat satu}.\\ &\color{blue}\textrm{Tetapi}\\ &\color{blue}\textrm{bukan fungsi injektif}\\ &\color{red}\textrm{bukan pula fungsi}\\ &\color{red}\textrm{surjektif} \end{aligned}\\\hline \textrm{b}&\textrm{Fungsi}&\textrm{Sama di atas}\\\hline \textrm{c}&\textrm{Bukan Fungsi}&\textrm{Tidak sesuai definisi}\\ &&\color{red}\textrm{hanya relasi saja}\\\hline \textrm{d}&\textrm{Fungsi}&\textrm{Sesuai definisi}\\ &&\color{blue}(\textrm{Fungsi bijektif})\\\hline \end{array}$
$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah daerah asal dari fungsi beberapa berikut}:\\ &\begin{array}{lllllllll}\\ \textrm{a}.&f(x)=x-3&\textrm{g}.&f(x)=\displaystyle \frac{\left | x \right |}{x}\\ \textrm{b}.&f(x)=\displaystyle \frac{6}{x^{2}-2x-8}&\textrm{h}.&f(x)=\left \lfloor x \right \rfloor \\ \textrm{c}.&f(x)=\displaystyle \frac{x^{2}-3x}{x^{2}-2x-15}&\textrm{i}.&f(x)=\left | x \right |+\left \lfloor x \right \rfloor\\ \textrm{d}.&y+2=x^{2}-5x+5&\textrm{j}.&f(x)=\sqrt{x^{2}-16}\\ \textrm{e}.&f(x)=\left | x-3 \right |&\textrm{k}.&f(x)=\sqrt{2x^{2}-50}\\ \textrm{f}.&f(x)=3-\left | 2x-1 \right |&\textrm{l}.&f(x)=\displaystyle \frac{2\sqrt{x}}{x-3} \end{array}\\\\ \end{array}$.
$.\qquad \begin{aligned}&\textrm{catatan}\: \: \left \lfloor x \right \rfloor\: \: \textrm{adalah }\\ &\textrm{bulat terbesar atau sama dengan}\: \: x \end{aligned}$.
$.\qquad \color{blue}\textrm{Jawab}$.
$.\qquad \begin{array}{|c|c|}\hline (\textrm{a})&(\textrm{b})\\\hline \begin{aligned}f(x)&=x-3\\ \textrm{selu}&\textrm{ruh bilangan real}\\ &x\: \: \textrm{akan terdefinisi}\\ &\textrm{atau tetap bernilai}\\ &\textrm{real}\\ \textrm{sehi}&\textrm{ngga},\\ D_{f}&=\left \{ x|x\in \mathbb{R} \right \}\\ & \end{aligned}&\begin{aligned}f(x)&=\displaystyle \frac{6}{x^{2}-2x-8}\\ \textrm{terd}&\textrm{efinisi ketika}\\ &\textrm{penyebut tidak sama}\\ &\textrm{dengan}\: \: 0, \: \: \textrm{yaitu}:\\ &\begin{aligned}x^{2}-2x-8&\neq 0\\ (x-4)(x+2)&\neq 0\\ x\neq 4\: \: \textrm{dan}\: \: x&\neq -2 \end{aligned}\\ D_{f}&=\left \{ x|x\in \mathbb{R},\: x\neq 4\: \: \textrm{dan}\: \: x\neq -2 \right \} \end{aligned}\\\hline \end{array}$
$.\qquad \begin{array}{|c|c|}\hline (\textrm{d})&(\textrm{e})\\\hline \begin{aligned}&y+2=x^{2}-5x+5\\ &y=x^{2}-5x+5-2\\ &f(x)=x^{2}-5x+3\\ &\textrm{Sehingga daerah}\\ &\textrm{asalnya}\\ &D_{f}=\left \{ x|x\in \mathbb{R} \right \}\\ & \end{aligned}&\begin{aligned}f(x)&=\left | x-3 \right |\\ D_{f}&=\left \{ x|x\in \mathbb{R} \right \}\\ \textrm{teta}&\textrm{pi pada \textit{range} fungsinya}\\ &\textrm{hanya akan berupa}\\ &\textrm{bilangan positif saja}.\\ \textrm{yait}&\textrm{u}:\\ R_{f}&=\left \{ y|y\in \mathbb{R},\: y\geq 0 \right \}\\ & \end{aligned}\\\hline \end{array}$
$.\qquad \begin{array}{|c|c|}\hline (\textrm{g})&(\textrm{h})\\\hline \begin{aligned} &f(x)=\displaystyle \frac{\left \lfloor x \right \rfloor}{x}\\ &\textrm{Sehingga daerah}\\ &\textrm{asalnya}\\ &D_{f}=\left \{ x|x\in \mathbb{R},\: x\neq 0 \right \}\\ & \end{aligned}&\begin{aligned}&f(x)=\left \lfloor x \right \rfloor\\ &\textrm{Sehingga daerah}\\ &\textrm{asalnya}\\ &D_{f}=\left \{ x|x\in \mathbb{R} \right \}\\ & \end{aligned}\\\hline \end{array}$.
$.\qquad \begin{array}{|c|c|}\hline (\textrm{i})&(\textrm{l})\\\hline \begin{aligned} &f(x)=\sqrt{x^{2}-16}\\ &\textrm{Sehingga daerah}\\ &\textrm{asalnya yaitu:}\\ &x^{2}-16\geq 0\\ &(x+4)(x-4)\geq 0\\ &x\leq -4\: \: \textrm{atau}\: \: x\geq 4\\ &D_{f}=\left \{ x|x\leq -4\: \: \textrm{atau}\: \: x\geq 4,\: x\in \mathbb{R} \right \}\\ & \end{aligned}&\begin{aligned}&f(x)=\displaystyle \frac{2\sqrt{x}}{x-3}\\ &\textrm{Sehingga daerah}\\ &\textrm{asalnya yaitu:}\\ &\begin{cases} \bullet & x\geq 0 \\ \bullet & x-3\neq 0 \end{cases}\\ &D_{f}=\left \{ x|x\geq 0,\: x\neq 3,\: x\in \mathbb{R} \right \}\\ & \end{aligned}\\\hline \end{array}$
$\begin{array}{ll}\\ 4.&\textrm{Jika}\: \left | x \right |\: \textrm{menyatakan nilai mutlak}\\ &\textrm{dan}\: \left \lfloor x \right \rfloor\: \textrm{menyatakan bilangan bulat terbesarnatau sama dengan}\: x\\ &\textrm{misalkan}\: \: \left \lfloor 1,6 \right \rfloor=1,\: \left \lfloor \pi \right \rfloor=3\\ &\textrm{Jika diberikan}\: \: f(x)=\left | x \right |+\left \lfloor x \right \rfloor,\: \textrm{maka tentukanlah nilai untuk}\\ &\textrm{a}.\quad f\left ( -3,5 \right )+f\left ( 2,5 \right )\\ &\textrm{b}.\quad f\left ( -1,5 \right )+f\left ( 3,5 \right )\\\\ &\textrm{Jawab}:\\ &\begin{array}{l}\\ \begin{aligned}\textrm{a}.f\left ( -3,5 \right )+f\left ( 2,5 \right )&=\left | -3,5 \right |+\left \lfloor -3,5 \right \rfloor+\left | 2,5 \right |+\left \lfloor 2,5 \right \rfloor\\ &=3,5+\left ( -4 \right )+2,5+2\\ &=4 \end{aligned}\\ \begin{aligned}\textrm{b}.f\left ( -1,5 \right )+f\left ( 3,5 \right )&=\left | -1,5 \right |+\left \lfloor -1,5 \right \rfloor+\left | 3,5 \right |+\left \lfloor 3,5 \right \rfloor\\ &=1,5+(-2)+3,5+3\\ &=6 \end{aligned} \end{array} \end{array}$
$\begin{array}{ll}\\ 5.&\textrm{Jika diketahui relasi}\: \: f\: \: \textrm{dengan kondisi}\\ &\begin{array}{l} (\textrm{a}).\quad f(1)=1\\ (\textrm{b}).\quad f(2x)=4f(x)+6\\ (\textrm{c}).\quad f(x+2)=f(x)+12x+12 \end{array}\\ &\textrm{maka nilai}\: \: f(14)\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(1)&=1\\ f(2.1)=f(2)&=4f(1)+6=4.1+6=10\\ f(1+2)=f(3)&=f(1)+12.1+12\\ f(3)&=1+12+12=25\\ f(3+2)=f(5)&=f(3)+12.3+12\\ f(5)&=25+36+12=73\\ f(5+2)=f(7)&= f(5)+12.5+12\\ f(7)&=73+60+12=145\\ f(7.2)=f(14)&=4.f(7)+6\\ f(14)&=4.145+6=580+6\\ &=586 \end{aligned} \end{array}$
$\begin{array}{ll}\\ 6.&\textbf{(OSK 2013)}\textrm{Fungsi}\: \: f\: \: \textrm{didefinisikan oleh}\\ & f(x)=\displaystyle \frac{kx}{2x+3},\quad x=-\displaystyle \frac{3}{2}.\\ & \textrm{Tentukanlah nilai}\: \: k\: \: \textrm{agar}\: \: f(f(x))=x\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f(x)&=\displaystyle \frac{kx}{2x+3},\quad x\neq -\frac{2}{3}\\ f\left ( f(x) \right )&=x\\ x&=f\left ( f(x) \right )\\ x&=\displaystyle \frac{k\left ( \displaystyle \frac{kx}{2x+3} \right )}{2\left ( \displaystyle \frac{kx}{2x+3} \right )+3}\\ x&=\displaystyle \frac{\displaystyle \frac{k^{2}x}{2x+3}}{\displaystyle \frac{2kx+3(2x+3)}{2x+3}}\\ x&=\displaystyle \frac{k^{2}x}{2kx+6x+9}\\ 2kx+6x+9&=k^{2}\\ 0&=k^{2}-2xk-6x-9\\ 0&=(k+3)(k-2x-3)\\ &\quad k=-3\: \: \textrm{atau}\: \: k=2x+3 \end{aligned} \end{array}$
$\color{blue}\textrm{E. Menggambar Grafik Fungsi}$
Untuk menggambar suatu fungsi $f(x)$ dengan kondisi rumusnya telah diketahui pada diagram Kartesius adalah sebagai berikut
- Menentukan titik-titik berupa pasangan terurut $(x,y)$ dalam tabel dengan $x$ anggota dari daerah asal (domain) dan $y$ adalah anggota dari daerah kawan (kodomain).
- mengkonversi titik-titik tadi ke dalam diagram kartesius
- menghubungkan titik-titik tersebut sehingga didapatkan grafik mulus
$\LARGE\colorbox{yellow}{CONTOH SOAL}$
$\begin{array}{ll}\\ 1.&\textrm{Gambarlah grafik fungsi}\\ &\textrm{a}.\quad f(x)=2x+5\\ &\textrm{b}.\quad g(x)=2x^{2}\\ &\textrm{c}.\quad h(x)=\displaystyle \frac{1}{x}\\\\ &\color{blue}\textrm{Jawab}:\\ & \end{array}$.
$.\qquad\begin{aligned}&\textrm{a. Menggambar grafik}\: \: f(x)=2x+5\\ &\begin{array}{|c|c|c|}\hline x&y=f(x)=2x+5&\textrm{Titik}\: (x,y)\\\hline -3&f(-3)=2(-3)+5=-1&(-3,-1)\\\hline -2&f(-2)=2(-2)+5=3&(-2,1)\\\hline -1&f(-1)=2(-1)+5=3&(-1,3)\\\hline 0&f(0)=2(0)+5=5&(0,5)\\\hline 1&f(1)=2(1)+5=7&(1,7)\\\hline 2&f(2)=2(2)+5=9&(2,9)\\\hline 3&f(3)=2(3)+5=11&(3,11)\\\hline \end{array} \end{aligned}$
$.\qquad\begin{aligned}&\textrm{b. Menggambar grafik}\: \: f(x)=2x^{2}\\ &\begin{array}{|c|c|c|}\hline x&y=f(x)=2x^{2}&\textrm{Titik}\: (x,y)\\\hline -3&f(-3)=2(-3)^{2}=18&(-3,18)\\\hline -2&f(-2)=2(-2)^{2}=8&(-2,8)\\\hline -1&f(-1)=2(-1)^{2}=2&(-1,2)\\\hline 0&f(0)=2(0)^{2}=0&(0,0)\\\hline 1&f(1)=2(1)^{2}=2&(1,2)\\\hline 2&f(2)=2(2)^{2}=8&(2,8)\\\hline 3&f(3)=2(3)^{2}=18&(3,18)\\\hline \end{array} \end{aligned}$
$.\qquad\begin{aligned}&\textrm{c. Menggambar grafik}\: \: f(x)=\displaystyle \frac{1}{x}\\ &\begin{array}{|c|c|c|}\hline x&y=f(x)=\displaystyle \frac{1}{x}&\textrm{Titik}\: (x,y)\\\hline -3&f(-3)=-\displaystyle \frac{1}{3}&(-3,-\displaystyle \frac{1}{3})\\\hline -2&f(-2)=-\displaystyle \frac{1}{2}&(-2,-\displaystyle \frac{1}{2})\\\hline -1&f(-1)=-1&(-1,-1)\\\hline 0&f(0)\: \: \color{blue}\textrm{tidak ada}&-\\\hline 1&f(1)=1&(1,1)\\\hline 2&f(2)=\displaystyle \frac{1}{2}&(2,\displaystyle \frac{1}{2})\\\hline 3&f(3)=\displaystyle \frac{1}{3}&(3,\displaystyle \frac{1}{3})\\\hline \end{array} \end{aligned}$
$\begin{array}{ll}\\ 2.&\textrm{Gambarlah grafik fungsi}\\ &\textrm{a}.\quad f(x)=\left \lfloor x \right \rfloor\\ &\textrm{b}.\quad g(x)=\left \lfloor x+1 \right \rfloor\\ &\textrm{c}.\quad h(x)=\left \{ x \right \}\\\\ &\textbf{Catatan}:\\ &\left \lfloor x \right \rfloor=\textrm{bilangan bulat terbesar tetapi}\\ &\: \: \: \, \quad\quad \textrm{lebih kecil atau sama dengan}\: \: x\\ &\left \{ x \right \}=\textrm{bagian pecahan dari}\: \: x\\\\ &\color{blue}\textrm{Jawab}:\\ & \end{array}$.
$.\qquad\begin{aligned}&\textrm{a. Menggambar grafik}\: \: f(x)=\left \lfloor x \right \rfloor\\ &\begin{array}{|c|c|c|}\hline x&y=f(x)=\left \lfloor x \right \rfloor&\textrm{Titik}\: (x,y)\\\hline \vdots &\vdots &\vdots \\\hline -1&f(-1)=\left \lfloor -1 \right \rfloor=-1&(-1,-1)\\\hline \vdots &\vdots &\vdots \\\hline -\displaystyle \frac{1}{2}&f(-\displaystyle \frac{1}{2})=\left \lfloor -\displaystyle \frac{1}{2} \right \rfloor=-1&(-\displaystyle \frac{1}{2},-1)\\\hline -\displaystyle \frac{1}{3}&f(-\displaystyle \frac{1}{3})=\left \lfloor -\displaystyle \frac{1}{3} \right \rfloor=-1&(-\displaystyle \frac{1}{3},-1)\\\hline -\displaystyle \frac{1}{4}&f(-\displaystyle \frac{1}{4})=\left \lfloor -\displaystyle \frac{1}{4} \right \rfloor=-1&(\displaystyle \frac{1}{4},-1)\\\hline \vdots &\vdots &\vdots \\\hline 0&f(0)=\left \lfloor 0 \right \rfloor=0&(0,0)\\\hline \vdots &\vdots &\vdots \\\hline \displaystyle \frac{1}{4}&f(\displaystyle \frac{1}{4})=\left \lfloor \displaystyle \frac{1}{4} \right \rfloor=0&(\displaystyle \frac{1}{4},0)\\\hline \displaystyle \frac{1}{3}&f(\displaystyle \frac{1}{3})=\left \lfloor \displaystyle \frac{1}{3} \right \rfloor=0&(\displaystyle \frac{1}{3},0)\\\hline \displaystyle \frac{1}{2}&f(\displaystyle \frac{1}{2})=\left \lfloor \displaystyle \frac{1}{2} \right \rfloor=0&(-\displaystyle \frac{1}{2},0)\\\hline \vdots &\vdots &\vdots \\\hline 1&f(1)=\left \lfloor 1 \right \rfloor=1&(1,1)\\\hline \vdots &\vdots &\vdots \\\hline 2\displaystyle \frac{1}{4}&f(2\displaystyle \frac{1}{4})=\left \lfloor 2\displaystyle \frac{1}{4} \right \rfloor=2&(2\displaystyle \frac{1}{4},2)\\\hline \vdots &\vdots &\vdots \\\hline 4\displaystyle \frac{1}{3}&f(4\displaystyle \frac{1}{3})=\left \lfloor 4\displaystyle \frac{1}{3} \right \rfloor=4&(4\displaystyle \frac{1}{3},4)\\\hline \vdots &\vdots &\vdots \\\hline \end{array} \end{aligned}$
$\begin{array}{ll}\\ 3.&\textbf{(OSK 2003)}\textrm{Jika}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{adalah bilangan real }\\ &\textrm{sedemikian sehingga}\: \: \displaystyle \left \lfloor \sqrt{x} \right \rfloor=9\: \: \textrm{dan}\: \: \displaystyle \left \lfloor \sqrt{y} \right \rfloor=12\: ,\\ &\textrm{maka nilai terkecil dari}\: \: \displaystyle \left \lfloor y-x \right \rfloor=....\\\\ &\textrm{Jawab}:\\ & \end{array}$.
$.\qquad \begin{aligned}&\textrm{Diketahui bahwa}\: \: \left \lfloor x \right \rfloor\: \: \textrm{adalah bilangan bulat terbesar }\\ &\textrm{yang lebih kecil atau sama dengan}\: \: x\\ &\textrm{Misal}\: \: \left \lfloor 3,2 \right \rfloor=3, \: \left \lfloor -2,47 \right \rfloor=-3,\: \: \left \lfloor 6 \right \rfloor=6,\: \: \textrm{dan lain-lain}.\\ &\textrm{Sehingga}\, \,\left \lfloor x \right \rfloor=a\Rightarrow a\leq x< a+1\: \: (\textrm{dengan}\: \: a\: \: \textrm{bilangan bulat}),\\ & \textrm{maka}\\ &\begin{cases} \left \lfloor \sqrt{x} \right \rfloor=9 & \Rightarrow 9\leq \sqrt{x}< 9+1\\ &\Leftrightarrow 9\leq \sqrt{x}< 10\Leftrightarrow \color{red}81\leq x< 100 \\ \left \lfloor \sqrt{y} \right \rfloor=12 & 12\leq \sqrt{y}< 12+1\\ &\Leftrightarrow 12\leq \sqrt{y}< 13\Leftrightarrow \color{red}144\leq y< 169 \end{cases}\\ \end{aligned}$.
$.\qquad \begin{aligned}&\bullet \: \: 144\leq y< 169\: \: \textrm{dan}\\ &\bullet \: \: \: \: 81\leq x< 100\: ,\: \textrm{dikalikan dengan}\: \: (-1)\\ &\quad \textrm{maka akan menjadi},\\ &\qquad -100< -x\leq -81,\\ &\textrm{sehingga}\: \: -99,\overline{999}\leq -x< -80,\overline{999}\\ &\textrm{Selanjutnya}\\ &\begin{array}{cll}\\ 144&\leq y< 169&\\ -99,\overline{999}&\leq -x< -80,\overline{999}&+\\\hline\\ 44,...&\leq y-x< 88,...\\ \end{array}\\ &\textrm{Jadi, nilai terkecil}\: \left \lfloor y-x \right \rfloor=\color{red}44 \end{aligned}$
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