Contoh Soal 3 Fungsi

$\begin{array}{ll}\\ 11.&\textrm{Fungsi berikut yang tidak mempunyai }\\ &\textrm{asimtot vertikal adalah}\: ....\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=\displaystyle \frac{x+2}{x^{2}-3}&\\ \textrm{b}.&f(x)=\displaystyle \frac{x}{(x-2)^{2}}\\ \color{red}\textrm{c}.&f(x)=\displaystyle \frac{x^{2}-9}{x+3}\\ \textrm{d}.&f(x)=\displaystyle \frac{-3}{x}\\ \textrm{e}.&\textrm{semuanya mempunyai asimtot vertikal} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Perh}&\: \textrm{atikanlah opsi jawaban}\: \: c,\: \textrm{yaitu}:\\ f(x)&=\displaystyle \frac{x^{2}-9}{x+3}\: \\ \textrm{Jika}&\: \textrm{disederhanakan akan menjadi }\\ &\textbf{fungsi linear}\: \: \textrm{yaitu}:\\ f(x)&=\displaystyle \frac{x^{2}-9}{x+3}=\displaystyle \frac{(x+3)(x-3)}{x+3}=x-3\\ \textrm{sehi}&\textrm{ngga fungsi pada opsi}\: \: c\\ & \textrm{adalah berupa persamaan linear}\\ &\textrm{yang secara otomatis }\\ &\textbf{tidak akan memiliki asimtot} \end{aligned} \end{array}$

Contoh Soal 2 Fungsi

$\begin{array}{ll}\\ 6.&\textrm{Diketahui bahwa}\\ &f(x)=\begin{cases} 0,&\textrm{untuk}\: x<0\\ x^{2},&\textrm{untuk}\: 0\leq x< 1\\ 2x-1,&\textrm{untuk}\: x\geq 1 \end{cases}\\ &\textrm{Nilai dari}\: f(-1)+f\left ( \displaystyle \frac{1}{2} \right )-f(3)\: \textrm{adalah}.... \\ &\begin{array}{llllll}\\ \textrm{a}.&-5\displaystyle \frac{1}{4}&&&\textrm{d}.&4\displaystyle \frac{3}{4}\\\\ \color{red}\textrm{b}.&-4\displaystyle \frac{3}{4}&\textrm{c}.&4&\textrm{e}.&5\displaystyle \frac{1}{4} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}\: &\\ f(x)&=\begin{cases} 0,&\textrm{untuk}\: x<0\\ x^{2},&\textrm{untuk}\: 0\leq x< 1\\ 2x-1,&\textrm{untuk}\: x\geq 1 \end{cases}\\ \textrm{maka nilai}&\: \: f(-1)+f\left ( \displaystyle \frac{1}{2} \right )-f(3)\\ &=0+\left ( \displaystyle \frac{1}{2} \right )^{2}-\left ( 2(3)-1 \right )\\ &=\displaystyle \frac{1}{4}-5\\ &=-4\displaystyle \frac{3}{4} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 7.&\textrm{Jika diketahui}\: \: f\left ( x+\displaystyle \frac{1}{x} \right )=x^{3}+\displaystyle \frac{1}{x^{3}}\: ,\\ &\textrm{maka nilai dari}\: \: f\left ( \displaystyle \frac{5}{2} \right )\: \: \textrm{adalah}.... \\ &\begin{array}{llllll}\\ \textrm{a}.&2\displaystyle \frac{1}{8}&&&\color{red}\textrm{d}.&8\displaystyle \frac{1}{8}\\\\ \textrm{b}.&2\displaystyle \frac{1}{2}&\textrm{c}.&4\displaystyle \frac{1}{8}&\textrm{e}.&12\displaystyle \frac{1}{8} \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Perhatikan}\, &\: \textrm{bahwa}:\\ (p+q)^{3}&=p^{3}+3p^{2}q+3pq^{2}+q^{3}\\ \textrm{Jika kita su}&\textrm{bstitusikan}\: \: p=x^{3}\: \: \textrm{dan}\: \: q=\displaystyle \frac{1}{x^{3}}\\ \left ( x+\displaystyle \frac{1}{x} \right )^{3}&=x^{3}+3x^{2}\left ( \displaystyle \frac{1}{x} \right )+3x\left ( \displaystyle \frac{1}{x} \right )^{2}+\left ( \displaystyle \frac{1}{x} \right )^{3}\\ &=x^{3}+\left ( \displaystyle \frac{1}{x} \right )^{3}+3x+\displaystyle \frac{3}{x}\\ &=\left ( x^{3}+\displaystyle \frac{1}{x^{3}} \right )+3\left ( x+\displaystyle \frac{1}{x} \right )\\ \textrm{sehingga}\quad&\\ f\left ( x+\displaystyle \frac{1}{x} \right )&=x^{3}+\displaystyle \frac{1}{x^{3}}\\ &=\left ( x+\displaystyle \frac{1}{x} \right )^{3}-3\left ( x+\displaystyle \frac{1}{x} \right )\\ f(u)&=u^{3}-3u,\: \: \textrm{maka}\\ f\left ( \displaystyle \frac{5}{2} \right )&=\left ( \displaystyle \frac{5}{2} \right )^{3}-3\left ( \displaystyle \frac{5}{2} \right )\\ &=\displaystyle \frac{125}{8}-\frac{15}{2}\\ &=\displaystyle \frac{65}{8}\\ &=8\displaystyle \frac{1}{8} \end{aligned} \end{array}$

$\begin{array}{ll}\\ 8.&\textrm{Misal fungsi}\: \: f\: \: \textrm{terdefinisi untuk }\\ &\textrm{seluruh bilangan real}\: \: x.\\ &\textrm{Jika}\: \: f(p+q)=f(pq)\: \: \textrm{untuk semua}\: \: p,\: q\\ & \textrm{bilangan bulat positif dan}\: \: f(1)=2,\\ &\textrm{maka nilai}\: \: f(2021)=....\\ &\begin{array}{llllll}\\ \textrm{a}.&0&&&\textrm{d}.&3\\\\ \textrm{b}.&1&\color{red}\textrm{c}.&2&\textrm{e}.&5 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}\: & \textrm{bahwa}\\ f(1)&=2\: \: \textrm{dan}\\ f(p+q)&=f(pq)\\ \textrm{maka}&\\ f(2)&=f(1+1)=f(1.1)=f(1)=2\\ f(3)&=f(1+2)=f(1.2)=f(2)=f(1)=2\\ f(4)&=f(1+3)=f(1.3)=f(3)=f(2)=f(1)=2\\ f(5)&=f(1+4)=f(1.4)=f(4)=f(3)=f(2)=f(1)=2\\ \vdots &\\ f(2021)&=\cdots =\cdots =\cdots =f(2)=f(1)=2 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 9.&\textrm{Jika}\: \: a_{_{0}}=\displaystyle \frac{2}{5}\: \: \textrm{dan}\: \: a_{n+1}=2\left | a_{n} \right |-1, \textrm{maka nilai}\: \: a_{_{2022}}\\ &\textrm{adalah}....\\ &\begin{array}{llllll}\\ \color{red}\textrm{a}.&-0,6&&&\textrm{d}.&0,4\\\\ \textrm{b}.&-0,2&\textrm{c}.&0,2&\textrm{e}.&0,6 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textrm{Diketahui}\: & \textrm{bahwa}\\ a_{_{0}}&=\displaystyle \frac{2}{5}=0,4\: \: \textrm{dan}\: \: a_{n+1}=2\left | a_{n} \right |-1,\: \textrm{maka}\\ a_{_{1}}&=2\left | a_{_{0}} \right |-1=2\left | 0,4 \right |-1=0,8-1=-0,2\\ a_{_{2}}&=2\left | a_{_{1}} \right |-1=2\left | -0,2 \right |-1=2(0,2)-1\\ &=0,4-1=-0,6\\ a_{_{3}}&=2\left | a_{_{2}} \right |-1=2\left | -0,6 \right |-1=1,2-1=0,2\\ a_{_{4}}&=2\left | a_{_{3}} \right |-1=2\left | 0,2 \right |-1=0,4-1=-0,6=\textbf{a}_{_{\textbf{2}}}\\ a_{_{5}}&=2\left | a_{_{4}} \right |-1=2\left | -0,6 \right |-1=1,2-1=0,2=\textbf{a}_{_{\textbf{3}}}\\ a_{_{6}}&=2\left | a_{_{5}} \right |-1=2\left | 0,2 \right |-1=0,4-1=-0,6=\textbf{a}_{_{\textbf{2}}}\\ a_{_{7}}&=2\left | a_{_{6}} \right |-1=2\left | -0,6 \right |-1=1,2-1=0,2=\textbf{a}_{_{\textbf{3}}}\\ \vdots &\\ a_{_{2022}}&=\cdots =\cdots =\textbf{a}_{_{\textbf{2}}}=-0,6 \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 10.&\textrm{Kurva}\: \: f(x)=\displaystyle \frac{10}{x^{2}-10x+25}\\ &\textrm{mempunyai asimtot vertikal pada}....\\ &\begin{array}{llllll}\\ \textrm{a}.&x=0\: \: \textrm{saja}&&&\\ \color{red}\textrm{b}.&x=5\: \: \textrm{saja}&\\ \textrm{c}.&x=10\: \: \textrm{saja}\\ \textrm{d}.&x=0\: \: \textrm{dan}\: \: x=5\: \: \textrm{saja}\\ \textrm{e}.&x=0,\: x=5,\: \: \textrm{dan}\: \: x=10 \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}\textbf{Asimtot vertikal}&(\textbf{tegak})\: \textrm{diperoleh saat}\\ x^{2}-10x+25&=0\\ (x-5)^{2}&=0\\ x-5&=0\\ x&=5\\\\ \textrm{Ilustrasinya gamba}&\textrm{rnya adalah sebagai berikut}: \end{aligned} \end{array}.$

Contoh Soal Fungsi

$\begin{array}{ll}\\ 1.&\textrm{Relasi berikut yang akan berupa fungsi adalah}....\\ &\begin{array}{llllll}\\ \textrm{a}.&f(x)=\sqrt{x}&&&\\ \textrm{b}.&f(x)=1-\sqrt{x}&\\ \textrm{c}.&f(x)=\sqrt{x}+1&\\ \textrm{d}.&f(x)=\sqrt{x}-1\\ \color{red}\textrm{e}.&f(x)=\left | x \right | \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|l|c|l|}\hline \textrm{No}&\: \: \qquad \textrm{Fungsi}&\textrm{Grafik}\\\hline 1.\textrm{a}&f(x)=y=\sqrt{x}&y^{2}=x\\\hline 1.\textrm{b}&f(x)=y=1-\sqrt{x}&(1-y)^{2}=x\\\hline 1.\textrm{c}&f(x)=y=1+\sqrt{x}&(y-1)^{2}=x\\\hline 1.\textrm{d}&f(x)=y=\sqrt{x}-1&(y+1)^{2}=x\\\hline 1.\textrm{e}&f(x)=y=\left | x \right |&y=\begin{cases} x & \text{ jika} \: \: x\geq 0 \\ -x & \text{ jika } \: \: x<0 \end{cases}\\\hline \end{array}\\ &\begin{aligned}&\textrm{Dengan prepeta yang berbeda}\\ &\textrm{akan menghasilkan peta yang}\\ &\textrm{berbeda pula (fungsi bijektif)} \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 2.&\textrm{Fungsi dari himpunan A ke himpunan B }\\ &\textrm{berikut termasuk jenis fungsi} \end{array}$.

$\begin{array}{ll}\\ .\: \quad.&\textrm{Relasi berikut yang akan berupa fungsi adalah}....\\ &\color{red}\textrm{a}.\quad \textrm{fungsi umum}\\ &\textrm{b}.\quad \textrm{fungsi satu-satu, tetapi bukan fungsi pada}\\ &\textrm{c}.\quad \textrm{fungsi pada, tetapi bukan fungsi satu-satu}\\ &\textrm{d}.\quad \textrm{fungsi pada dan satu-satu}\\ &\textrm{e}.\quad \textrm{tidak ada jawaban yang benar}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|l|}\hline \textrm{No}&\textrm{Keterangan}&\qquad\qquad \textrm{Alasan}\\\hline 2.\textrm{a}&\textrm{Sesuai}&\textrm{Sesuai definisi fungsi}\\\hline 2.\textrm{b}&\textrm{Salah}&\begin{aligned}&\textrm{Karena bukan fungsi }\\ &\textrm{satu-satu(fungsi injektif)}\\ &\textrm{walau benar dikatakan bukan}\\ &\textrm{fungsi pada (fungsi surjektif)} \end{aligned}\\\hline 2.\textrm{c}&\textrm{Salah}&\begin{aligned}&\textrm{Karena bukan fungsi }\\ &\textrm{pada(fungsi surjektif)}\\ &\textrm{walau benar dikatakan bukan }\\ &\textrm{fungsi satu-satu (fungsi injektif)} \end{aligned}\\\hline 2.\textrm{d}&\textrm{Salah}&\begin{aligned}&\textrm{Jelas bukan fungsi pada }\\ &\textrm{dan satu-satu(fungsi bijektif)} \end{aligned}\\\hline 2.\textrm{e}&\textrm{Salah}&\textrm{Tidak sesuai}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Himpunan pasangan terurut }\\ &\textrm{yang ditunjukkan oleh fungsi}\\ &f:x \mapsto 2-\left ( x+1 \right )^{2}\\ &\textrm{dari domain}\: \: \left \{ -1,0,1,2 \right \}\: \: \textrm{adalah}.... \\ &\begin{array}{ll}\\ \textrm{a}.&\left \{ (-1,2),(0,3),(1,5),(2,7) \right \}\\ \color{red}\textrm{b}.&\left \{ (-1,2),(0,1),(1,-2),(2,-7) \right \}\\ \textrm{c}.&\left \{ (-1,1),(0,-1),(1,-4),(2,7) \right \}\\ \textrm{d}.&\left \{ (-1,0),(0,3),(1,-2),(2,7) \right \}\\ \textrm{e}.&\left \{ (-1,0),(0,-4),(1,5),(2,-7) \right \} \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}f:x& \mapsto 2-\left ( x+1 \right )^{2}\\ -1&\mapsto 2-\left ( -1+1 \right )^{2}=2-0=2&....(-1,2)\\ 0&\mapsto 2-\left ( 0+1 \right )^{2}=2-1=1&....(0,1)\\ 1&\mapsto 2-\left ( 1+1 \right )^{2}=2-4=-2&....(1,-2)\\ 2&\mapsto 2-\left ( 2+1 \right )^{2}=2-9=-7&....(2,-7) \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Dari beberapa fungsi berikut yang }\\ &\textrm{merupakan fungsi genap adalah}\: ....\\ &\begin{array}{ll}\\ \color{red}\textrm{a}.&f(x)=x^{2}+\left | x \right |-1\\ \textrm{b}.&f(x)=x^{3}-\left | x \right |+x\\ \textrm{c}.&f(x)=x\left | x \right |+x\\ \textrm{d}.&f(x)=\sqrt{x-1}\\ \textrm{e}.&f(x)=4-2x \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\\ \textrm{Suatu fungsi}&\: \color{red}\textrm{dinamakan fungsi genap}\\ &\textrm{jika}\: f(x)=f(-x)\\ & \end{aligned}\\ &\begin{array}{|c|l|l|c|}\hline \textrm{No}&\qquad f(x)&\qquad f(-x)&\textrm{Keterangan}\\\hline 4.\textrm{a}&x^{2}+\left | x \right |-1&x^{2}+\left | x \right |-1&f(x)=f(-x)\\\hline 4.\textrm{b}&x^{3}-\left | x \right |+x&-x^{3}-\left | x \right |-x&f(x)\neq f(-x)\\\hline 4.\textrm{c}&x\left | x \right |+x&-x\left | x \right |-x&f(x)\neq f(-x)\\\hline 4.\textrm{d}&\sqrt{x-1}&\sqrt{-x-1}&f(x)\neq f(-x)\\\hline 4.\textrm{e}.&4-2x&4+2x&f(x)\neq f(-x)\\\hline \end{array} \end{array}$

$\begin{array}{ll}\\ 5.&\textrm{Diketahui himpunan}\\ &A=\left \{ x|x\: \textrm{adalah faktor prima dari}\: 16 \right \}\\ &B=\left \{ x|x\: \textrm{adalah faktor dari}\: 16 \right \}\\ &\textrm{Banyaknya pemetaan dari}\: A\: ke\: B\: \textrm{adalah}.... \\ &\begin{array}{llllll}\\ \textrm{a}.&1&&&\textrm{d}.&25\\\\ \textrm{b}.&2 \quad &\color{red}\textrm{c}.&5&\textrm{e}.&32 \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{aligned}A&=\left \{ x|x\: \textrm{adalah faktor prima dari}\: 16 \right \}\\ &=\left \{ 2 \right \}\Rightarrow n(A)=1\\ B&=\left \{ x|x\: \textrm{adalah faktor dari}\: 16 \right \}\\ &=\left \{ 1,2,4,8,16 \right \} \Rightarrow n(B)=5\\ \textrm{B}&\textrm{anyaknya pemetaan dari}\: A\: \textrm{ke}\: B\: \textrm{adalah}:\\ &=n(B)^{n(A)}\\ &=5^{1}\\ &=5\end{aligned} \end{array}$

Lanjutan Materi Fungsi

$\color{blue}\textrm{F. Domain, Kodomain dan Range Fungsi}$

Suatu fungsi  $\color{red}f$ dari himpunan $\color{red}\textrm{A}$ ke himpunan $\color{red}\textrm{B}$  dituliskan dengan bentuk  $f:\textrm{A}\rightarrow \textrm{B}$. Jika fungsi  $\color{red}f$  memetakan  $\color{blue}x\color{black}\in \textrm{A}$  ke  $\color{blue}y\color{black}\in \textrm{B}$, maka dituliskan dengan  $f:x\rightarrow y$  atau  $f:x\rightarrow f(x)$.

Perhatikan gambar berikut sebagai ilustrasinya

• Himpunan  $\textrm{A}$  sebagai Domain/daerah asal/prapeta dari fungsi  $\color{red}f$
• Himpunan  $\textrm{B}$  sebagai Kodomain/daerah kawan dari fungsi  $\color{red}f$
• Himpunan semua bayangan (bagian dari peta) disebut sebagai Range/daerah hasil  dari fungsi  $\color{red}f$.

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

1. Perhatikanlah gambar berikut

dan tentukanlah domain, kodomain, serta range fungsinya

$\begin{aligned}\textrm{Dari}\: &\textrm{ilustrasi di atas diperoleh bahwa}:\\ &\textrm{Domain}\qquad :\quad \color{blue}D_{_{f}}\color{black}=A=\left \{ a,b,c,d \right \}\\ &\textrm{Kodomain}\: \: \: \: :\quad \color{red}K_{_{f}}\color{black}=B=\left \{ 1,2,3,4,5 \right \}\\ &\textrm{Range}\: \: \: \qquad :\quad R_{_{f}}=\left \{ 1,2,3,5 \right \}\subseteq B \end{aligned}$.

$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah domain dan range dari fungsi}\\ & f(x)=\sqrt{x^{2}-x}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \qquad\qquad\qquad\qquad\textrm{Domain}&\qquad\quad\textrm{Range}\\\hline \begin{aligned}&\textrm{Kumpulan nilai}\: \: x\\ &\textrm{yang mungkin, yaitu:}\\ &x^{2}-x\geq 0\\ &x(x-1)\geq 0\\ &\color{blue}\textrm{dengan garis bilangan}\\ &\begin{array}{llllllllll}\\ +&+&+&-&-&-&-&+&+&+\\\hline &&0&&&&&1&& \end{array}\\ &\textrm{Jadi},\: \color{red}D_{_{f}}\color{black}=\left \{ x|x\leq 0\: \textrm{atau}\: x\geq 1 ,\: \: x\in \mathbb{R}\right \} \end{aligned}&\begin{aligned}&\textrm{Hasil akar pangkat 2}\\ &\textrm{tidak pernah negatif}\\ &\textrm{Jadi},\: \color{red}R_{_{f}}\color{black}=\left \{ y|y\geq 0 \right \}\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}$.

$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah domain dari}\\ &\begin{array}{ll}\\ \textrm{a}.\quad f(x)=2x+3&\\ \textrm{b}.\quad f(x)=\displaystyle \frac{2}{3x-15}&\\ \textrm{c}.\quad g(x)=\displaystyle \frac{x-1}{x^{2}-x-6}&\\ \textrm{d}.\quad g(x)=\sqrt{x^{2}-1} &\\ \textrm{e}.\quad h(x)=\sqrt{3x+2}\\ \textrm{f}.\quad h(x)=\sqrt{\displaystyle \frac{x-1}{x^{2}-x-6}}\\ \textrm{g}.\quad k(x)=\: ^{^{2}}\log x^{2}-2x-15\\ \textrm{h}.\quad k(x)=\: ^{^{^{\textbf{(x+2)}}}}\log (x^{2}-2x-3)\end{array}\\\\ &\color{blue}\textrm{Jawab}: \end{array}.$

$. \qquad \begin{array}{|l|}\hline \begin{aligned}\textrm{a}.\quad f(x)&=2x+3\\ \color{red}D_{_{f}}&=\left \{ x|x\in \mathbb{R} \right \}\\  \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad f(x)&=\displaystyle \frac{2}{3x-15}\\ &\textrm{supaya terdefinisi}\\ &\textrm{maka},\\ &3x-15\neq 0\\ &x\neq 5,\: \: \textrm{sehingga}\\ \color{red}D_{_{f}}&=\left \{ x|x\neq 5,\: x\in \mathbb{R} \right \}\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad g(x)&=\displaystyle \frac{x-1}{x^{2}-x-6}\\ &\textrm{supaya terdefinisi}\\ &\textrm{maka},\\ &x^{2}-x-6\neq 0\\ &x\neq 3\: \textrm{dan}\: x\neq -2,\\ & \textrm{sehingga}\\ \color{red}D_{_{g}}&=\left \{ x|x\neq3\: \textrm{dan}\: x\neq -2,\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad g(x)&=\sqrt{x^{2}-1}\\ \textrm{ma}&\textrm{ka}\: \: x^{2}-1\geq 0\\ &(x+1)(x-1)\geq 0\\ \color{red}D_{_{g}}&=\left \{ x|x\leq -1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \}\\  \end{aligned}\\\hline \begin{aligned}\textrm{e}.\quad h(x)&=\sqrt{3x+2}\\ \textrm{ma}&\textrm{ka}\: \: 3x+2\geq 0\\ &\: \: x\geq -\displaystyle \frac{2}{3}\\ \color{red}D_{_{h}}&=\left \{ x|x\geq -\displaystyle \frac{2}{3},\: x\in \mathbb{R} \right \}\\  \end{aligned}\\\hline \begin{aligned}\textrm{f}.\quad h(x)&=\sqrt{\displaystyle \frac{x-1}{x^{2}-x-6}}\\ &=\sqrt{\displaystyle \frac{(x-1)}{(x-3)(x+2)}}\\ &=\sqrt{\displaystyle \frac{x-1}{(x-3)(x+2)}}\\ &\textrm{maka},\: \: \displaystyle \frac{x-1}{(x-3)(x+2)}\geq 0\\ \color{red}D_{_{h}}&=\left \{ x|-2<x\leq 1\: \textrm{atau}\: \: x>3,\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \end{array}$.

$. \qquad \begin{array}{|l|}\hline \begin{aligned}\textrm{g}.\quad k(x)&=\: ^{^{^{2}}}\log (x^{2}-2x-15)\\ \textrm{sya}&\textrm{rat}\: \: (x^{2}-2x-15)>0\\ &\: \: \: \: \: \: \: \: (x-5)(x+3)>0\\ \color{red}D_{_{k}}&=\left \{ x|x<-3\: \: \textrm{atau}\: \: x>5,\: \: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{h}.\quad k(x)&=\: ^{^{^{\textbf{(x+2)}}}}\log (x^{2}-2x-3)\\ \textrm{sya}&\textrm{rat}\: \: 1.\: \begin{cases} (x+2) & >0\Rightarrow x>-2\\ (x+2) & \neq 0 \Rightarrow x\neq -2 \end{cases}\\ &\qquad 2.\: \: (x^{2}-2x-3)>0\Rightarrow (x-3)(x+1)>0\\ \color{red}D_{_{k}}&=\left \{ x|-2< x< -1\: \: \textrm{atau}\: \: x>3,\: \: x\in \mathbb{R}\right \}\end{aligned}\\\hline \end{array}$.

$\begin{array}{ll}\\ 4.&\textrm{Diketahui bahwa 2 buah fungsi}\\ &f(x)=2x+1\: \: \textrm{dan}\: \: g(x)=\sqrt{1-x}\\ &\begin{array}{ll}\\ \textrm{a}.\quad (f+g)(x)&\textrm{c}.\quad (f.g)(x)\\ \textrm{b}.\quad (f-g)(x)&\textrm{d}.\quad \left ( \displaystyle \frac{f}{g} \right )(x)\\ \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad (f+g)(x)&=f(x)+g(x)\\ &=(2x+1)+\sqrt{1-x}\\ D_{_{(f+g)}}&=\left \{ x|x\leq 1,\: \: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad (f-g)(x)&=f(x)-g(x)\\ &=(2x+1)-\sqrt{1-x}\\ D_{_{(f-g)}}&=\left \{ x|x\leq 1,\: \: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad (f.g)(x)&=f(x).g(x)\\ &=(2x+1)\sqrt{1-x}\\ &=\sqrt{(2x+1)^{2}(1-x)}\\ &\: \: \: \: \: \: (2x+1)^{2}(1-x)\geq 0\\ D_{_{(f.g)}}&=\left \{ x|x\leq 1,\: \: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad \left ( \displaystyle \frac{f}{g} \right )(x)&=....................\\ &....................\\ &.................... \end{aligned}\\\hline \end{array} \end{array}$

DAFTAR PUSTAKA

1. Sodyarto. Nugroho, Maryanto. 2008. Matematika 2 untuk SMA dan MA Kelas XI Program IPA. Jakarta: Pusat Perbukuan Departemen Pendidikan Nasional.
2. Sunardi, Waluyo, S., Sutrisno, & Subagya. 2005. Matematika 2 untuk SMA Kelas 2 IPA. Jakarta: BUMI AKSARA.

Fungsi (Matematika Wajib Kelas X)

$\color{blue}\textrm{A. Pendahuluan}$

$\begin{aligned}&\\\textrm{Fungsi}\: &\textrm{atau pemetaan dari A ke B adalah}\\ &\textrm{suatu relasi khusus yang memasangkan}\\ &\textrm{setiap} \: \: x\in A\: \: \textrm{ke tepat satu}\: \: y\in B.\\ & \end{aligned}$

$\color{red}\begin{array}{|l|l|}\hline \textrm{Notasi}&\color{black}f:x \rightarrow y\: \: \: \color{red}\textrm{atau}\: \: \: \color{black}f:x \rightarrow f(x) \\\hline \textrm{Dibaca}&\textrm{fungsi}\: \: \color{black}f\: \: \color{red}\textrm{memetakan}\: \: \color{black}x\in A\: \: \color{red}\textrm{ke}\: \: \color{black}y\in B\\\hline \: \: \: \: \color{black}A&\textrm{Domain atau daerah asal fungsi atau}\quad \color{black}D_{f}\\\hline \: \: \: \: \, \color{black}x&\textrm{prapeta(sebelum dipetakan)}\\\hline \: \: \: \: \color{black}B&\textrm{Kodomain atau daerah kawan fungsi atau}\quad \color{black}K_{f}\\\hline \: \: \: \: \, \color{black}y&\textrm{peta(bayangan dari prapeta) adalah Range}\\ &\textrm{atau}\quad \color{black}R_{f}\\\hline \end{array}$

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

$\LARGE\colorbox{yellow}{CONTOH SOAL}$

$\begin{array}{ll}\\ 1.&\textrm{Gambarlah grafik fungsi}\\ &\textrm{a}.\quad f(x)=2x+5\\ &\textrm{b}.\quad g(x)=2x^{2}\\ &\textrm{c}.\quad h(x)=\displaystyle \frac{1}{x}\\\\ &\color{blue}\textrm{Jawab}:\\ & \end{array}$.

$.\qquad\begin{aligned}&\textrm{a.  Menggambar grafik}\: \: f(x)=2x+5\\ &\begin{array}{|c|c|c|}\hline x&y=f(x)=2x+5&\textrm{Titik}\: (x,y)\\\hline -3&f(-3)=2(-3)+5=-1&(-3,-1)\\\hline -2&f(-2)=2(-2)+5=3&(-2,1)\\\hline -1&f(-1)=2(-1)+5=3&(-1,3)\\\hline 0&f(0)=2(0)+5=5&(0,5)\\\hline 1&f(1)=2(1)+5=7&(1,7)\\\hline 2&f(2)=2(2)+5=9&(2,9)\\\hline 3&f(3)=2(3)+5=11&(3,11)\\\hline \end{array} \end{aligned}$

$.\qquad\begin{aligned}&\textrm{b. Menggambar grafik}\: \: f(x)=2x^{2}\\ &\begin{array}{|c|c|c|}\hline x&y=f(x)=2x^{2}&\textrm{Titik}\: (x,y)\\\hline -3&f(-3)=2(-3)^{2}=18&(-3,18)\\\hline -2&f(-2)=2(-2)^{2}=8&(-2,8)\\\hline -1&f(-1)=2(-1)^{2}=2&(-1,2)\\\hline 0&f(0)=2(0)^{2}=0&(0,0)\\\hline 1&f(1)=2(1)^{2}=2&(1,2)\\\hline 2&f(2)=2(2)^{2}=8&(2,8)\\\hline 3&f(3)=2(3)^{2}=18&(3,18)\\\hline \end{array} \end{aligned}$

$.\qquad\begin{aligned}&\textrm{c. Menggambar grafik}\: \: f(x)=\displaystyle \frac{1}{x}\\ &\begin{array}{|c|c|c|}\hline x&y=f(x)=\displaystyle \frac{1}{x}&\textrm{Titik}\: (x,y)\\\hline -3&f(-3)=-\displaystyle \frac{1}{3}&(-3,-\displaystyle \frac{1}{3})\\\hline -2&f(-2)=-\displaystyle \frac{1}{2}&(-2,-\displaystyle \frac{1}{2})\\\hline -1&f(-1)=-1&(-1,-1)\\\hline 0&f(0)\: \: \color{blue}\textrm{tidak ada}&-\\\hline 1&f(1)=1&(1,1)\\\hline 2&f(2)=\displaystyle \frac{1}{2}&(2,\displaystyle \frac{1}{2})\\\hline 3&f(3)=\displaystyle \frac{1}{3}&(3,\displaystyle \frac{1}{3})\\\hline \end{array} \end{aligned}$

$\begin{array}{ll}\\ 2.&\textrm{Gambarlah grafik fungsi}\\ &\textrm{a}.\quad f(x)=\left \lfloor x \right \rfloor\\ &\textrm{b}.\quad g(x)=\left \lfloor x+1 \right \rfloor\\ &\textrm{c}.\quad h(x)=\left \{ x \right \}\\\\ &\textbf{Catatan}:\\ &\left \lfloor x \right \rfloor=\textrm{bilangan bulat terbesar tetapi}\\ &\: \: \: \, \quad\quad \textrm{lebih kecil atau sama dengan}\: \: x\\ &\left \{ x \right \}=\textrm{bagian pecahan dari}\: \: x\\\\ &\color{blue}\textrm{Jawab}:\\ & \end{array}$.

$.\qquad\begin{aligned}&\textrm{a. Menggambar grafik}\: \: f(x)=\left \lfloor x \right \rfloor\\ &\begin{array}{|c|c|c|}\hline x&y=f(x)=\left \lfloor x \right \rfloor&\textrm{Titik}\: (x,y)\\\hline \vdots &\vdots &\vdots \\\hline -1&f(-1)=\left \lfloor -1 \right \rfloor=-1&(-1,-1)\\\hline \vdots &\vdots &\vdots \\\hline -\displaystyle \frac{1}{2}&f(-\displaystyle \frac{1}{2})=\left \lfloor -\displaystyle \frac{1}{2} \right \rfloor=-1&(-\displaystyle \frac{1}{2},-1)\\\hline -\displaystyle \frac{1}{3}&f(-\displaystyle \frac{1}{3})=\left \lfloor -\displaystyle \frac{1}{3} \right \rfloor=-1&(-\displaystyle \frac{1}{3},-1)\\\hline -\displaystyle \frac{1}{4}&f(-\displaystyle \frac{1}{4})=\left \lfloor -\displaystyle \frac{1}{4} \right \rfloor=-1&(\displaystyle \frac{1}{4},-1)\\\hline \vdots &\vdots &\vdots \\\hline 0&f(0)=\left \lfloor 0 \right \rfloor=0&(0,0)\\\hline \vdots &\vdots &\vdots \\\hline \displaystyle \frac{1}{4}&f(\displaystyle \frac{1}{4})=\left \lfloor \displaystyle \frac{1}{4} \right \rfloor=0&(\displaystyle \frac{1}{4},0)\\\hline \displaystyle \frac{1}{3}&f(\displaystyle \frac{1}{3})=\left \lfloor \displaystyle \frac{1}{3} \right \rfloor=0&(\displaystyle \frac{1}{3},0)\\\hline \displaystyle \frac{1}{2}&f(\displaystyle \frac{1}{2})=\left \lfloor \displaystyle \frac{1}{2} \right \rfloor=0&(-\displaystyle \frac{1}{2},0)\\\hline \vdots &\vdots &\vdots \\\hline 1&f(1)=\left \lfloor 1 \right \rfloor=1&(1,1)\\\hline \vdots &\vdots &\vdots \\\hline 2\displaystyle \frac{1}{4}&f(2\displaystyle \frac{1}{4})=\left \lfloor 2\displaystyle \frac{1}{4} \right \rfloor=2&(2\displaystyle \frac{1}{4},2)\\\hline \vdots &\vdots &\vdots \\\hline 4\displaystyle \frac{1}{3}&f(4\displaystyle \frac{1}{3})=\left \lfloor 4\displaystyle \frac{1}{3} \right \rfloor=4&(4\displaystyle \frac{1}{3},4)\\\hline \vdots &\vdots &\vdots \\\hline \end{array} \end{aligned}$

$\begin{array}{ll}\\ 3.&\textbf{(OSK 2003)}\textrm{Jika}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{adalah bilangan real }\\ &\textrm{sedemikian sehingga}\: \: \displaystyle \left \lfloor \sqrt{x} \right \rfloor=9\: \: \textrm{dan}\: \: \displaystyle \left \lfloor \sqrt{y} \right \rfloor=12\: ,\\ &\textrm{maka nilai terkecil dari}\: \: \displaystyle \left \lfloor y-x \right \rfloor=....\\\\ &\textrm{Jawab}:\\ & \end{array}$.

$.\qquad \begin{aligned}&\textrm{Diketahui bahwa}\: \: \left \lfloor x \right \rfloor\: \: \textrm{adalah bilangan bulat terbesar }\\ &\textrm{yang lebih kecil atau sama dengan}\: \: x\\ &\textrm{Misal}\: \: \left \lfloor 3,2 \right \rfloor=3, \: \left \lfloor -2,47 \right \rfloor=-3,\: \: \left \lfloor 6 \right \rfloor=6,\: \: \textrm{dan lain-lain}.\\ &\textrm{Sehingga}\, \,\left \lfloor x \right \rfloor=a\Rightarrow a\leq x< a+1\: \: (\textrm{dengan}\: \: a\: \: \textrm{bilangan bulat}),\\ & \textrm{maka}\\ &\begin{cases} \left \lfloor \sqrt{x} \right \rfloor=9 & \Rightarrow 9\leq \sqrt{x}< 9+1\\ &\Leftrightarrow 9\leq \sqrt{x}< 10\Leftrightarrow \color{red}81\leq x< 100 \\ \left \lfloor \sqrt{y} \right \rfloor=12 & 12\leq \sqrt{y}< 12+1\\ &\Leftrightarrow 12\leq \sqrt{y}< 13\Leftrightarrow \color{red}144\leq y< 169 \end{cases}\\  \end{aligned}$.

$.\qquad \begin{aligned}&\bullet \: \: 144\leq y< 169\: \: \textrm{dan}\\ &\bullet \: \: \: \: 81\leq x< 100\: ,\: \textrm{dikalikan dengan}\: \: (-1)\\ &\quad \textrm{maka akan menjadi},\\ &\qquad -100< -x\leq -81,\\ &\textrm{sehingga}\: \: -99,\overline{999}\leq -x< -80,\overline{999}\\ &\textrm{Selanjutnya}\\ &\begin{array}{cll}\\ 144&\leq y< 169&\\ -99,\overline{999}&\leq -x< -80,\overline{999}&+\\\hline\\ 44,...&\leq y-x< 88,...\\ \end{array}\\ &\textrm{Jadi, nilai terkecil}\: \left \lfloor y-x \right \rfloor=\color{red}44 \end{aligned}$