$\color{blue}\textrm{F. Domain, Kodomain dan Range Fungsi}$
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 $f:x\rightarrow f(x)$.
- 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 $\color{red}f$.
$\begin{aligned}\textrm{Dari}\: &\textrm{ilustrasi di atas diperoleh bahwa}:\\ &\textrm{Domain}\qquad :\quad \color{blue}D_{_{f}}\color{black}=A=\left \{ a,b,c,d \right \}\\ &\textrm{Kodomain}\: \: \: \: :\quad \color{red}K_{_{f}}\color{black}=B=\left \{ 1,2,3,4,5 \right \}\\ &\textrm{Range}\: \: \: \qquad :\quad R_{_{f}}=\left \{ 1,2,3,5 \right \}\subseteq B \end{aligned}$.
$\begin{array}{ll}\\ 2.&\textrm{Tentukanlah domain dan range dari fungsi}\\ & f(x)=\sqrt{x^{2}-x}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \qquad\qquad\qquad\qquad\textrm{Domain}&\qquad\quad\textrm{Range}\\\hline \begin{aligned}&\textrm{Kumpulan nilai}\: \: x\\ &\textrm{yang mungkin, yaitu:}\\ &x^{2}-x\geq 0\\ &x(x-1)\geq 0\\ &\color{blue}\textrm{dengan garis bilangan}\\ &\begin{array}{llllllllll}\\ +&+&+&-&-&-&-&+&+&+\\\hline &&0&&&&&1&& \end{array}\\ &\textrm{Jadi},\: \color{red}D_{_{f}}\color{black}=\left \{ x|x\leq 0\: \textrm{atau}\: x\geq 1 ,\: \: x\in \mathbb{R}\right \} \end{aligned}&\begin{aligned}&\textrm{Hasil akar pangkat 2}\\ &\textrm{tidak pernah negatif}\\ &\textrm{Jadi},\: \color{red}R_{_{f}}\color{black}=\left \{ y|y\geq 0 \right \}\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}$.
$\begin{array}{ll}\\ 3.&\textrm{Tentukanlah domain dari}\\ &\begin{array}{ll}\\ \textrm{a}.\quad f(x)=2x+3&\\ \textrm{b}.\quad f(x)=\displaystyle \frac{2}{3x-15}&\\ \textrm{c}.\quad g(x)=\displaystyle \frac{x-1}{x^{2}-x-6}&\\ \textrm{d}.\quad g(x)=\sqrt{x^{2}-1} &\\ \textrm{e}.\quad h(x)=\sqrt{3x+2}\\ \textrm{f}.\quad h(x)=\sqrt{\displaystyle \frac{x-1}{x^{2}-x-6}}\\ \textrm{g}.\quad k(x)=\: ^{^{2}}\log x^{2}-2x-15\\ \textrm{h}.\quad k(x)=\: ^{^{^{\textbf{(x+2)}}}}\log (x^{2}-2x-3)\end{array}\\\\ &\color{blue}\textrm{Jawab}: \end{array}.$
$. \qquad \begin{array}{|l|}\hline \begin{aligned}\textrm{a}.\quad f(x)&=2x+3\\ \color{red}D_{_{f}}&=\left \{ x|x\in \mathbb{R} \right \}\\ \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad f(x)&=\displaystyle \frac{2}{3x-15}\\ &\textrm{supaya terdefinisi}\\ &\textrm{maka},\\ &3x-15\neq 0\\ &x\neq 5,\: \: \textrm{sehingga}\\ \color{red}D_{_{f}}&=\left \{ x|x\neq 5,\: x\in \mathbb{R} \right \}\\ & \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad g(x)&=\displaystyle \frac{x-1}{x^{2}-x-6}\\ &\textrm{supaya terdefinisi}\\ &\textrm{maka},\\ &x^{2}-x-6\neq 0\\ &x\neq 3\: \textrm{dan}\: x\neq -2,\\ & \textrm{sehingga}\\ \color{red}D_{_{g}}&=\left \{ x|x\neq3\: \textrm{dan}\: x\neq -2,\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad g(x)&=\sqrt{x^{2}-1}\\ \textrm{ma}&\textrm{ka}\: \: x^{2}-1\geq 0\\ &(x+1)(x-1)\geq 0\\ \color{red}D_{_{g}}&=\left \{ x|x\leq -1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \}\\ \end{aligned}\\\hline \begin{aligned}\textrm{e}.\quad h(x)&=\sqrt{3x+2}\\ \textrm{ma}&\textrm{ka}\: \: 3x+2\geq 0\\ &\: \: x\geq -\displaystyle \frac{2}{3}\\ \color{red}D_{_{h}}&=\left \{ x|x\geq -\displaystyle \frac{2}{3},\: x\in \mathbb{R} \right \}\\ \end{aligned}\\\hline \begin{aligned}\textrm{f}.\quad h(x)&=\sqrt{\displaystyle \frac{x-1}{x^{2}-x-6}}\\ &=\sqrt{\displaystyle \frac{(x-1)}{(x-3)(x+2)}}\\ &=\sqrt{\displaystyle \frac{x-1}{(x-3)(x+2)}}\\ &\textrm{maka},\: \: \displaystyle \frac{x-1}{(x-3)(x+2)}\geq 0\\ \color{red}D_{_{h}}&=\left \{ x|-2<x\leq 1\: \textrm{atau}\: \: x>3,\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \end{array}$.
$. \qquad \begin{array}{|l|}\hline \begin{aligned}\textrm{g}.\quad k(x)&=\: ^{^{^{2}}}\log (x^{2}-2x-15)\\ \textrm{sya}&\textrm{rat}\: \: (x^{2}-2x-15)>0\\ &\: \: \: \: \: \: \: \: (x-5)(x+3)>0\\ \color{red}D_{_{k}}&=\left \{ x|x<-3\: \: \textrm{atau}\: \: x>5,\: \: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{h}.\quad k(x)&=\: ^{^{^{\textbf{(x+2)}}}}\log (x^{2}-2x-3)\\ \textrm{sya}&\textrm{rat}\: \: 1.\: \begin{cases} (x+2) & >0\Rightarrow x>-2\\ (x+2) & \neq 0 \Rightarrow x\neq -2 \end{cases}\\ &\qquad 2.\: \: (x^{2}-2x-3)>0\Rightarrow (x-3)(x+1)>0\\ \color{red}D_{_{k}}&=\left \{ x|-2< x< -1\: \: \textrm{atau}\: \: x>3,\: \: x\in \mathbb{R}\right \}\end{aligned}\\\hline \end{array}$.
$\begin{array}{ll}\\ 4.&\textrm{Diketahui bahwa 2 buah fungsi}\\ &f(x)=2x+1\: \: \textrm{dan}\: \: g(x)=\sqrt{1-x}\\ &\begin{array}{ll}\\ \textrm{a}.\quad (f+g)(x)&\textrm{c}.\quad (f.g)(x)\\ \textrm{b}.\quad (f-g)(x)&\textrm{d}.\quad \left ( \displaystyle \frac{f}{g} \right )(x)\\ \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{array}{|l|l|}\hline \begin{aligned}\textrm{a}.\quad (f+g)(x)&=f(x)+g(x)\\ &=(2x+1)+\sqrt{1-x}\\ D_{_{(f+g)}}&=\left \{ x|x\leq 1,\: \: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{b}.\quad (f-g)(x)&=f(x)-g(x)\\ &=(2x+1)-\sqrt{1-x}\\ D_{_{(f-g)}}&=\left \{ x|x\leq 1,\: \: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{c}.\quad (f.g)(x)&=f(x).g(x)\\ &=(2x+1)\sqrt{1-x}\\ &=\sqrt{(2x+1)^{2}(1-x)}\\ &\: \: \: \: \: \: (2x+1)^{2}(1-x)\geq 0\\ D_{_{(f.g)}}&=\left \{ x|x\leq 1,\: \: x\in \mathbb{R} \right \} \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad \left ( \displaystyle \frac{f}{g} \right )(x)&=....................\\ &....................\\ &.................... \end{aligned}\\\hline \end{array} \end{array}$
DAFTAR PUSTAKA
- Sodyarto. Nugroho, Maryanto. 2008. Matematika 2 untuk SMA dan MA Kelas XI Program IPA. Jakarta: Pusat Perbukuan Departemen Pendidikan Nasional.
- Sunardi, Waluyo, S., Sutrisno, & Subagya. 2005. Matematika 2 untuk SMA Kelas 2 IPA. Jakarta: BUMI AKSARA.
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