Lanjutan Materi Fungsi

$\text{F. Domain, Kodomain dan Range Fungsi}$

Suatu fungsi  $f$ dari himpunan $\text{A}$ ke himpunan $\text{B}$  dituliskan dengan bentuk  $f:\text{A}\to \text{B}$. Jika fungsi  $f$  memetakan  $x\in \text{A}$  ke  $y\in \text{B}$, maka dituliskan dengan  $f:x\to y$  atau  $f:x\to f(x)$.

Perhatikan gambar berikut sebagai ilustrasinya

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

$\text{CONTOH SOAL}$

1. Perhatikanlah gambar berikut

dan tentukanlah domain, kodomain, serta range fungsinya

$\begin{array}{rl}\text{Dari}& \text{ilustrasi di atas diperoleh bahwa}:\\ & \text{Domain}:D_{{}_f}=A=\{a,b,c,d\}\\ & \text{Kodomain}:K_{{}_f}=B=\{1,2,3,4,5\}\\ & \text{Range}:R_{{}_f}=\{1,2,3,5\}\subseteq B\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah domain dan range dari fungsi}\\ & f(x)=\sqrt{x^2-x}\\ \\ & \text{Jawab}:\\ & \begin{array}{|ll|}\hline \text{Domain}& \text{Range}\\ \begin{array}{rl}& \text{Kumpulan nilai}x\\ & \text{yang mungkin, yaitu:}\\ & x^2-x\ge 0\\ & x(x-1)\ge 0\\ & \text{dengan garis bilangan}\\ & \begin{array}{l}\\ +& +& +& -& -& -& -& +& +& +\\ & & 0& & & & & 1& & \end{array}\\ & \text{Jadi},D_{{}_f}=\{x|x\le 0\text{atau}x\ge 1,x\in \mathbb{R}\}\end{array}& \begin{array}{rl}& \text{Hasil akar pangkat 2}\\ & \text{tidak pernah negatif}\\ & \text{Jadi},R_{{}_f}=\{y|y\ge 0\}\\ & \\ & \\ & \\ & \\ & \\ & \end{array}\\ \hline\end{array}\end{array}$.

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

$.\begin{array}{|l|}\hline \begin{array}{rl}\text{a}.f(x)& =2x+3\\ D_{{}_f}& =\{x|x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{b}.f(x)& ={\displaystyle \frac{2}{3x-15}}\\ & \text{supaya terdefinisi}\\ & \text{maka},\\ & 3x-15\ne 0\\ & x\ne 5,\text{sehingga}\\ D_{{}_f}& =\{x|x\ne 5,x\in \mathbb{R}\}\\ & \end{array}\\ \begin{array}{rl}\text{c}.g(x)& ={\displaystyle \frac{x-1}{x^2-x-6}}\\ & \text{supaya terdefinisi}\\ & \text{maka},\\ & x^2-x-6\ne 0\\ & x\ne 3\text{dan}x\ne -2,\\ & \text{sehingga}\\ D_{{}_g}& =\{x|x\ne 3\text{dan}x\ne -2,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{d}.g(x)& =\sqrt{x^2-1}\\ \text{ma}& \text{ka}{x}^2-1\ge 0\\ & (x+1)(x-1)\ge 0\\ D_{{}_g}& =\{x|x\le -1\text{atau}x\ge 1,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{e}.h(x)& =\sqrt{3x+2}\\ \text{ma}& \text{ka}3x+2\ge 0\\ & x\ge -{\displaystyle \frac{2}{3}}\\ D_{{}_h}& =\{x|x\ge -{\displaystyle \frac{2}{3},x\in \mathbb{R}}\}\end{array}\\ \begin{array}{rl}\text{f}.h(x)& =\sqrt{{\displaystyle \frac{x-1}{x^2-x-6}}}\\ & =\sqrt{{\displaystyle \frac{(x-1)}{(x-3)(x+2)}}}\\ & =\sqrt{{\displaystyle \frac{x-1}{(x-3)(x+2)}}}\\ & \text{maka},{\displaystyle \frac{x-1}{(x-3)(x+2)}\ge 0}\\ D_{{}_h}& =\{x|-2<x\le 1\text{atau}x>3,x\in \mathbb{R}\}\end{array}\\ \hline\end{array}$.

$.\begin{array}{|l|}\hline \begin{array}{rl}\text{g}.k(x)& ={}^{{}^{{}^2}}\log (x^2-2x-15)\\ \text{sya}& \text{rat}(x^2-2x-15)>0\\ & (x-5)(x+3)>0\\ D_{{}_k}& =\{x|x<-3\text{atau}x>5,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{h}.k(x)& ={}^{{}^{{}^{\mathbf{\text{(x+2)}}}}}\log (x^2-2x-3)\\ \text{sya}& \text{rat}1.\{\begin{array}{ll}(x+2)& >0\Rightarrow x>-2\\ (x+2)& \ne 0\Rightarrow x\ne -2\end{array}\\ & 2.(x^2-2x-3)>0\Rightarrow (x-3)(x+1)>0\\ D_{{}_k}& =\{x|-2<x<-1\text{atau}x>3,x\in \mathbb{R}\}\end{array}\\ \hline\end{array}$.

$\begin{array}{l}\\ 4.& \text{Diketahui bahwa 2 buah fungsi}\\ & f(x)=2x+1\text{dan}g(x)=\sqrt{1-x}\\ & \begin{array}{l}\\ \text{a}.(f+g)(x)& \text{c}.(f.g)(x)\\ \text{b}.(f-g)(x)& \text{d}.\left({\displaystyle \frac{f}{g}}\right)(x)\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{|l|}\hline \begin{array}{rl}\text{a}.(f+g)(x)& =f(x)+g(x)\\ & =(2x+1)+\sqrt{1-x}\\ D_{{}_{(f+g)}}& =\{x|x\le 1,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{b}.(f-g)(x)& =f(x)-g(x)\\ & =(2x+1)-\sqrt{1-x}\\ D_{{}_{(f-g)}}& =\{x|x\le 1,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{c}.(f.g)(x)& =f(x).g(x)\\ & =(2x+1)\sqrt{1-x}\\ & =\sqrt{(2x+1{)}^2(1-x)}\\ & (2x+1{)}^2(1-x)\ge 0\\ D_{{}_{(f.g)}}& =\{x|x\le 1,x\in \mathbb{R}\}\end{array}\\ \begin{array}{rl}\text{d}.\left({\displaystyle \frac{f}{g}}\right)(x)& =....................\\ & ....................\\ & ....................\end{array}\\ \hline\end{array}\end{array}$

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.