Fungsi (Matematika Wajib Kelas X)

$\text{A. Pendahuluan}$

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

$\begin{array}{|ll|}\hline \text{Notasi}& f:x\to y\text{atau}f:x\to f(x)\\ \text{Dibaca}& \text{fungsi}f\text{memetakan}x\in A\text{ke}y\in B\\ A& \text{Domain atau daerah asal fungsi atau}{D}_f\\ x& \text{prapeta(sebelum dipetakan)}\\ B& \text{Kodomain atau daerah kawan fungsi atau}{K}_f\\ y& \text{peta(bayangan dari prapeta) adalah Range}\\ & \text{atau}{R}_f\\ \hline\end{array}$

Sebagai ilustrasi perhatikanlah gambar berikut!

Sebagai misal, diberikan 

$\begin{array}{rl}& \boxed{{\displaystyle f:x\to f(x)=3x+2}}\\ & \mathit{\text{dibaca}}:\\ & \text{sebuah fungsi}f\text{memetakan}xke3x+2\end{array}$

$\text{B. Sifat-Sifat Fungsi}$

$\begin{array}{|ccl|}\hline \text{Injektif}& \text{Surjektif}& \text{Bijektif}\\ (\text{satu-satu})& (\text{pada})& (\text{korespondensi satu-satu})\\ \begin{array}{rl}& \text{Jika setiap anggota}\\ & \text{himpunan A memiliki}\\ & \text{bayangan berbeda di}\\ & \text{himpunan B}\end{array}& \begin{array}{rl}& \text{Jika setiap anggota}\\ & \text{himpunan di B}\\ & \text{mempunyai prapeta}\\ & \text{di himpunan A}\end{array}& \begin{array}{rl}& \text{Jika fungsi yang injektif}\\ & \text{sekaligus juga surjektif}\\ & \\ & \end{array}\\ \hline\end{array}$

$\text{C. Operasi Aljabar Fungsi}$

$\begin{array}{|ll|}\hline \text{Aljabar Fungsi}& \text{Daerah Asal}\\ (f+g)(x)=f(x)+g(x)& D_{{}_{(f+g)}}=D_{{}_f}\cap D_{{}_g}\\ (f-g)(x)=f(x)-g(x)& D_{{}_{(f-g)}}=D_{{}_f}\cap D_{{}_g}\\ (f.g)(x)=f(x).g(x)& D_{{}_{(f.g)}}=D_{{}_f}\cap D_{{}_g}\\ \left({\displaystyle \frac{f}{g}}\right)(x)={\displaystyle \frac{f(x)}{g(x)}}& D_{{}_{\left(\frac{f}{g}\right)}}=D_{{}_f}\cap D_{{}_g},\\ & \text{dengan}g(x)\ne 0\\ \hline\end{array}$

$\text{D. Macam-Macam Fungsi}$

$\begin{array}{|ll|}\hline \text{Fungsi Konstan}& \text{Berupa konstanta}\\ & f(x)=c\\ \text{Fungsi Identitas}& \text{Nilainya dirinya sendiri}\\ & f(x)=x\\ \text{Fungsi linear}& \text{Fungsi berupa garis lurus}\\ & f(x)=ax+b\\ \text{Fungsi Kuadrat}& \text{Fungsi Kuadrat/parabola}\\ & f(x)=a{x}^2+bx+c,a\ne 0\\ \text{Fungsi Rasional}& \text{Fungsi Pecahan}\\ & f(x)={\displaystyle \frac{p(x)}{q(x)}}\\ \text{Fungsi Khusus 1}& \text{Fungsi Modulus(nilai mutlak)}\\ & f(x)=\left|x\right|\\ \text{Fungsi Khusus 2}& \text{Fungsi tangga}\\ & f(x)=\lfloor x\rfloor \\ \text{Fungsi Khusus 3}& \text{Fungsi genap dan ganjil}\\ & \{\begin{array}{ll}\text{Fungsi ganjil}& f(-x)=-f(x)\\ \text{Fungsi genap}& f(-x)=f(x)\end{array}\\ \hline\end{array}$

$\text{CONTOH SOAL}$

$\begin{array}{l}\\ 1.& \text{Diketahui 2 humpuan sebagai berikut}:\\ & \{\begin{array}{ll}\text{P}& =\{-2,-1,0,1,2\}\\ \text{Q}& =\{0,1,2,5,7\}\end{array}\\ & \text{Di antara relasi dari P ke Q berikut manakah }\\ & \text{yang merupakan fungsi}\\ & \text{a}.\text{A}=\{(-2,0),(-1,0),(0,0),(1,0),(2,0)\}\\ & \text{b}.\text{B}=\{(-2,1),(-1,2),(0,5),(1,7),(-2,2)\}\\ & \text{c}.\text{C}=\{(-2,0),(-1,1),(0,2),(1,5),(2,7)\}\\ \\ & \text{Jawab}:\\ & \text{Semuanya Fungsi kecuali}\mathbf{\text{ poin b)}}\end{array}$

$\begin{array}{l}\\ 2.& \text{Relasi berikut yang merupakan fungsi adalah}....\end{array}$

$.\begin{array}{|cll|}\hline \text{Poin}& \text{Jenis}& \text{Keterangan}\\ \text{a}& \text{Fungsi}& \begin{array}{rl}& \text{Sesuai definisi}\end{array}\\ & & \begin{array}{rl}& \text{yaitu}:\\ & \text{Setiap prepeta}\\ & \text{(anggota himpunan A) }\\ & \text{memiliki peta di }\\ & \text{himpunan B tepat satu}.\\ & \text{Tetapi}\\ & \text{bukan fungsi injektif}\\ & \text{bukan pula fungsi}\\ & \text{surjektif}\end{array}\\ \text{b}& \text{Fungsi}& \text{Sama di atas}\\ \text{c}& \text{Bukan Fungsi}& \text{Tidak sesuai definisi}\\ & & \text{hanya relasi saja}\\ \text{d}& \text{Fungsi}& \text{Sesuai definisi}\\ & & (\text{Fungsi bijektif})\\ \hline\end{array}$

$\begin{array}{l}\\ 3.& \text{Tentukanlah daerah asal dari fungsi beberapa berikut}:\\ & \begin{array}{l}\\ \text{a}.& f(x)=x-3& \text{g}.& f(x)={\displaystyle \frac{\left|x\right|}{x}}\\ \text{b}.& f(x)={\displaystyle \frac{6}{x^2-2x-8}}& \text{h}.& f(x)=\lfloor x\rfloor \\ \text{c}.& f(x)={\displaystyle \frac{x^2-3x}{x^2-2x-15}}& \text{i}.& f(x)=\left|x\right|+\lfloor x\rfloor \\ \text{d}.& y+2=x^2-5x+5& \text{j}.& f(x)=\sqrt{x^2-16}\\ \text{e}.& f(x)=|x-3|& \text{k}.& f(x)=\sqrt{2x^2-50}\\ \text{f}.& f(x)=3-|2x-1|& \text{l}.& f(x)={\displaystyle \frac{2\sqrt{x}}{x-3}}\end{array}\\ \end{array}$.

$.\begin{array}{rl}& \text{catatan}\lfloor x\rfloor \text{adalah }\\ & \text{bulat terbesar atau sama dengan}x\end{array}$.

$.\text{Jawab}$.

$.\begin{array}{|cc|}\hline (\text{a})& (\text{b})\\ \begin{array}{rl}f(x)& =x-3\\ \text{selu}& \text{ruh bilangan real}\\ & x\text{akan terdefinisi}\\ & \text{atau tetap bernilai}\\ & \text{real}\\ \text{sehi}& \text{ngga},\\ D_f& =\{x|x\in \mathbb{R}\}\\ & \end{array}& \begin{array}{rl}f(x)& ={\displaystyle \frac{6}{x^2-2x-8}}\\ \text{terd}& \text{efinisi ketika}\\ & \text{penyebut tidak sama}\\ & \text{dengan}0,\text{yaitu}:\\ & \begin{array}{rl}{x}^2-2x-8& \ne 0\\ (x-4)(x+2)& \ne 0\\ x\ne 4\text{dan}x& \ne -2\end{array}\\ D_f& =\{x|x\in \mathbb{R},x\ne 4\text{dan}x\ne -2\}\end{array}\\ \hline\end{array}$

$.\begin{array}{|cc|}\hline (\text{d})& (\text{e})\\ \begin{array}{rl}& y+2=x^2-5x+5\\ & y=x^2-5x+5-2\\ & f(x)=x^2-5x+3\\ & \text{Sehingga daerah}\\ & \text{asalnya}\\ & D_f=\{x|x\in \mathbb{R}\}\\ & \end{array}& \begin{array}{rl}f(x)& =|x-3|\\ D_f& =\{x|x\in \mathbb{R}\}\\ \text{teta}& \text{pi pada textit{range} fungsinya}\\ & \text{hanya akan berupa}\\ & \text{bilangan positif saja}.\\ \text{yait}& \text{u}:\\ R_f& =\{y|y\in \mathbb{R},y\ge 0\}\\ & \end{array}\\ \hline\end{array}$

$.\begin{array}{|cc|}\hline (\text{g})& (\text{h})\\ \begin{array}{rl}& f(x)={\displaystyle \frac{\lfloor x\rfloor }{x}}\\ & \text{Sehingga daerah}\\ & \text{asalnya}\\ & D_f=\{x|x\in \mathbb{R},x\ne 0\}\\ & \end{array}& \begin{array}{rl}& f(x)=\lfloor x\rfloor \\ & \text{Sehingga daerah}\\ & \text{asalnya}\\ & D_f=\{x|x\in \mathbb{R}\}\\ & \end{array}\\ \hline\end{array}$.

$.\begin{array}{|cc|}\hline (\text{i})& (\text{l})\\ \begin{array}{rl}& f(x)=\sqrt{x^2-16}\\ & \text{Sehingga daerah}\\ & \text{asalnya yaitu:}\\ & x^2-16\ge 0\\ & (x+4)(x-4)\ge 0\\ & x\le -4\text{atau}x\ge 4\\ & D_f=\{x|x\le -4\text{atau}x\ge 4,x\in \mathbb{R}\}\\ & \end{array}& \begin{array}{rl}& f(x)={\displaystyle \frac{2\sqrt{x}}{x-3}}\\ & \text{Sehingga daerah}\\ & \text{asalnya yaitu:}\\ & \{\begin{array}{ll}\bullet & x\ge 0\\ \bullet & x-3\ne 0\end{array}\\ & D_f=\{x|x\ge 0,x\ne 3,x\in \mathbb{R}\}\\ & \end{array}\\ \hline\end{array}$

$\begin{array}{l}\\ 4.& \text{Jika}\left|x\right|\text{menyatakan nilai mutlak}\\ & \text{dan}\lfloor x\rfloor \text{menyatakan bilangan bulat terbesarnatau sama dengan}x\\ & \text{misalkan}\lfloor 1,6\rfloor =1,\lfloor \pi \rfloor =3\\ & \text{Jika diberikan}f(x)=\left|x\right|+\lfloor x\rfloor ,\text{maka tentukanlah nilai untuk}\\ & \text{a}.f(-3,5)+f(2,5)\\ & \text{b}.f(-1,5)+f(3,5)\\ \\ & \text{Jawab}:\\ & \begin{array}{c}\\ \begin{array}{rl}\text{a}.f(-3,5)+f(2,5)& =|-3,5|+\lfloor -3,5\rfloor +|2,5|+\lfloor 2,5\rfloor \\ & =3,5+(-4)+2,5+2\\ & =4\end{array}\\ \begin{array}{rl}\text{b}.f(-1,5)+f(3,5)& =|-1,5|+\lfloor -1,5\rfloor +|3,5|+\lfloor 3,5\rfloor \\ & =1,5+(-2)+3,5+3\\ & =6\end{array}\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Jika diketahui relasi}f\text{dengan kondisi}\\ & \begin{array}{c}(\text{a}).f(1)=1\\ (\text{b}).f(2x)=4f(x)+6\\ (\text{c}).f(x+2)=f(x)+12x+12\end{array}\\ & \text{maka nilai}f(14)\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(1)& =1\\ f(2.1)=f(2)& =4f(1)+6=4.1+6=10\\ f(1+2)=f(3)& =f(1)+12.1+12\\ f(3)& =1+12+12=25\\ f(3+2)=f(5)& =f(3)+12.3+12\\ f(5)& =25+36+12=73\\ f(5+2)=f(7)& =f(5)+12.5+12\\ f(7)& =73+60+12=145\\ f(7.2)=f(14)& =4.f(7)+6\\ f(14)& =4.145+6=580+6\\ & =586\end{array}\end{array}$

$\begin{array}{l}\\ 6.& \mathbf{\text{(OSK 2013)}}\text{Fungsi}f\text{didefinisikan oleh}\\ & f(x)={\displaystyle \frac{kx}{2x+3},x=-{\displaystyle \frac{3}{2}.}}\\ & \text{Tentukanlah nilai}k\text{agar}f(f(x))=x\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f(x)& ={\displaystyle \frac{kx}{2x+3},x\ne -\frac{2}{3}}\\ f(f(x))& =x\\ x& =f(f(x))\\ x& ={\displaystyle \frac{k\left({\displaystyle \frac{kx}{2x+3}}\right)}{2\left({\displaystyle \frac{kx}{2x+3}}\right)+3}}\\ x& ={\displaystyle \frac{{\displaystyle \frac{k^{2}x}{2x+3}}}{{\displaystyle \frac{2kx+3(2x+3)}{2x+3}}}}\\ x& ={\displaystyle \frac{k^{2}x}{2kx+6x+9}}\\ 2kx+6x+9& =k^2\\ 0& =k^2-2xk-6x-9\\ 0& =(k+3)(k-2x-3)\\ & k=-3\text{atau}k=2x+3\end{array}\end{array}$

$\text{E. Menggambar Grafik Fungsi}$

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

$\text{CONTOH SOAL}$

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

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

$.\begin{array}{rl}& \text{b. Menggambar grafik}f(x)=2x^2\\ & \begin{array}{|ccc|}\hline x& y=f(x)=2x^2& \text{Titik}(x,y)\\ -3& f(-3)=2(-3{)}^2=18& (-3,18)\\ -2& f(-2)=2(-2{)}^2=8& (-2,8)\\ -1& f(-1)=2(-1{)}^2=2& (-1,2)\\ 0& f(0)=2(0{)}^2=0& (0,0)\\ 1& f(1)=2(1{)}^2=2& (1,2)\\ 2& f(2)=2(2{)}^2=8& (2,8)\\ 3& f(3)=2(3{)}^2=18& (3,18)\\ \hline\end{array}\end{array}$

$.\begin{array}{rl}& \text{c. Menggambar grafik}f(x)={\displaystyle \frac{1}{x}}\\ & \begin{array}{|ccc|}\hline x& y=f(x)={\displaystyle \frac{1}{x}}& \text{Titik}(x,y)\\ -3& f(-3)=-{\displaystyle \frac{1}{3}}& (-3,-{\displaystyle \frac{1}{3})}\\ -2& f(-2)=-{\displaystyle \frac{1}{2}}& (-2,-{\displaystyle \frac{1}{2})}\\ -1& f(-1)=-1& (-1,-1)\\ 0& f(0)\text{tidak ada}& -\\ 1& f(1)=1& (1,1)\\ 2& f(2)={\displaystyle \frac{1}{2}}& (2,{\displaystyle \frac{1}{2})}\\ 3& f(3)={\displaystyle \frac{1}{3}}& (3,{\displaystyle \frac{1}{3})}\\ \hline\end{array}\end{array}$

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

$.\begin{array}{rl}& \text{a. Menggambar grafik}f(x)=\lfloor x\rfloor \\ & \begin{array}{|ccc|}\hline x& y=f(x)=\lfloor x\rfloor & \text{Titik}(x,y)\\ ⋮& ⋮& ⋮\\ -1& f(-1)=\lfloor -1\rfloor =-1& (-1,-1)\\ ⋮& ⋮& ⋮\\ -{\displaystyle \frac{1}{2}}& f(-{\displaystyle \frac{1}{2})=\lfloor -{\displaystyle \frac{1}{2}}\rfloor =-1}& (-{\displaystyle \frac{1}{2},-1)}\\ -{\displaystyle \frac{1}{3}}& f(-{\displaystyle \frac{1}{3})=\lfloor -{\displaystyle \frac{1}{3}}\rfloor =-1}& (-{\displaystyle \frac{1}{3},-1)}\\ -{\displaystyle \frac{1}{4}}& f(-{\displaystyle \frac{1}{4})=\lfloor -{\displaystyle \frac{1}{4}}\rfloor =-1}& ({\displaystyle \frac{1}{4},-1)}\\ ⋮& ⋮& ⋮\\ 0& f(0)=\lfloor 0\rfloor =0& (0,0)\\ ⋮& ⋮& ⋮\\ {\displaystyle \frac{1}{4}}& f({\displaystyle \frac{1}{4})=\lfloor {\displaystyle \frac{1}{4}}\rfloor =0}& ({\displaystyle \frac{1}{4},0)}\\ {\displaystyle \frac{1}{3}}& f({\displaystyle \frac{1}{3})=\lfloor {\displaystyle \frac{1}{3}}\rfloor =0}& ({\displaystyle \frac{1}{3},0)}\\ {\displaystyle \frac{1}{2}}& f({\displaystyle \frac{1}{2})=\lfloor {\displaystyle \frac{1}{2}}\rfloor =0}& (-{\displaystyle \frac{1}{2},0)}\\ ⋮& ⋮& ⋮\\ 1& f(1)=\lfloor 1\rfloor =1& (1,1)\\ ⋮& ⋮& ⋮\\ 2{\displaystyle \frac{1}{4}}& f(2{\displaystyle \frac{1}{4})=\lfloor 2{\displaystyle \frac{1}{4}}\rfloor =2}& (2{\displaystyle \frac{1}{4},2)}\\ ⋮& ⋮& ⋮\\ 4{\displaystyle \frac{1}{3}}& f(4{\displaystyle \frac{1}{3})=\lfloor 4{\displaystyle \frac{1}{3}}\rfloor =4}& (4{\displaystyle \frac{1}{3},4)}\\ ⋮& ⋮& ⋮\\ \hline\end{array}\end{array}$

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

$.\begin{array}{rl}& \text{Diketahui bahwa}\lfloor x\rfloor \text{adalah bilangan bulat terbesar }\\ & \text{yang lebih kecil atau sama dengan}x\\ & \text{Misal}\lfloor 3,2\rfloor =3,\lfloor -2,47\rfloor =-3,\lfloor 6\rfloor =6,\text{dan lain-lain}.\\ & \text{Sehingga}\lfloor x\rfloor =a\Rightarrow a\le x<a+1(\text{dengan}a\text{bilangan bulat}),\\ & \text{maka}\\ & \{\begin{array}{ll}\lfloor \sqrt{x}\rfloor =9& \Rightarrow 9\le \sqrt{x}<9+1\\ & \iff 9\le \sqrt{x}<10\iff 81\le x<100\\ \lfloor \sqrt{y}\rfloor =12& 12\le \sqrt{y}<12+1\\ & \iff 12\le \sqrt{y}<13\iff 144\le y<169\end{array}\end{array}$.

$.\begin{array}{rl}& \bullet 144\le y<169\text{dan}\\ & \bullet 81\le x<100,\text{dikalikan dengan}(-1)\\ & \text{maka akan menjadi},\\ & -100<-x\le -81,\\ & \text{sehingga}-99,\overline{999}\le -x<-80,\overline{999}\\ & \text{Selanjutnya}\\ & \begin{array}{c}\\ 144& \le y<169& \\ -99,\overline{999}& \le -x<-80,\overline{999}& +\\ \\ 44,...& \le y-x<88,...\end{array}\\ & \text{Jadi, nilai terkecil}\lfloor y-x\rfloor =44\end{array}$