Contoh Soal 3 Fungsi

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

Contoh Soal 2 Fungsi

$\begin{array}{l}\\ 6.& \text{Diketahui bahwa}\\ & f(x)=\{\begin{array}{ll}0,& \text{untuk}x<0\\ x^2,& \text{untuk}0\le x<1\\ 2x-1,& \text{untuk}x\ge 1\end{array}\\ & \text{Nilai dari}f(-1)+f\left({\displaystyle \frac{1}{2}}\right)-f(3)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -5{\displaystyle \frac{1}{4}}& & & \text{d}.& 4{\displaystyle \frac{3}{4}}\\ \\ \text{b}.& -4{\displaystyle \frac{3}{4}}& \text{c}.& 4& \text{e}.& 5{\displaystyle \frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \\ f(x)& =\{\begin{array}{ll}0,& \text{untuk}x<0\\ x^2,& \text{untuk}0\le x<1\\ 2x-1,& \text{untuk}x\ge 1\end{array}\\ \text{maka nilai}& f(-1)+f\left({\displaystyle \frac{1}{2}}\right)-f(3)\\ & =0+{\left({\displaystyle \frac{1}{2}}\right)}^2-(2(3)-1)\\ & ={\displaystyle \frac{1}{4}-5}\\ & =-4{\displaystyle \frac{3}{4}}\end{array}\end{array}$.

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

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

$\begin{array}{l}\\ 9.& \text{Jika}{a}_{{}_0}={\displaystyle \frac{2}{5}\text{dan}{a}_{n+1}=2\left|a_n\right|-1,\text{maka nilai}{a}_{{}_{2022}}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -0,6& & & \text{d}.& 0,4\\ \\ \text{b}.& -0,2& \text{c}.& 0,2& \text{e}.& 0,6\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}\\ a_{{}_0}& ={\displaystyle \frac{2}{5}=0,4\text{dan}{a}_{n+1}=2\left|a_n\right|-1,\text{maka}}\\ a_{{}_1}& =2\left|a_{{}_0}\right|-1=2|0,4|-1=0,8-1=-0,2\\ a_{{}_2}& =2\left|a_{{}_1}\right|-1=2|-0,2|-1=2(0,2)-1\\ & =0,4-1=-0,6\\ a_{{}_3}& =2\left|a_{{}_2}\right|-1=2|-0,6|-1=1,2-1=0,2\\ a_{{}_4}& =2\left|a_{{}_3}\right|-1=2|0,2|-1=0,4-1=-0,6={\mathbf{\text{a}}}_{{}_{\mathbf{\text{2}}}}\\ a_{{}_5}& =2\left|a_{{}_4}\right|-1=2|-0,6|-1=1,2-1=0,2={\mathbf{\text{a}}}_{{}_{\mathbf{\text{3}}}}\\ a_{{}_6}& =2\left|a_{{}_5}\right|-1=2|0,2|-1=0,4-1=-0,6={\mathbf{\text{a}}}_{{}_{\mathbf{\text{2}}}}\\ a_{{}_7}& =2\left|a_{{}_6}\right|-1=2|-0,6|-1=1,2-1=0,2={\mathbf{\text{a}}}_{{}_{\mathbf{\text{3}}}}\\ ⋮& \\ a_{{}_{2022}}& =\cdots =\cdots ={\mathbf{\text{a}}}_{{}_{\mathbf{\text{2}}}}=-0,6\end{array}\end{array}$.

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

Contoh Soal Fungsi

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

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

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

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

$\begin{array}{l}\\ 4.& \text{Dari beberapa fungsi berikut yang }\\ & \text{merupakan fungsi genap adalah}....\\ & \begin{array}{l}\\ \text{a}.& f(x)=x^2+\left|x\right|-1\\ \text{b}.& f(x)=x^3-\left|x\right|+x\\ \text{c}.& f(x)=x\left|x\right|+x\\ \text{d}.& f(x)=\sqrt{x-1}\\ \text{e}.& f(x)=4-2x\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \\ \text{Suatu fungsi}& \text{dinamakan fungsi genap}\\ & \text{jika}f(x)=f(-x)\\ & \end{array}\\ & \begin{array}{|cllc|}\hline \text{No}& f(x)& f(-x)& \text{Keterangan}\\ 4.\text{a}& x^2+\left|x\right|-1& x^2+\left|x\right|-1& f(x)=f(-x)\\ 4.\text{b}& x^3-\left|x\right|+x& -x^3-\left|x\right|-x& f(x)\ne f(-x)\\ 4.\text{c}& x\left|x\right|+x& -x\left|x\right|-x& f(x)\ne f(-x)\\ 4.\text{d}& \sqrt{x-1}& \sqrt{-x-1}& f(x)\ne f(-x)\\ 4.\text{e}.& 4-2x& 4+2x& f(x)\ne f(-x)\\ \hline\end{array}\end{array}$

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

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.

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

$\text{CONTOH SOAL}$

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

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

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

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

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

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

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

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

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