Fungsi (Matematika Wajib Kelas X)

A. Pendahuluan fungsi

Pada awal bab ini kita akan ingatkan kembali tentang notasi, domain(daerah asal), kodomain(daerah kawan), serta range(daerah hasil) fungsi.

Perhatikan tabel berikut

\color{blue}\begin{array}{|l|l|}\hline \textrm{Fungsi}&\textrm{Fungsi atau Pemetaan dari A ke B adalah}\\ &\textrm{sebuah relasi khusus yang memasangkan setiap}\: \: x\in A\\ &\textrm{ke tepat satu}\: \: y\in B\\\hline \textrm{Notasi}&f:x \rightarrow y\: \: \: \textrm{atau}\: \: \: f:x \rightarrow f(x) \\\hline \textrm{Dibaca}&\textrm{fungsi}\: \: f\: \: \textrm{memetakan}\: \: x\in A\: \: \textrm{ke}\: \: y\in B\\\hline \: \: \: \: \textrm{A}&\textrm{Domain atau daerah asal fungsi atau}\quad D_{f}\\\hline \: \: \: \: \, x&\textrm{prapeta(sebelum dipetakan)}\\\hline \: \: \: \: \textrm{B}&\textrm{Kodomain atau daerah kawan fungsi atau}\quad K_{f}\\\hline \: \: \: \: \, y&\textrm{peta(bayangan dari prapeta) adalah Range atau}\quad R_{f}\\\hline \end{array}.

B. Sifat-Sifat Fungsi

\color{blue}\begin{array}{|c|c|l|}\hline \textrm{Injektif(satu-satu)}&\textrm{Surjektif(pada)}&\textrm{Bijektif(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 dan}\\ &\textrm{sekaligus juga surjektif}\\ &\\ & \end{aligned}\\\hline \end{array}.

C. Operasi Aljabar Fungsi

\color{blue}\begin{array}{|c|c|}\hline \textrm{Aljabar Fungsi}&\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}.

D. Macam-Macam Fungsi

\color{blue}\begin{array}{|l|l|l|}\hline \textrm{Fungsi Linear}&\textrm{Fungsi Konstan}&f(x)=c\\\hline &\textrm{Fungsi Identitas}&f(x)=x\\\hline &\textrm{Fungsi linear/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 Modulus(nilai mutlak)}&f(x)=\left | x \right |\\\hline \textrm{Fungsi Khusus} &\textrm{Fungsi tangga}&f(x)=\left \lfloor x \right \rfloor\\\hline &\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}{\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}}.

\color{fuchsia}\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 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 \}\\\\ &\textrm{Jawab}:\\ &\textrm{Semuanya Fungsi kecuali}\textbf{ poin b)} \end{array}

\color{fuchsia}\begin{array}{ll}\\ 2.&\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\qquad \textrm{catatan}\: \: \left \lfloor x \right \rfloor\: \: \textrm{adalah bulat terbesar atau sama dengan}\: \: x\\ \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}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|c|}\hline (\textrm{a})&(\textrm{b})&(\textrm{c})\\\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 \textit{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}&\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} \end{array}.

\color{fuchsia}\begin{array}{ll}\\ 3.&\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}.

\color{green}\begin{array}{ll}\\ 4.&\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}.

\color{blue}\begin{array}{ll}\\ 5.&\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}.

Tinggalkan Balasan

Alamat email Anda tidak akan dipublikasikan. Ruas yang wajib ditandai *