fungsi komposisi invers, Materi kelas XI, materi kelas XI semester II, Uncategorized

Fungsi Komposisi dan Invers


A. Fungsi (Pemetaan) 

A. 1.  Pengartian Fungsi

Sebuah fungsi  f   dari himpunan  A  ke himpunan  B  adalah relasi dengan memasangkan setiap unsur himpunan  A  berpasangan dengan tepat satu unsur pada himpunan  B.

\begin{array}{|l|l|l|l|}\hline \multicolumn{4}{|c|}{\textrm{Contoh Relasi atau Fungsi dinyatakan dengan 4 macam cara, yaitu:}}\\\hline \textrm{Pasangan berurutan}&\textrm{Diagram panah}&\textrm{Peramaan}&\textrm{Grafik fungsi Kartesius}\\\hline \begin{aligned}&\textrm{misal}:\\ &\left \{ (a,1),(b,2),(c,3) \right \} \end{aligned}&\begin{array}{|ll|ll|ll|}\cline{1-2}\cline{5-6} a&&&&&1\\\cline{2-5} &&&&&\\ b&&&&&2\\\cline{2-5} &&&&&\\ c&&&&&3\\\cline{2-5} &&&&&\\\cline{1-2}\cline{5-6} \multicolumn{3}{l}{\textbf{A}}&\multicolumn{3}{r}{\textbf{B}} \end{array}&\begin{aligned}&\textrm{misal}:\\ &\circ y=3x\\ &\circ y=\displaystyle \frac{1}{x}\\ &\circ y=x^{2}-2\\ &\circ y=\: ^{^{^{2}}}\log x\\ &\textrm{dan lain lain }\end{aligned}&\begin{array}{ll|llllll}\\ &\textbf{Y}&&&&&&\\ &4&&&\bullet &&&\\ &3&&&&&&\\ &2&&&&&&\\ &1&&\bullet &&&&\\ &&&&&&&\textbf{X}\\\cline{1-7} &0&&1&2&3&4 \end{array}\\\hline \end{array}.

A. 2.  Domain, Kodomain dan Range Suatu Fungsi

Sebuah fungsi  f  dari himpunan  A  ke himpunan  B sering dituliskan sebagai  f  :  A → B. Jika fungsi  f  memetakan  x ∈ A ke  y ∈ B dapat dituliskan dengan  f  :  x  →  y   atau  f  :  x →  f(x).

Perhatikanlah ilustrasi berikut

382

  • Himpunan A disebut daerah asal (prapeta) atau domain dari fungsi  f.
  • Himpunan B disebut daerah kawan atau kodomain dari fungsi  f.
  • Himpunan semua bayangan (peta) disebut daerah hasil atau range dari fungsi  f.

Sebagai Contoh ilustrasi berikut:

383

\begin{aligned}\textrm{Dari}\: &\textrm{ilustrasi di atas diperoleh bahwa}:\\ &\textrm{Domain}\qquad :\quad D_{_{f}}=A=\left \{ a,b,c,d \right \}\\ &\textrm{Kodomain}\: \: \: \: :\quad B=\left \{ 1,2,3,4,5 \right \}\\ &\textrm{Range}\: \: \: \qquad :\quad R_{_{f}}=\left \{ 1,2,3,5 \right \}\subseteq B \end{aligned}.

A. 3. Fungsi-Fungsi Khusus

\begin{array}{|l|l|l|}\hline \multicolumn{3}{|c|}{\textrm{Fungsi Khusus}}\\\hline \textrm{Fungsi Konstan}&\textrm{Fungsi Identitas}&\textrm{Fungsi Linear}\\\hline \begin{aligned}&f(x)=c \end{aligned}&\begin{aligned}&f(x)=x \end{aligned}&\begin{aligned}&f(x)=ax+b \end{aligned}\\\hline \textrm{Fungsi Kuadrat}&\textrm{Fungsi Mutlak}&\textrm{Fungsi Tangga}\\\hline \begin{aligned}&f(x)=ax^{2}+bx+c \end{aligned}&\begin{aligned}&f(x)=\left | x \right | \end{aligned}&\begin{aligned}&f(x)=\displaystyle \left \lfloor x \right \rfloor \end{aligned}\\\hline \textrm{Fungsi Ganjil dan Genap}&\multicolumn{2}{c|}{.}\\\cline{1-1} \multicolumn{1}{|c|}{\begin{aligned}&\\ &\begin{cases} \textrm{Ganjil}: & f(-x)=-f(x) \\ \textrm{Genap}: & f(-x)=f(x) \end{cases}\\ & \end{aligned}}&\multicolumn{2}{|c|}{\begin{aligned}&\textrm{Untuk mendapatkan keterangan}\\ &\textrm{lebih lanjut silahkan Anda merujuk ke referensi berikut}:\\ &\textrm{Buku Matematika BSE untuk SMA/MA kls XI IPA, penulis}\\ &\textrm{Nugroho Soedyarto dan Maryanto} \end{aligned}}\\\hline \end{array}.

A. 4. Sifat-Sifat Fungsi

\begin{array}{|c|c|l|}\hline \multicolumn{3}{|c|}{\textrm{Sifat-sifat fungsi}\: \: f : A\rightarrow B}\\\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}.

A. 5. Aljabar Fungsi

\begin{array}{|l|l|}\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}.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Tentukanlah domain dan range dari fungsi}\: \: f(x)=\sqrt{x^{2}-x} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|l|l|}\hline \textrm{Domain}&\textrm{Range}\\\hline \begin{aligned}&\textrm{Kumpulan nilai}\: \: x\\ &\textrm{yang mungkin, yaitu:}\\ &x^{2}-x\geq 0\\ &x(x-1)\geq 0\\ &\textrm{dengan garis bilangan}\\ &\begin{array}{llllllllll}\\ +&+&+&-&-&-&-&+&+&+\\\cline{1-10} &&0&&&&&1&& \end{array}\\ &\textrm{Jadi},\: D_{_{f}}=\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},\: R_{_{f}}=\left \{ y|y\geq 0 \right \}\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}\\\hline \end{array}.

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

Jawab:

\begin{array}{|l|l|l|}\hline \begin{aligned}\textrm{a}.\quad f(x)&=2x+3\\ D_{_{f}}&=\left \{ x|x\in \mathbb{R} \right \}\\ &\\ &\\ &\\ &\\ & \end{aligned}&\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}\\ D_{_{f}}&=\left \{ x|x\neq 5,\: x\in \mathbb{R} \right \}\\ & \end{aligned}&\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}\\ 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\\ D_{_{g}}&=\left \{ x|x\leq -1\: \: \textrm{atau}\: \: x\geq 1,\: x\in \mathbb{R} \right \}\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{e}.\quad h(x)&=\sqrt{3x+2}\\ \textrm{ma}&\textrm{ka}\: \: 3x+2\geq 0\\ &\: \: x\geq -\displaystyle \frac{2}{3}\\ D_{_{h}}&=\left \{ x|x\geq -\displaystyle \frac{2}{3},\: x\in \mathbb{R} \right \}\\ &\\ &\\ &\\ & \end{aligned}&\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\\ D_{_{h}}&=\left \{ x|-2<x\leq 1\: \textrm{atau}\: \: x>3,\: x\in \mathbb{R} \right \} \end{aligned}\\\hline \end{array}.

\begin{array}{|l|l|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\\ D_{_{k}}&=\left \{ x|x<-3\: \: \textrm{atau}\: \: x>5,\: \: x\in \mathbb{R} \right \}\\ & \end{aligned}&\multicolumn{2}{|l|}{\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\\ D_{_{k}}&=\left \{ x|-2< x< -1\: \: \textrm{atau}\: \: x>3,\: \: x\in \mathbb{R}\right \}\end{aligned}}\\\hline \end{array}.

\begin{array}{ll}\\ 3.&\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} \end{array}\\\\\\ \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}&\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{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}&\begin{aligned}\textrm{d}.\quad \left ( \displaystyle \frac{f}{g} \right )(x)&=....................\\ &....................\\ &.................... \end{aligned}\\\hline \end{array}.

B. Fungsi Komposisi

Perhatikanlah ilustrasi berikut ini!

384

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{Komposisi Fungsi}}\\\hline \textrm{Syarat}&\textrm{Sifat-sifat}\\\hline \begin{aligned}&R_{_{f}}\cap D_{_{g}}\neq \left \{ \: \right \} \end{aligned}&\begin{aligned}1.\: \: &\textrm{Tidak komutatif}\quad (f\circ g)(x)\neq (g\circ f)(x)\\ 2.\: \: &\textrm{Bersifat asosiatif}\quad f\circ (g\circ h)(x)= (f\circ g)\circ h(x)\\ 3.\: \: &\textrm{Adanya unsur dentitas}\quad (f\circ I)(x)=(I\circ f)(x)=f(x) \end{aligned}\\\hline \end{array}.

C. Fungsi Invers

385

\begin{aligned}&\bullet \quad \textrm{Suatu fungsi}\: \: f:A\rightarrow B\: \: \textrm{memiliki fungsi invers} \: \: g:B\rightarrow A\: \: \textsl{jika dan hanya jika}\: \: f\: \: \textrm{merupakan fungsi}\: \textbf{bijektif}\\ &\bullet \quad \textrm{Jika fungsi}\: \: g\: \: \textrm{ada, maka}\: \: g\: \: \textrm{dinyatakan dengan}\: \: f^{-1}\: \: (\textrm{dibaca}:\: \: f\: \: \textrm{invers})\end{aligned}.

\begin{array}{ll}\\ \textrm{Perlu}&\textrm{diingat bahwa pada invers fungsi komposisi berlaku ketentuan sebagai berikut}\\ \blacklozenge &\left ( g\circ f \right )^{-1}(x)=\left ( f^{-1}\circ g^{-1} \right )(x)\\ \blacklozenge &\left ( f\circ g \right )^{-1}(x)=\left ( g^{-1}\circ f^{-1} \right )(x)\\ \blacklozenge &f(x)=\left ( \left (f^{-1} \right )^{-1}(x) \right )\\ \blacklozenge &x=f^{-1}\left ( f(x) \right )=\left ( f^{-1}\circ f \right )(x)=\left ( f\circ f^{-1} \right )(x)=f\left ( f^{-1}(x) \right ) \end{array}.

\LARGE\fbox{\LARGE\fbox{LANJUTAN CONTOH SOAL}}.

\begin{array}{ll}\\ 4.&\textrm{Tentukanlah}\: \: (f\circ g)(x)\: \: \textrm{dan}\: \: (g\circ f)(x)\: \: \textrm{Jika}:\\ &\textrm{a}.\quad f(x)=2-x\: \: \textrm{dah}\: \: g(x)=5x+3\\ &\textrm{b}.\quad f(x)=2x+1\: \: \textrm{dah}\: \: g(x)=x^{2}-4\\ &\textrm{c}.\quad f(x)=\displaystyle \frac{5}{x-4}\: \: \textrm{dah}\: \: g(x)=3x^{2}\\ &\textrm{d}.\quad f(x)=\sqrt{4-x}\: \: \textrm{dah}\: \: g(x)=x^{2}+x\\ &\textrm{e}.\quad f(x)=x^{3}+1\: \: \textrm{dah}\: \: g(x)=\displaystyle \frac{x}{x-1}\\ &\textrm{f}.\quad f(x)=\displaystyle \frac{3}{x-2}\: \: \textrm{dah}\: \: g(x)=\sqrt{x-4}\\ \end{array}.

Jawab: untuk No. 4 a)

\begin{aligned}(f\circ g)(x)&=f\left ( g(x) \right )\\ &=f\left ( 5x+3 \right )\\ &=2-\left ( 5x+3 \right )\\ &=-5x-1\\ &\\ \textrm{dan}\: \: \: \: \: \: \: \: \: \: &\\ (g\circ f)(x)&=g\left ( f(x) \right )\\ &=g(2-x)\\ &=5(2-x)+3\\ &=10-5x+3\\ &=13-10x \end{aligned}.

\begin{array}{ll}\\ 5.&\textrm{Diketahui bahwa}\: \: g(x)=3x+2\: \: \textrm{dan}\: \: (g\circ f)(x)=4x-5.\: \: \textrm{Tentukanlah}\: \: f(x) \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}(g\circ f)(x)&=4x-5\\ g(f(x))&=4x-5\\ 3.f(x)+2&=4x-5\\ 3.f(x)&=4x-7\\ f(x)&=\displaystyle \frac{4x-7}{3} \end{aligned}.

\begin{array}{ll}\\ 6.&\textrm{Diketahui bahwa}\: \: g(x)=x+4\: \: \textrm{dan}\: \: (f\circ g)(x)=2x^{2}+3.\: \: \textrm{Tentukanlah}\: \: f(x) \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}(f\circ g)(x)&=2x^{2}+3\\ f(g(x))&=2x^{2}+3\\ f(x+4)&=2x^{2}+3,\qquad \textrm{misalkan}\: \: x+4=a\Rightarrow x=a-4,\\ \textrm{sehingga}&,\\ f(a)&=2(a-4)^{2}+3\\ f(a)&=2(a^{2}-8a+16)+3\\ &=2a^{2}-16a+35\\ f(x)&=2x^{2}-16x+35 \end{aligned}.

\begin{array}{ll}\\ 7.&\textrm{Diketahui}\: \: f(x)=3x\: \: \textrm{dan}\: \: g(x)=3^{x}.\: \: \textrm{Tentukanlah rumus untuk}\: \: ^{^{27}}\log (g\circ f)(x) \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}^{^{27}}\log (g\circ f)(x)&=\: ^{^{27}}\log g(f(x))\\ &=\: ^{^{27}}\log 3^{3x}\\ &=\: ^{^{3^{3}}}\log 3^{3x}\\ &=\: ^{^{^{\left (3^{3} \right )}}}\log \left ( 3^{3} \right )^{x}\\ &=x \end{aligned}.

\begin{array}{ll}\\ 8.&\textrm{Tentukanlah invers dari}\: \: f(x)=6^{2x} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}f(x)&=6^{2x}\\ y&=6^{2x}\\ \log y&=\log 6^{2x}\\ \log y&=2x\log 6\\ \displaystyle \frac{\log y}{\log 6}&=2x\\ \displaystyle \frac{\log y}{2\log 6}&=x\\ \displaystyle \frac{\log y}{\log 6^{2}}&=x\\ \displaystyle \frac{\log y}{\log 36}&=x\\ x&=\displaystyle \frac{\log y}{\log 36}&\\ x&=\: ^{^{36}}\log y\\ f^{-1}(x)&=\: ^{^{36}}\log x \end{aligned}.

\begin{array}{ll}\\ 9.&\textrm{Tentukanlah inver dari}\: \: f(x)=\displaystyle \frac{2x+3}{4x-5},\: \: \: x\neq \displaystyle \frac{5}{4} \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}f(x)&=\displaystyle \frac{2x+3}{4x-5}\\ y&=\displaystyle \frac{2x+3}{4x-5}\\ (4x-5)y&=2x+3\\ 4xy-5y&=2x+3\\ 4xy-2x&=5y+3\\ x(4y-2)&=5y+3\\ x&=\displaystyle \frac{5y+3}{4y-2}\\ f^{-1}(y)&=\displaystyle \frac{5y+3}{4y-2}\\ \textrm{maka},&\\ f^{-1}(x)&=\displaystyle \frac{5x+3}{4x-2},\: \: \: x\neq \displaystyle \frac{1}{2} \end{aligned}.

\begin{array}{ll}\\ 10.&\textrm{Jika}\: \: f(x)=-2x-4\: \: \textrm{dan}\: \: g(x)=\displaystyle \frac{20-3x}{2},\: \: \textrm{maka nilai dari}\: \: (f\circ g)^{-1}(2)=.... \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ &(f\circ g)^{-1}(x)=\left ( g^{-1}\circ f^{-1} \right )(x)\\ & \end{aligned}}\\\hline (f\circ g)^{-1}(x)&\left ( g^{-1}\circ f^{-1} \right )(x)\\\hline \begin{aligned}(f\circ g)(x)&=f(g(x))\\ y&=-2\left ( \displaystyle \frac{20-3x}{2} \right )-4\\ y&=3x-20-4\\ y&=3x-24\\ y+24&=3x\\ x&=\displaystyle \frac{y+24}{3}\\ (f\circ g)^{-1}y&=\displaystyle \frac{y+24}{3}\\ (f\circ g)^{-1}(2)&=\displaystyle \frac{2+24}{3}\\ &=\displaystyle \frac{26}{3} \end{aligned}&\begin{aligned}\left ( g^{-1}\circ f^{-1} \right )(x)&=g^{-1}\left ( f^{-1}(x) \right )\\ &=......\\ &=....\\ &=....\\ &=....\\ &=....\\ &=....\\ &=....\\ &=....\\ &=....\\ \left ( g^{-1}\circ f^{-1} \right )(2)&=....\\ &=.... \end{aligned}\\\hline \end{array}

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