MATERI FUNGSI KOMPOSISI DAN FUNGSI INVERS TERLENGKAP KELAS 10 KURIKULUM 2013 | MATEMATIKA

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MATERI FUNGSI KOMPOSISI DAN FUNGSI INVERS TERLENGKAP KELAS 10 KURIKULUM 2013 | MATEMATIKA 

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 : Arightarrow 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}.
A.6. Identitas Fungsi
I(x) = X
CONTOH SOAL
Diketahui f(x) = 3x +5
                   g(x)= x
Tentukan: 
  1. (I o f)(x)
Jawab:
    1. (I o f)(x)= I(f(x))
                            = I (3x + 5)
                            = 3x + 5
        
    LARGEfbox{LARGEfbox{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}-xgeq 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|xleq 0: textrm{atau}: xgeq 1 ,: : xin mathbb{R}right } end{aligned}&begin{aligned}&textrm{Hasil akar pangkat 2}\ &textrm{tidak pernah negatif}\ &textrm{Jadi},: R_{_{f}}=left { y|ygeq 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|xin mathbb{R} right }\ &\ &\ &\ &\ & end{aligned}&begin{aligned}textrm{b}.quad f(x)&=displaystyle frac{2}{3x-15}\ &textrm{supaya terdefinisi}\ &textrm{maka},\ &3x-15neq 0\ &xneq 5,: : textrm{sehingga}\ D_{_{f}}&=left { x|xneq 5,: xin 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-6neq 0\ &xneq 3: textrm{dan}: xneq -2,\ & textrm{sehingga}\ D_{_{g}}&=left { x|xneq3: textrm{dan}: xneq -2,: xin mathbb{R} right } end{aligned}\hline begin{aligned}textrm{d}.quad g(x)&=sqrt{x^{2}-1}\ textrm{ma}&textrm{ka}: : x^{2}-1geq 0\ &(x+1)(x-1)geq 0\ D_{_{g}}&=left { x|xleq -1: : textrm{atau}: : xgeq 1,: xin mathbb{R} right }\ &\ &\ &\ &\ & end{aligned}&begin{aligned}textrm{e}.quad h(x)&=sqrt{3x+2}\ textrm{ma}&textrm{ka}: : 3x+2geq 0\ &: : xgeq -displaystyle frac{2}{3}\ D_{_{h}}&=left { x|xgeq -displaystyle frac{2}{3},: xin 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<xleq 1: textrm{atau}: : x>3,: xin 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,: : xin 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) & >0Rightarrow x>-2\ (x+2) & neq 0 Rightarrow xneq -2 end{cases}\ &qquad 2.: : (x^{2}-2x-3)>0Rightarrow (x-3)(x+1)>0\ D_{_{k}}&=left { x|-2< x< -1: : textrm{atau}: : x>3,: : xin 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|xleq 1,: : xin 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|xleq 1,: : xin 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|xleq 1,: : xin 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 (fcirc g)(x)neq (gcirc f)(x)\ 2.: : &textrm{Bersifat asosiatif}quad fcirc (gcirc h)(x)= (fcirc g)circ h(x)\ 3.: : &textrm{Adanya unsur dentitas}quad (fcirc I)(x)=(Icirc f)(x)=f(x) end{aligned}\hline end{array}.

    c. Fungsi Invers

    385
    begin{aligned}&bullet quad textrm{Suatu fungsi}: : f:Arightarrow B: : textrm{memiliki fungsi invers} : : g:Brightarrow 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 ( gcirc f right )^{-1}(x)=left ( f^{-1}circ g^{-1} right )(x)\ blacklozenge &left ( fcirc 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 ( fcirc f^{-1} right )(x)=fleft ( f^{-1}(x) right ) end{array}.
    LARGEfbox{LARGEfbox{LANJUTAN CONTOH SOAL}}.
    begin{array}{ll}\ 4.&textrm{Tentukanlah}: : (fcirc g)(x): : textrm{dan}: : (gcirc 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}(fcirc g)(x)&=fleft ( g(x) right )\ &=fleft ( 5x+3 right )\ &=2-left ( 5x+3 right )\ &=-5x-1\ &\ textrm{dan}: : : : : : : : : : &\ (gcirc f)(x)&=gleft ( 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}: : (gcirc f)(x)=4x-5.: : textrm{Tentukanlah}: : f(x) end{array}\\\ textrm{Jawab}:\\ begin{aligned}(gcirc 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}: : (fcirc g)(x)=2x^{2}+3.: : textrm{Tentukanlah}: : f(x) end{array}\\\ textrm{Jawab}:\\ begin{aligned}(fcirc g)(x)&=2x^{2}+3\ f(g(x))&=2x^{2}+3\ f(x+4)&=2x^{2}+3,qquad textrm{misalkan}: : x+4=aRightarrow 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 (gcirc f)(x) end{array}\\\ textrm{Jawab}:\\ begin{aligned}^{^{27}}log (gcirc 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&=2xlog 6\ displaystyle frac{log y}{log 6}&=2x\ displaystyle frac{log y}{2log 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},: : : xneq 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},: : : xneq 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}: : (fcirc g)^{-1}(2)=.... end{array}\\\ textrm{Jawab}:\\ begin{array}{|c|c|}hline multicolumn{2}{|c|}{begin{aligned}&\ &(fcirc g)^{-1}(x)=left ( g^{-1}circ f^{-1} right )(x)\ & end{aligned}}\hline (fcirc g)^{-1}(x)&left ( g^{-1}circ f^{-1} right )(x)\hline begin{aligned}(fcirc g)(x)&=f(g(x))\ y&=-2left ( displaystyle frac{20-3x}{2} right )-4\ y&=3x-20-4\ y&=3x-24\ y+24&=3x\ x&=displaystyle frac{y+24}{3}\ (fcirc g)^{-1}y&=displaystyle frac{y+24}{3}\ (fcirc 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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