Love Blooming Rose FUNGSI ~ BELAJAR ONLINE BERSAMA STELLA

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Senin, 06 Mei 2019

FUNGSI

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A. Operasi 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}.
B. Fungsi Komposisi
Perhatikanlah ilustrasi berikut ini!
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\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
Perhatikan pula islustrasi berikut
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\\ &\qquad \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{CONTOH SOAL}}..
\begin{array}{ll}\\ 1.&\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}\\\\ &\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} \end{array}.
\begin{array}{ll}\\ 2.&\textrm{Diketahui fungsi}\: \: f\: \: \textrm{dan}\: \: g\: \: \textrm{yang dinyatakan berupa pasangan berurut seperti}\\ &\textrm{berikut}:\\ &\begin{array}{ll}\\ f&=\left \{ (4,1),(0,3),(1,4),(3,6),(2,10) \right \}\\ g&=\left \{ (1,0),(3,1),(4,2),(6,3),(10,4) \right \}\\ \end{array}\\\\ &\textrm{Tentukanlah}\: \: f\circ g,\: g\circ f,\: (f\circ g)(3),\: (f\circ g)(6), (g\circ f)(1),\: \textrm{dan}\: (g\circ f)(0)\\\\ &\textrm{Jawab}:\\ &\bullet \quad f\circ g=\left \{ (...,...),(...,...),(...,...),(...,...),(...,...) \right \}\\ &\bullet \quad g\circ f=\left \{ (0,1),(1,2),(2,4),(3,3),(4,0) \right \}\\ &\bullet \quad (f\circ g)(3)=f(g(3))=....=....\\ &\bullet \quad (f\circ g)(6)=f(g(6))=....=....\\ &\bullet \quad (g\circ f)(1)=g(f(1))=g(4)=2\\ &\bullet \quad (g\circ f)(0)=g(f(0))=g(3)=1\\ &\textrm{Perhatikanlah ilustrasi berikut}\\ \end{array}.
\begin{array}{ll}\\ 3.&\textrm{Diketahui fungsi}\: \: f,\: g,\: \textrm{dan}\: \: h\: \: \textrm{dengan}\: \: f(x)=x-4,\: g(x)=3x+2\\ &\textrm{dan}\: \: h(x)=x^{2}-1\\ &\begin{array}{ll}\\ \textrm{a}.&(f\circ g\circ h)(x)\\ \textrm{b}.&(f\circ h\circ g)(x)\\ \textrm{c}.&(h\circ g\circ f)(x)\\ \textrm{d}.&(g\circ f\circ h)(x)\\ \textrm{e}.&(g\circ h\circ f)(x)\\ \textrm{f}.&(h\circ f\circ g)(x) \end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|}\hline \begin{aligned}\textrm{a}.\quad \textrm{Misal}\: \: z(x)&=(f\circ g\circ h)(x)\\ &=(f\circ g)\left (x^{2}-1 \right )\\ &=f\left ( 3\left ( x^{2}-1 \right )+2 \right )\\ &=f\left ( 3x^{2}-1 \right )\\ &=\left ( 3x^{2}-1 \right )-4\\ &=3x^{2}-5\\ & \end{aligned}&\begin{aligned}\textrm{b}.\quad \textrm{Misal}\: \: y(x)&=(f\circ h\circ g)(x)\\ &=(f\circ h)\left (3x+2 \right )\\ &=f\left ( \left (3x+2 \right )^{2}-1 \right )\\ &=f\left ( 9x^{2}+12x+4-1 \right )\\ &=f\left ( 9x^{2}+12x+3 \right )\\ &=\left ( 9x^{2}+12x+3 \right )-4\\ &=9x^{2}+12x-1 \end{aligned} \\\hline \begin{aligned}\textrm{c}.\quad \textrm{Misal}\: \: w(x)&=(h\circ g\circ f)(x)\\ &=(h\circ g)\left (x-4 \right )\\ &=h\left ( 3\left ( x-4 \right )+2 \right )\\ &=h\left ( 3x-2 \right )\\ &=\left ( 3x-2 \right )^{2}-1\\ &=9x^{2}-12x+3 \end{aligned} &\begin{aligned}&\textrm{Untuk jawaban}\\ &\textrm{yang lain silahkan}\\ &\textrm{di coba sendiri}\\ &\textrm{sebagai latihan}\\ &\textrm{mandiri}\\ & \end{aligned} \\\hline \end{array} \end{array}.
\begin{array}{ll}\\ 4.&\textrm{Tentukanlah \textbf{invers} dari fungsi-fungsi berikut ini}\\ &\begin{array}{lllllll}\\ \textrm{a}.&f(x)=3x-4&\textrm{f}.&m(x)=\sqrt[5]{4x-1},\: x\neq \displaystyle \frac{1}{4}\\ \textrm{b}.&g(x)=5-3x&\textrm{g}.&n(x)=x^{3}-4\\ \textrm{c}.&j(x)=(x-2)^{2}&\textrm{h}&o(x)=\left ( 1-x^{3} \right )^{^{\frac{1}{3}}}-3\\ \textrm{d}.&k(x)=3x^{2}-2&\textrm{i}&p(x)=\displaystyle \frac{3x-2}{x-3},\: x\neq 3\\ \textrm{e}.&l(x)=x^{2}-4x+3&\textrm{j}&q(x-2)=\displaystyle \frac{x-4}{2x-4},\: x\neq 2 \end{array}\\\\ &\textrm{Jawab}:\\ &\textrm{Soal yang belum diselesaikan silahkan dicoba sendiri untuk latihan} \end{array}.
.\qquad \begin{array}{|l|l|l|}\hline \begin{aligned}\textrm{a}.\quad f(x)=y&=3x-4\\ 3x-4&=y\\ 3x&=y+4\\ x&=\displaystyle \frac{y+4}{3}\\ f^{-1}(y)&=\displaystyle \frac{y+4}{3}\\ f^{-1}(x)&=\displaystyle \frac{x+4}{3} \end{aligned}&\begin{aligned}\textrm{b}.\quad g(x)=y&=5-3x\\ 5-3x&=y\\ -3x&=y-5\\ x&=\displaystyle \frac{y-5}{-3}=\frac{5-y}{3}\\ g^{-1}(y)&=\displaystyle \frac{5-y}{3}\\ g^{-1}(x)&=\displaystyle \frac{5-x}{3} \end{aligned}&\begin{aligned}\textrm{c}.\quad j(x)=y&=(x-2)^{2}\\ (x-2)^{2}&=y\\ (x-2)&=y^{^{\frac{1}{2}}}=\pm \sqrt{y}\\ x&=\pm \sqrt{y}+2\\ j^{-1}(y)&=\pm \sqrt{y}+2\\ j^{-1}(x)&=\pm \sqrt{x}+2\\ \textrm{dengan}&\: \: x\geq 0\\ & \end{aligned}\\\hline \begin{aligned}\textrm{d}.\quad k(x)=y&=3x^{2}-2\\ 3x^{2}-2&=y\\ 3x^{2}&=y+2\\ x^{2}&=\displaystyle \frac{y+2}{3}\\ x&=\pm \sqrt{\displaystyle \frac{y+2}{3}}\\ k^{-1}(y)&=\pm \sqrt{\displaystyle \frac{y+2}{3}}\\ k^{-1}(x)&=\pm \sqrt{\displaystyle \frac{x+2}{3}}\\ \textrm{dengan}&\: x\geq -2\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{f}.\quad m(x)=y&=\sqrt[5]{4x-1}\\ \sqrt[5]{4x-1}&=y\\ 4x-1&=y^{5}\\ 4x&=y^{5}+1\\ x&=\displaystyle \frac{y^{5}+1}{4}\\ m^{-1}(y)&=\displaystyle \frac{y^{5}+1}{4}\\ m^{-1}(x)&=\displaystyle \frac{x^{5}+1}{4}\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{j}.\quad q(x-2)&=\displaystyle \frac{x-4}{2x-4}\\ q(x)&=q((x+2)-2)\\ q(x)=y&=\displaystyle \frac{(x+2)-4}{2(x+2)-4}\\ q(x)=y&=\displaystyle \frac{x-2}{2x},\: x\neq 0\\ 2xy&=x-2\\ 2xy-x&=-2\\ x(2y-1)&=-2\\ x&=\displaystyle \frac{-2}{2y-1}=\frac{2}{1-2y}\\ q^{-1}(y)&=\frac{2}{1-2y}\\ q^{-1}(x)&=\frac{2}{1-2x}\\ \textrm{dengan}&\: x\neq \displaystyle \frac{1}{2} \end{aligned}\\\hline \end{array}

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