Contoh Soal Fungsi Komposisi dan Fungsi Invers

\begin{array}{ll}\\ \fbox{1}.&\textrm{Grafik berikut yang merupakan fungsi}\: \: y=f(x)\: ,\: \textrm{yaitu}:\end{array}.

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\begin{array}{ll}\\ \: .&\textrm{Pernyataan berikut yang benar adalah}....\\ &\begin{array}{ll}\\ \textrm{a}.\quad (1),(2),\: \textrm{dan}\: (3)\: \: \textrm{saja}.\\ \textrm{b}.\quad (1),\: \textrm{dan}\: (3)\: \: \textrm{saja}.\\ \textrm{c}.\quad (2),\: \textrm{dan}\: (4)\: \: \textrm{saja}.\\ \textrm{d}.\quad (4)\: \: \textrm{saja}.\\ \textrm{e}.\quad \textrm{Benar semuanya}.\\ \end{array}\end{array}\\\\\\ \textrm{Jawab}:\textbf{b}\\\\ \begin{aligned}&\textrm{Yang perlu Anda lakukan adalah cukup dengan menarik}\\ &\textrm{sembarang garis yang sejajar dengan sumbu Y. Jika garis}\\ &\textrm{tersebut memotong grafik lebih dari 1 titik, maka grafik}\\ &\textrm{tersebut bukanlah garfik fungsi yang dimaksud}. \end{aligned}.

\begin{array}{ll}\\ \fbox{2}.&\textrm{Diketahui}\: \: f(x)=x+1.\: \: \textrm{Nilai}\: \: f(2x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2f(x)&&\textrm{d}.\quad f(x)+2\\ \textrm{b}.\quad 2f(x)-1&\textrm{c}.\quad 1-2f(x)&\textrm{e}.\quad 2-f(x) \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{b}\\\\ \begin{aligned}f(x)&=x+1\\ f(2x)&=(2x)+1\\ f(2x)&=2x+2-1\\ &=2(x+1)-1\\ &=2f(x)-1 \end{aligned}.

\begin{array}{ll}\\ \fbox{3}.&\textrm{Diketahui}\\\\ &f(x)=\begin{cases} 2x-1 & \textrm{untuk}\: \: \: 0<x<1 \\ x^{2}+1 & \textrm{untuk}\: \: x\: \: \textrm{yang lainnya} \end{cases}\\\\ &\textrm{Nilai}\: \: f(2).f(-4)+f(\frac{1}{2}).f(3)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 52&&\textrm{d}.\quad 105\\ \textrm{b}.\quad 55&\textrm{c}.\quad 85&\textrm{e}.\quad 210 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{aligned}f(2).f(-4)+f(\frac{1}{2}).f(3)&=\left ( 2^{2}+1 \right ).\left ( (-4)^{2}+1 \right )+\left ( 2\left ( \displaystyle \frac{1}{2} \right )-1 \right ).\left ( 3^{2}+1 \right )\\ &=5.17+0.10\\ &=85 \end{aligned}.

\begin{array}{ll}\\ \fbox{4}.&\textrm{Diketahui}\: \: f(x)=3^{x}.\: \: \textrm{untuk tiap nilai}\: \: x\: \: \textrm{berlaku}\\\\ &\textrm{Nilai}\: \: f(x+1)-f(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad f(x)&&\textrm{d}.\quad f(x-1)\\ \textrm{b}.\quad 2f(x)&\textrm{c}.\quad 3f(x)&\textrm{e}.\quad 3f(x+1) \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{b}\\\\ \begin{aligned}f(x+1)-f(x)&=3^{(x+1)}-3^{x}\\ &=3^{x}.3-3^{x}\\ &=2.3^{x}\\ &=2f(x) \end{aligned}.

\begin{array}{ll}\\ \fbox{5}.&\textrm{Diketahui}\: \: f(x)=\displaystyle \frac{x}{x-1}.\: \textrm{Nilai}\: \: f\left ( 3x \right )\: \: \textrm{jika dinyatakan dalam}\: \: f(x)\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad \displaystyle \frac{3f(x)}{2f(x)-1}&&\textrm{d}.\quad \displaystyle \frac{3f(x)}{f(x)+1}\\ \textrm{b}.\quad \displaystyle \frac{f(x)}{2f(x)+1}&\textrm{c}.\quad \displaystyle \frac{3f(x)}{2f(x)+5}&\textrm{e}.\quad \displaystyle \frac{3f(x)}{2f(x)+1} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{e}\\\\ \begin{aligned}f(3x)&=\displaystyle \frac{3x}{3x-1}\times \displaystyle \frac{\displaystyle \frac{1}{x-1}}{\displaystyle \frac{1}{x-1}}\\ &=\displaystyle \frac{\displaystyle \frac{3x}{x-1}}{\displaystyle \frac{3x-1}{x-1}}\\ &=\displaystyle \frac{3\left ( \displaystyle \frac{x}{x-1} \right )}{\displaystyle \frac{2x}{x-1}+\frac{x-1}{x-1}}\\ &=\displaystyle \frac{3f(x)}{2f(x)+1} \end{aligned}.

\begin{array}{ll}\\ \fbox{6}.&\textrm{Jika}\: \: f\left ( \displaystyle \frac{x-1}{2} \right )=2x+3,\: \: \textrm{maka nilai}\: \: f\left ( \displaystyle x-\frac{1}{4} \right )=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 4x+1&&\textrm{d}.\quad x-1\\ \textrm{b}.\quad 4(x+1)&\textrm{c}.\quad x+4&\textrm{e}.\quad x-4 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{b}\\\\ \begin{aligned}f\left ( \displaystyle \frac{x-1}{2} \right )&=2x+3\\ f\left ( \displaystyle \frac{(2x)-1}{2} \right )&=2(2x)+3\\ f\left ( \displaystyle x-\frac{1}{2} \right )&=4x+3\\ f\left ( \displaystyle \left ( x+\displaystyle \frac{1}{2} \right )-\frac{1}{2} \right )&=4\left ( x+\displaystyle \frac{1}{2} \right )+3\\ f(x)&=4x+5\\ f\left ( \displaystyle x-\frac{1}{4} \right )&=4\left ( \displaystyle x-\frac{1}{4} \right )+5\\ &=4x-1+5\\ &=4x+4\\ &=4(x+1) \end{aligned}.

\begin{array}{ll}\\ \fbox{7}.&\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 \end{array}\\\\\\ \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}.

\begin{array}{ll}\\ \fbox{8}.&\textbf{(OSK)}\textrm{Sebuah fungsi memenuhi persamaan}\: \: f\left ( \displaystyle \frac{1}{x} \right )+\displaystyle \frac{1}{x}f(-x)=2x,\\ &\textrm{untuk setiap bilangan real}\: \: x\: \: \textrm{dengan}\: \: x\neq 0. \: \: \textrm{Nilai}\: \: f(2)=....\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\textrm{Diketahui bahwa}:&\\ f\left ( \displaystyle \frac{1}{x} \right )+\displaystyle \frac{1}{x}f(-x)&=2x\\ \textrm{maka},\qquad\qquad\quad \: \: &\\ x.f\left ( \displaystyle \frac{1}{x} \right )+f(-x)&=2x^{2},\quad \textrm{saat dikalikan dengan}\: \: x,\quad \left ( x\neq 0 \right )\\ &............\textcircled{1}\\ f\left ( \displaystyle -x \right )-\displaystyle xf\left ( \displaystyle \frac{1}{x} \right )&=-\displaystyle \frac{2}{x},\quad \textrm{saat dikalikan dengan}\: \: \displaystyle \frac{1}{x}=-x,\quad \left ( x\neq 0 \right )\\ &............\textcircled{2}\\ \textrm{Saat persamaan}\: \textcircled{1}&\: \textrm{dan}\: \textcircled{2}\: \textrm{dijumlahklan, maka akan diperoleh}\\ 2f(-x)&=2x^{2}-\displaystyle \frac{2}{x}\\ f(-x)&=x^{2}-\displaystyle \frac{1}{x}\\ f(-(-2))&=(-2)^{2}-\displaystyle \frac{1}{(-2)}\\ f(2)&=4+\displaystyle \frac{1}{2}\\ &=4\displaystyle \frac{1}{2} \end{aligned}.

\begin{array}{ll}\\ \fbox{9}.&\textbf{(OSK)}\textrm{Jika}\: \: x\: \: \textrm{dan}\: \: y\: \: \textrm{adalah bilangan real sedemikian sehingga}\: \: \displaystyle \left \lfloor \sqrt{x} \right \rfloor=9\: \: \textrm{dan}\: \: \displaystyle \left \lfloor \sqrt{y} \right \rfloor=12\: ,\\ &\textrm{maka nilai terkecil dari}\: \: \displaystyle \left \lfloor y-x \right \rfloor=....\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\textrm{Diketahui bahwa}\: &\: \left \lfloor x \right \rfloor\: \: \textrm{adalah bilangan bulat terbesar yang lebih kecil atau sama dengan}\: \: x\\ \textrm{Misal}\: \: \left \lfloor 3,2 \right \rfloor=3, \: &\left \lfloor -2,47 \right \rfloor=-3,\: \: \left \lfloor 6 \right \rfloor=6,\: \: \textrm{dan lain-lain}.\\ \textrm{Sehingga}\, \,\left \lfloor x \right \rfloor=a&\Rightarrow a\leq x< a+1\: \: (\textrm{dengan}\: \: a\: \: \textrm{bilangan bulat}),\: \: \textrm{maka}\\ &\begin{cases} \left \lfloor \sqrt{x} \right \rfloor=9 & \Rightarrow 9\leq \sqrt{x}< 9+1\\ &\Leftrightarrow 9\leq \sqrt{x}< 10\Leftrightarrow 81\leq x< 100 \\ \left \lfloor \sqrt{y} \right \rfloor=12 & 12\leq \sqrt{y}< 12+1\\ &\Leftrightarrow 12\leq \sqrt{y}< 13\Leftrightarrow 144\leq y< 169 \end{cases}\\ \end{aligned}\\\\\\ \begin{aligned}\bullet \: \: 144\leq y< 169&\: \: \textrm{dan}\\ \bullet \: \: \: \: 81\leq x< 100&\: ,\: \textrm{dikalikan dengan}\: \: (-1)\: \: \textrm{maka akan menjadi},\: -100< -x\leq -81,\\ \textrm{maka}\: \: &-99,\overline{999}\leq -x< -80,\overline{999}\\ \textrm{Selanjutnya}&\\ &\begin{array}{cll}\\ 144&\leq y< 169&\\ -99,\overline{999}&\leq -x< -80,\overline{999}&+\\\cline{1-2}\\ 44,...&\leq y-x< 88,...\\ \multicolumn{3}{l}{\begin{aligned}\textrm{atau}\: \: \: \: \: \: \: \: \: \: \: &\\ \left \lfloor y-x \right \rfloor&=44 \end{aligned}}\\ \end{array}\\\\ \textrm{Jadi, nilai terkecil}&\: \left \lfloor y-x \right \rfloor=44\\ & \end{aligned}.

\begin{array}{ll}\\ \fbox{10}.&\textbf{(OSP)}\textrm{Jika}\: \: y=f(x)\: \: \textrm{adalah fungi yang memenuhi persamaan}\: \: \displaystyle \frac{x}{\left | x \right |}+\displaystyle \frac{\left | y \right |}{y}=2y\\ &\textrm{maka daerah hasil dari fungsi tersebut adalah}\: ....\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}\textrm{Diketahui bahwa}:&\\ &\displaystyle \frac{x}{\left | x \right |}+\displaystyle \frac{\left | y \right |}{y}=2y\\ \textrm{jelas bahwa}\: \: &x,\: y\: \neq 0,\: \: \textrm{sehingga untuk} \end{aligned}}\\\hline x,\: y> 0\: \: \begin{cases} \left | x \right | &=x \\ \left | y \right | & =y \end{cases}&x> 0,\: y< 0\: \: \begin{cases} \left | x \right | &=x \\ \left | y \right | & =-y \end{cases}\\\hline \begin{aligned}\displaystyle \frac{x}{\left | x \right |}+\displaystyle \frac{\left | y \right |}{y}&=2y\\ \displaystyle \frac{x}{x}+\frac{y}{y}&=2y\\ 1+1&=2y\\ 1&=y\\ \textrm{memenu}&\textrm{hi} \end{aligned}&\begin{aligned}\displaystyle \frac{x}{\left | x \right |}+\displaystyle \frac{\left | y \right |}{y}&=2y\\ \displaystyle \frac{x}{x}+\frac{-y}{y}&=2y\\ 1-1&=2y\\ 0&=y\\ \textrm{tidak m}&\textrm{emenuhi} \end{aligned} \\\hline x,\: y< 0\: \: \begin{cases} \left | x \right | &=-x \\ \left | y \right | & =-y \end{cases}& x< 0,\: y> 0\: \: \begin{cases} \left | x \right | &=-x \\ \left | y \right | & =y \end{cases}\\\hline \begin{aligned}\displaystyle \frac{x}{\left | x \right |}+\displaystyle \frac{\left | y \right |}{y}&=2y\\ \displaystyle \frac{x}{-x}+\frac{-y}{y}&=2y\\ -1-1&=2y\\ -1&=y\\ \textrm{memenu}&\textrm{hi} \end{aligned} &\begin{aligned}\displaystyle \frac{x}{\left | x \right |}+\displaystyle \frac{\left | y \right |}{y}&=2y\\ \displaystyle \frac{x}{-x}+\frac{y}{y}&=2y\\ -1+1&=2y\\ 0&=y\\ \textrm{tidak m}&\textrm{emenuhi} \end{aligned} \\\hline \multicolumn{2}{|c|}{\textrm{Jadi, daerah hasil yang memenuhi}:\left \{ -1,\: 1 \right \}}\\\hline \end{array}.

 

 

Tentang ahmadthohir1089

Nama saya Ahmad Thohir asli orang Purwodadi, Jawa Tengah, lahir di Grobogan 02 Februari 1980. Pendidikan : Tingkat dasar lulus dari MI Nahdlatut Thullab di desa Manggarwetan,kecamatan Godong lulus tahun 1993. dan untuk tingkat menengah saya tempuh di MTs Nahdlatut Thullab Manggar Wetan lulus tahun 1996. Sedang untuk tingkat SMA saya menamatkannya di MA Futuhiyyah-2 Mranggen, Demak lulus tahun 1999. Setelah itu saya Kuliah di IKIP PGRI Semarang pada fakultas FPMIPA Pendidikan Matematika lulus tahun 2004. Pekerjaan : Sebagai guru (PNS DPK Kemenag) mapel matematika di MA Futuhiyah Jeketro, Gubug. Pengalaman mengajar : 1. GTT di MTs Miftahul Mubtadiin Tambakan Gubug tahun 2003 s/d 2005 2. GTT di SMK Negeri 3 Semarang 2005 s/d 2009 3. GT di MA Futuhiyah Jeketro Gubug sejak 1 September 2009
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