Contoh Soal 3 Fungsi Komposisi dan Fungsi Invers

 $\begin{array}{l}\\ 11.& \text{Diketahui beberapa fungsi memiliki }\\ & \text{sifat-sifat sebagaimana berikut ini}:\\ & (i)\mathrm{\Phi }(-x)=-\mathrm{\Phi }(x)\text{untuk setiap}x\\ & (ii)\mathrm{\Phi }(-x)=\mathrm{\Phi }(x)\text{untuk setiap}x\\ & \text{Jika diketahui fungsi}f\text{dan}g\\ & \text{memiliki sifat}(i)\text{dan fungsi}h\text{dan}k\\ & \text{memiliki sifat}(ii),\text{maka pernyataan }\\ & \text{berikut yang salah adalah}....\\ & (1)(f+g)(-x)=-(f+g)(x)\\ & (2)(f.k)(-x)=-(f.k)(x)\\ & (3)(h-k)(-x)=(h-k)(x)\\ & (4)(h-g)(-x)=(h-g)(x)\\ & \begin{array}{l}\\ \text{a}.& (1),(2)\text{dan}(3)\\ \text{b}.& (1)\text{dan}(3)\\ \text{c}.& (2)\text{dan}(4)\\ \text{d}.& (4)\text{saja}\\ \text{e}.& \text{semuanya benar}\end{array}\\ & (\mathbf{\text{SIMAK UI 2014 Mat Das}})\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui}\text{bahwa}:\\ & \{\begin{array}{l}\mathrm{\Phi }(-x)=-\mathrm{\Phi }(x)\\ (\text{fungsi ganjil})\{\begin{array}{ll}f& \text{ misal }f(x)=x\\ g& \text{ misal }g(x)=2x\end{array}\\ \\ \mathrm{\Phi }(-x)=\mathrm{\Phi }(x)\\ (\text{fungsi genap})\{\begin{array}{ll}h& \text{ misal }h(x)=x^2\\ k& \text{ misal }k(x)=2x^2\end{array}\end{array}\\ & \end{array}\\ & \begin{array}{|c|}\hline \begin{array}{rl}(1)(f+g)(-x)& =-(f+g)(x)\\ & \text{benar}\\ (2)(f.k)(-x)& =-(f.k)(x)\\ & \text{benar}\end{array}\\ \begin{array}{rl}(3)(h-k)(-x)& =(h-k)(x)\\ & \text{benar}\\ (4)(h-g)(-x)& =(h-g)(x)\\ & \text{salah}\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Jika}f(x)={\displaystyle \frac{1}{2x-1}\text{dan}(f\circ g)(x)={\displaystyle \frac{x}{3x-2}}}\\ & \text{maka}g(x)=....\\ & (\mathbf{\text{UMPTN 1998}})\\ & \begin{array}{l}\\ \text{a}.& x+{\displaystyle \frac{1}{2}}& & & \text{d}.& 1-{\displaystyle \frac{2}{x}}\\ \\ \text{b}.& x-{\displaystyle \frac{1}{2}}& \text{c}.& 2-{\displaystyle \frac{1}{x}}& \text{e}.& 2-{\displaystyle \frac{1}{2x}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 1}}\\ & \text{Diketahui bahwa}f(x)={\displaystyle \frac{1}{2x-1},\text{dan}}\\ & (f\circ g)(x)={\displaystyle \frac{x}{3x-2},\text{maka}}\\ & \iff (f\circ g)(x)=f(g(x))={\displaystyle \frac{x}{3x-2}}\\ & \iff {\displaystyle \frac{1}{2g(x)-1}={\displaystyle \frac{x}{3x-2}}}\\ & \iff {\displaystyle \frac{1}{2g(x)-1}={\displaystyle \frac{1}{\left({\displaystyle \frac{3x-2}{x}}\right)}}}\\ & \text{Dari bentuk di atas didapatkan}\\ & 2g(x)-1={\displaystyle \frac{3x-2}{x}}\\ & \iff 2g(x)=1+(3-{\displaystyle \frac{2}{x}})\\ & \iff 2g(x)=4-{\displaystyle \frac{2}{x}}\\ & \iff g(x)=2-{\displaystyle \frac{1}{x}}\end{array}\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 2}}\\ & \text{Diketahui bahwa}f(x)={\displaystyle \frac{1}{2x-1},\text{dengan}}\\ & f^{-1}(x)={\displaystyle \frac{x+1}{2x}.........(\text{tunjukkan sendiri})}\\ & \text{serta}(f\circ g)(x)={\displaystyle \frac{x}{3x-2},\text{maka}}\\ & g(x)=(f^{-1}\circ f\circ g)(x)=(I\circ g)(x)=g(x)\\ & \iff g(x)={\displaystyle \frac{f(g(x))+1}{2(f(g(x)))}}\\ & \iff g(x)={\displaystyle \frac{\left({\displaystyle \frac{x}{3x-2}}\right)+1}{2\left({\displaystyle \frac{x}{3x-2}}\right)}}\\ & \iff g(x)={\displaystyle \frac{{\displaystyle \frac{4x-2}{3x-2}}}{{\displaystyle \frac{2x}{3x-2}}}}\\ & \iff g(x)={\displaystyle \frac{4x-2}{2x}}\\ & \iff g(x)=2-{\displaystyle \frac{1}{x}}\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Jika}f(x)=x^2-9\text{dan}(f\circ g)(x)=x(x-6)\\ & \text{rumus fungsi}g(x)=....\\ & \begin{array}{l}\\ \text{a}.& x+3& & & \text{d}.& 3x+1\\ \text{b}.& x-3& \text{c}.& -x& \text{e}.& x\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 1}}\\ & \text{Diketahui bahwa}f(x)=x^2-9,\text{dan}\\ & (f\circ g)(x)=x(x-6)=x^2-6x,\text{maka}\\ & \iff (f\circ g)(x)=f(g(x))=x^2-6x\\ & \iff {(g(x))}^2-9=x^2-6x\\ & \iff {(g(x))}^2-9=x^2-6x+9-9\\ & \iff {(g(x))}^2-9=(x-3{)}^2-9\\ & \text{Dari bentuk di atas didapatkan}\\ & g(x)=x-3\end{array}\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 2}}\\ & \text{Diketahui bahwa}f(x)=x^2-9,\text{dengan}\\ & f^{-1}(x)=\sqrt{x+9}.......(\text{tunjukkan sendiri})\\ & \text{serta}(f\circ g)(x)=x^2-6x,\text{maka}\\ & g(x)=(f^{-1}\circ f\circ g)(x)=(I\circ g)(x)=g(x)\\ & \iff g(x)=\sqrt{(f(g(x)))+9}\\ & \iff g(x)=\sqrt{x^2-6x+9}\\ & \iff g(x)=\sqrt{(x-3{)}^2}\\ & \iff g(x)=x-3\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Jika}f(x)={\displaystyle \frac{1}{\sqrt{x^2-2}}\text{dan}}\\ & (f\circ g)(x)={\displaystyle \frac{1}{\sqrt{x^2+6x+7}},}\\ & \text{maka}g(x+2)=....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{x+3}}& & & \text{d}.& x+3\\ \text{b}.& {\displaystyle \frac{1}{x-2}}& \text{c}.& x-2& \text{e}.& x+5\end{array}\\ & (\mathbf{\text{UM UGM 2010 Mat Das}})\\ \\ & \text{Jawab}:\\ & \mathbf{\text{Alternatif 1}}\\ & \begin{array}{rl}(f\circ g)(x)& ={\displaystyle \frac{1}{\sqrt{x^2+6x+7}}}\\ f(g(x))& ={\displaystyle \frac{1}{\sqrt{x^2+6x+7}}}\\ {\displaystyle \frac{1}{\sqrt{{(g(x))}^2-2}}}& ={\displaystyle \frac{1}{\sqrt{x^2+6x+7}}}\\ {(g(x))}^2-2& =x^2+6x+7\\ {(g(x))}^2& =x^2+6x+9\\ g(x)& =\sqrt{x^2+6x+9}=\sqrt{(x+3{)}^2}\\ g(x)& =x+3\\ g(x+2)& =(x+2)+3\\ & =x+5\end{array}\\ & \begin{array}{rl}& \mathbf{\text{Alternatif 2}}\\ & \text{Diketahui bahwa}f(x)={\displaystyle \frac{1}{\sqrt{x^2-2}},\text{dengan}}\\ & f^{-1}(x)=\sqrt{{\displaystyle \frac{1}{x^2}+2}}.......(\text{akan ditunjukkan})\\ & \text{serta}(f\circ g)(x)={\displaystyle \frac{1}{\sqrt{x^2+6x+7}},\text{maka}}\\ & \begin{array}{|cc|}\hline \begin{array}{rl}f(x)=y& ={\displaystyle \frac{1}{\sqrt{x^2-2}}}\\ y^2& ={\displaystyle \frac{1}{x^2-2}}\\ x^2-2& ={\displaystyle \frac{1}{y^2}}\\ x^2& ={\displaystyle \frac{1}{y^2}+2}\\ x& =\sqrt{{\displaystyle \frac{1}{y^2}+2}}\\ f^{-1}(y)& =\sqrt{{\displaystyle \frac{1}{y^2}+2}}\\ f^{-1}(x)& =\sqrt{{\displaystyle \frac{1}{x^2}+2}}\end{array}& \begin{array}{rl}g(x)& =(f^{-1}\circ f\circ g)(x)\\ & =\sqrt{{\displaystyle \frac{1}{{\left({\displaystyle \frac{1}{\sqrt{x^2+6x+7}}}\right)}^2}+2}}\\ & =\sqrt{(x^2+6x+7)+2}\\ & =\sqrt{(x^2+6x+9)}\\ & =\sqrt{(x+3{)}^2}\\ & =x+3\\ g(x+2)& =(x+2)+3\\ & =x+5\\ & \end{array}\\ \hline\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Jika}g(x)=2x+4\text{dan}(f\circ g)(x)=4x^2+8x-3,\\ & \text{maka}{f}^{-1}(x)=....\\ & \begin{array}{l}\\ \text{a}.& x+9\\ \text{b}.& \sqrt{x}+2\\ \text{c}.& x^2-4x-3\\ \text{d}.& \sqrt{x+1}+2\\ \text{e}.& \sqrt{x+7}+2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{|ccc|}\hline \text{Sintak 1}& \text{ Sintak 2}& \text{Hasil Invers}\\ \begin{array}{rl}g(x)=y& =2x+4\\ y-4& =2x\\ x& ={\displaystyle \frac{y-4}{2}}\\ f^{-1}(y)& ={\displaystyle \frac{y-4}{2}}\\ f^{-1}(x)& ={\displaystyle \frac{x-4}{2}}\\ & \end{array}& \begin{array}{rl}f(x)& =(f\circ g\circ g^{-1})(x)\\ & =4{(g^{{}^{-1}}(x))}^2+8(g^{-1}(x))-3\\ & =4{\left({\displaystyle \frac{x-4}{2}}\right)}^2+8\left({\displaystyle \frac{x-4}{2}}\right)-3\\ & =\left({\displaystyle x^2-8x+16}\right)+4x-16-3\\ & =x^2-4x-3\\ & =x^2-4x+4-7\\ & =(x-2{)}^2-7\end{array}& \begin{array}{rl}f(x)=y& =(x-2{)}^2-7\\ y+7& =(x-2{)}^2\\ \sqrt{y+7}& =(x-2)\\ (x-2)& =\sqrt{y+7}\\ x& =\sqrt{y+7}+2\\ f^{-1}(y)& =\sqrt{y+7}+2\\ f^{-1}(x)& =\sqrt{x+7}+2\end{array}\\ \hline\end{array}\end{array}$