Contoh Soal Fungsi

$\begin{array}{l}\\ 1.& \text{Relasi berikut yang akan berupa fungsi adalah}....\\ & \begin{array}{l}\\ \text{a}.& f(x)=\sqrt{x}& & & \\ \text{b}.& f(x)=1-\sqrt{x}& \\ \text{c}.& f(x)=\sqrt{x}+1& \\ \text{d}.& f(x)=\sqrt{x}-1\\ \text{e}.& f(x)=\left|x\right|\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{|clc|}\hline \text{No}& \text{Fungsi}& \text{Grafik}\\ 1.\text{a}& f(x)=y=\sqrt{x}& y^2=x\\ 1.\text{b}& f(x)=y=1-\sqrt{x}& (1-y{)}^2=x\\ 1.\text{c}& f(x)=y=1+\sqrt{x}& (y-1{)}^2=x\\ 1.\text{d}& f(x)=y=\sqrt{x}-1& (y+1{)}^2=x\\ 1.\text{e}& f(x)=y=\left|x\right|& y=\{\begin{array}{ll}x& \text{ jika}x\ge 0\\ -x& \text{ jika }x<0\end{array}\\ \hline\end{array}\\ & \begin{array}{rl}& \text{Dengan prepeta yang berbeda}\\ & \text{akan menghasilkan peta yang}\\ & \text{berbeda pula (fungsi bijektif)}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Fungsi dari himpunan A ke himpunan B }\\ & \text{berikut termasuk jenis fungsi}\end{array}$.

$\begin{array}{l}\\ ..& \text{Relasi berikut yang akan berupa fungsi adalah}....\\ & \text{a}.\text{fungsi umum}\\ & \text{b}.\text{fungsi satu-satu, tetapi bukan fungsi pada}\\ & \text{c}.\text{fungsi pada, tetapi bukan fungsi satu-satu}\\ & \text{d}.\text{fungsi pada dan satu-satu}\\ & \text{e}.\text{tidak ada jawaban yang benar}\\ \\ & \text{Jawab}:\\ & \begin{array}{|ccl|}\hline \text{No}& \text{Keterangan}& \text{Alasan}\\ 2.\text{a}& \text{Sesuai}& \text{Sesuai definisi fungsi}\\ 2.\text{b}& \text{Salah}& \begin{array}{rl}& \text{Karena bukan fungsi }\\ & \text{satu-satu(fungsi injektif)}\\ & \text{walau benar dikatakan bukan}\\ & \text{fungsi pada (fungsi surjektif)}\end{array}\\ 2.\text{c}& \text{Salah}& \begin{array}{rl}& \text{Karena bukan fungsi }\\ & \text{pada(fungsi surjektif)}\\ & \text{walau benar dikatakan bukan }\\ & \text{fungsi satu-satu (fungsi injektif)}\end{array}\\ 2.\text{d}& \text{Salah}& \begin{array}{rl}& \text{Jelas bukan fungsi pada }\\ & \text{dan satu-satu(fungsi bijektif)}\end{array}\\ 2.\text{e}& \text{Salah}& \text{Tidak sesuai}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Himpunan pasangan terurut }\\ & \text{yang ditunjukkan oleh fungsi}\\ & f:x\mapsto 2-{(x+1)}^2\\ & \text{dari domain}\{-1,0,1,2\}\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& \{(-1,2),(0,3),(1,5),(2,7)\}\\ \text{b}.& \{(-1,2),(0,1),(1,-2),(2,-7)\}\\ \text{c}.& \{(-1,1),(0,-1),(1,-4),(2,7)\}\\ \text{d}.& \{(-1,0),(0,3),(1,-2),(2,7)\}\\ \text{e}.& \{(-1,0),(0,-4),(1,5),(2,-7)\}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}f:x& \mapsto 2-{(x+1)}^2\\ -1& \mapsto 2-{(-1+1)}^2=2-0=2& ....(-1,2)\\ 0& \mapsto 2-{(0+1)}^2=2-1=1& ....(0,1)\\ 1& \mapsto 2-{(1+1)}^2=2-4=-2& ....(1,-2)\\ 2& \mapsto 2-{(2+1)}^2=2-9=-7& ....(2,-7)\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Dari beberapa fungsi berikut yang }\\ & \text{merupakan fungsi genap adalah}....\\ & \begin{array}{l}\\ \text{a}.& f(x)=x^2+\left|x\right|-1\\ \text{b}.& f(x)=x^3-\left|x\right|+x\\ \text{c}.& f(x)=x\left|x\right|+x\\ \text{d}.& f(x)=\sqrt{x-1}\\ \text{e}.& f(x)=4-2x\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \\ \text{Suatu fungsi}& \text{dinamakan fungsi genap}\\ & \text{jika}f(x)=f(-x)\\ & \end{array}\\ & \begin{array}{|cllc|}\hline \text{No}& f(x)& f(-x)& \text{Keterangan}\\ 4.\text{a}& x^2+\left|x\right|-1& x^2+\left|x\right|-1& f(x)=f(-x)\\ 4.\text{b}& x^3-\left|x\right|+x& -x^3-\left|x\right|-x& f(x)\ne f(-x)\\ 4.\text{c}& x\left|x\right|+x& -x\left|x\right|-x& f(x)\ne f(-x)\\ 4.\text{d}& \sqrt{x-1}& \sqrt{-x-1}& f(x)\ne f(-x)\\ 4.\text{e}.& 4-2x& 4+2x& f(x)\ne f(-x)\\ \hline\end{array}\end{array}$

$\begin{array}{l}\\ 5.& \text{Diketahui himpunan}\\ & A=\left\{x|x\text{adalah faktor prima dari}16\right\}\\ & B=\left\{x|x\text{adalah faktor dari}16\right\}\\ & \text{Banyaknya pemetaan dari}AkeB\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1& & & \text{d}.& 25\\ \\ \text{b}.& 2& \text{c}.& 5& \text{e}.& 32\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}A& =\left\{x|x\text{adalah faktor prima dari}16\right\}\\ & =\left\{2\right\}\Rightarrow n(A)=1\\ B& =\left\{x|x\text{adalah faktor dari}16\right\}\\ & =\{1,2,4,8,16\}\Rightarrow n(B)=5\\ \text{B}& \text{anyaknya pemetaan dari}A\text{ke}B\text{adalah}:\\ & =n(B{)}^{n(A)}\\ & =5^1\\ & =5\end{array}\end{array}$