Contoh Soal Fungsi

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

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

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

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

$\begin{array}{ll}\\ 4.&\textrm{Dari beberapa fungsi berikut yang }\\ &\textrm{merupakan fungsi genap adalah}\: ....\\ &\begin{array}{ll}\\ \color{red}\textrm{a}.&f(x)=x^{2}+\left | x \right |-1\\ \textrm{b}.&f(x)=x^{3}-\left | x \right |+x\\ \textrm{c}.&f(x)=x\left | x \right |+x\\ \textrm{d}.&f(x)=\sqrt{x-1}\\ \textrm{e}.&f(x)=4-2x \end{array}\\\\ &\color{blue}\textrm{Jawab}:\\ &\begin{aligned}&\\ \textrm{Suatu fungsi}&\: \color{red}\textrm{dinamakan fungsi genap}\\ &\textrm{jika}\: f(x)=f(-x)\\ & \end{aligned}\\ &\begin{array}{|c|l|l|c|}\hline \textrm{No}&\qquad f(x)&\qquad f(-x)&\textrm{Keterangan}\\\hline 4.\textrm{a}&x^{2}+\left | x \right |-1&x^{2}+\left | x \right |-1&f(x)=f(-x)\\\hline 4.\textrm{b}&x^{3}-\left | x \right |+x&-x^{3}-\left | x \right |-x&f(x)\neq f(-x)\\\hline 4.\textrm{c}&x\left | x \right |+x&-x\left | x \right |-x&f(x)\neq f(-x)\\\hline 4.\textrm{d}&\sqrt{x-1}&\sqrt{-x-1}&f(x)\neq f(-x)\\\hline 4.\textrm{e}.&4-2x&4+2x&f(x)\neq f(-x)\\\hline \end{array} \end{array}$

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