Contoh Soal 2 Fungsi

$\begin{array}{l}\\ 6.& \text{Diketahui bahwa}\\ & f(x)=\{\begin{array}{ll}0,& \text{untuk}x<0\\ x^2,& \text{untuk}0\le x<1\\ 2x-1,& \text{untuk}x\ge 1\end{array}\\ & \text{Nilai dari}f(-1)+f\left({\displaystyle \frac{1}{2}}\right)-f(3)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -5{\displaystyle \frac{1}{4}}& & & \text{d}.& 4{\displaystyle \frac{3}{4}}\\ \\ \text{b}.& -4{\displaystyle \frac{3}{4}}& \text{c}.& 4& \text{e}.& 5{\displaystyle \frac{1}{4}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \\ f(x)& =\{\begin{array}{ll}0,& \text{untuk}x<0\\ x^2,& \text{untuk}0\le x<1\\ 2x-1,& \text{untuk}x\ge 1\end{array}\\ \text{maka nilai}& f(-1)+f\left({\displaystyle \frac{1}{2}}\right)-f(3)\\ & =0+{\left({\displaystyle \frac{1}{2}}\right)}^2-(2(3)-1)\\ & ={\displaystyle \frac{1}{4}-5}\\ & =-4{\displaystyle \frac{3}{4}}\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Jika diketahui}f(x+{\displaystyle \frac{1}{x}})=x^3+{\displaystyle \frac{1}{x^3},}\\ & \text{maka nilai dari}f\left({\displaystyle \frac{5}{2}}\right)\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 2{\displaystyle \frac{1}{8}}& & & \text{d}.& 8{\displaystyle \frac{1}{8}}\\ \\ \text{b}.& 2{\displaystyle \frac{1}{2}}& \text{c}.& 4{\displaystyle \frac{1}{8}}& \text{e}.& 12{\displaystyle \frac{1}{8}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Perhatikan}& \text{bahwa}:\\ (p+q{)}^3& =p^3+3p^{2}q+3p{q}^2+q^3\\ \text{Jika kita su}& \text{bstitusikan}p=x^3\text{dan}q={\displaystyle \frac{1}{x^3}}\\ {(x+{\displaystyle \frac{1}{x}})}^3& =x^3+3x^2\left({\displaystyle \frac{1}{x}}\right)+3x{\left({\displaystyle \frac{1}{x}}\right)}^2+{\left({\displaystyle \frac{1}{x}}\right)}^3\\ & =x^3+{\left({\displaystyle \frac{1}{x}}\right)}^3+3x+{\displaystyle \frac{3}{x}}\\ & =(x^3+{\displaystyle \frac{1}{x^3}})+3(x+{\displaystyle \frac{1}{x}})\\ \text{sehingga}& \\ f(x+{\displaystyle \frac{1}{x}})& =x^3+{\displaystyle \frac{1}{x^3}}\\ & ={(x+{\displaystyle \frac{1}{x}})}^3-3(x+{\displaystyle \frac{1}{x}})\\ f(u)& =u^3-3u,\text{maka}\\ f\left({\displaystyle \frac{5}{2}}\right)& ={\left({\displaystyle \frac{5}{2}}\right)}^3-3\left({\displaystyle \frac{5}{2}}\right)\\ & ={\displaystyle \frac{125}{8}-\frac{15}{2}}\\ & ={\displaystyle \frac{65}{8}}\\ & =8{\displaystyle \frac{1}{8}}\end{array}\end{array}$

$\begin{array}{l}\\ 8.& \text{Misal fungsi}f\text{terdefinisi untuk }\\ & \text{seluruh bilangan real}x.\\ & \text{Jika}f(p+q)=f(pq)\text{untuk semua}p,q\\ & \text{bilangan bulat positif dan}f(1)=2,\\ & \text{maka nilai}f(2021)=....\\ & \begin{array}{l}\\ \text{a}.& 0& & & \text{d}.& 3\\ \\ \text{b}.& 1& \text{c}.& 2& \text{e}.& 5\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}\\ f(1)& =2\text{dan}\\ f(p+q)& =f(pq)\\ \text{maka}& \\ f(2)& =f(1+1)=f(1.1)=f(1)=2\\ f(3)& =f(1+2)=f(1.2)=f(2)=f(1)=2\\ f(4)& =f(1+3)=f(1.3)=f(3)=f(2)=f(1)=2\\ f(5)& =f(1+4)=f(1.4)=f(4)=f(3)=f(2)=f(1)=2\\ ⋮& \\ f(2021)& =\cdots =\cdots =\cdots =f(2)=f(1)=2\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Jika}{a}_{{}_0}={\displaystyle \frac{2}{5}\text{dan}{a}_{n+1}=2\left|a_n\right|-1,\text{maka nilai}{a}_{{}_{2022}}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -0,6& & & \text{d}.& 0,4\\ \\ \text{b}.& -0,2& \text{c}.& 0,2& \text{e}.& 0,6\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\text{Diketahui}& \text{bahwa}\\ a_{{}_0}& ={\displaystyle \frac{2}{5}=0,4\text{dan}{a}_{n+1}=2\left|a_n\right|-1,\text{maka}}\\ a_{{}_1}& =2\left|a_{{}_0}\right|-1=2|0,4|-1=0,8-1=-0,2\\ a_{{}_2}& =2\left|a_{{}_1}\right|-1=2|-0,2|-1=2(0,2)-1\\ & =0,4-1=-0,6\\ a_{{}_3}& =2\left|a_{{}_2}\right|-1=2|-0,6|-1=1,2-1=0,2\\ a_{{}_4}& =2\left|a_{{}_3}\right|-1=2|0,2|-1=0,4-1=-0,6={\mathbf{\text{a}}}_{{}_{\mathbf{\text{2}}}}\\ a_{{}_5}& =2\left|a_{{}_4}\right|-1=2|-0,6|-1=1,2-1=0,2={\mathbf{\text{a}}}_{{}_{\mathbf{\text{3}}}}\\ a_{{}_6}& =2\left|a_{{}_5}\right|-1=2|0,2|-1=0,4-1=-0,6={\mathbf{\text{a}}}_{{}_{\mathbf{\text{2}}}}\\ a_{{}_7}& =2\left|a_{{}_6}\right|-1=2|-0,6|-1=1,2-1=0,2={\mathbf{\text{a}}}_{{}_{\mathbf{\text{3}}}}\\ ⋮& \\ a_{{}_{2022}}& =\cdots =\cdots ={\mathbf{\text{a}}}_{{}_{\mathbf{\text{2}}}}=-0,6\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Kurva}f(x)={\displaystyle \frac{10}{x^2-10x+25}}\\ & \text{mempunyai asimtot vertikal pada}....\\ & \begin{array}{l}\\ \text{a}.& x=0\text{saja}& & & \\ \text{b}.& x=5\text{saja}& \\ \text{c}.& x=10\text{saja}\\ \text{d}.& x=0\text{dan}x=5\text{saja}\\ \text{e}.& x=0,x=5,\text{dan}x=10\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}\mathbf{\text{Asimtot vertikal}}& (\mathbf{\text{tegak}})\text{diperoleh saat}\\ x^2-10x+25& =0\\ (x-5{)}^2& =0\\ x-5& =0\\ x& =5\\ \\ \text{Ilustrasinya gamba}& \text{rnya adalah sebagai berikut}:\end{array}\end{array}.$