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Lanjutan Limit Fungsi

B. Sifat-Sifat Limit Fungsi Aljabar

\begin{array}{ll}\\ 1.&\underset{x\rightarrow a}{\textrm{lim}}\: k=k\\ 2.&\underset{x\rightarrow a}{\textrm{lim}}\: x=a\\ 3.&\underset{x\rightarrow a}{\textrm{lim}}\: k.f(x)=k.\underset{x\rightarrow a}{\textrm{lim}}\: f(x)\\ 4.&\underset{x\rightarrow a}{\textrm{lim}}\: \left ( f(x)+g(x) \right )=\underset{x\rightarrow a}{\textrm{lim}}\: f(x)+\underset{x\rightarrow a}{\textrm{lim}}\: g(x)\\ 5.&\underset{x\rightarrow a}{\textrm{lim}}\: \left ( f(x)-g(x) \right )=\underset{x\rightarrow a}{\textrm{lim}}\: f(x)-\underset{x\rightarrow a}{\textrm{lim}}\: g(x)\\ 6.&\underset{x\rightarrow a}{\textrm{lim}}\: \left ( f(x)\times g(x) \right )=\underset{x\rightarrow a}{\textrm{lim}}\: f(x)\times \underset{x\rightarrow a}{\textrm{lim}}\: g(x)\\ 7.&\underset{x\rightarrow a}{\textrm{lim}}\: \displaystyle \frac{f(x)}{g(x)}=\displaystyle \frac{\underset{x\rightarrow a}{\textrm{lim}}\: f(x)}{\underset{x\rightarrow a}{\textrm{lim}}\: g(x)},\quad \textrm{dengan}\: \: \: g(x)\neq 0\\ 8.&\underset{x\rightarrow a}{\textrm{lim}}\: \left (f(x) \right )^{n}= \left [\underset{x\rightarrow a}{\textrm{lim}}\: f(x)) \right ]^{n}\\ 9.&\underset{x\rightarrow a}{\textrm{lim}}\: \sqrt[n]{f(x)}=\sqrt[n]{\underset{x\rightarrow a}{\textrm{lim}}\: f(x)},\quad \textrm{dengan}\: \: \underset{x\rightarrow a}{\textrm{lim}}\: f(x)\geq 0,\: \textrm{saat}\: \: n\: \: \textrm{genap} \end{array}.

Langkah penyelesaian limit fungsi aljabar adalah sebagaimana berikut ini:

  • Substitusi langsung
  • Faktorisasi
  • Mengalikan dengan faktor sekawan

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Substitusi Langsung}}\\\hline \begin{aligned}&\\ \underset{x\rightarrow -1}{\textrm{lim}}\: x^{2}+3&=(-1)^{2}+3\\ &=1+3\\ &=4\\ & \end{aligned}&\begin{aligned}\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{x^{3}+1}{x+1}&=\displaystyle \frac{0^{3}+1}{0+1}\\ &=\displaystyle \frac{1}{1}\\ &=1 \end{aligned}\\\hline \end{array}.

Perhatikan juga contoh soal berikut!

\begin{aligned}&\textrm{Diberikan polinom}\: \: f(x)=a_{_{n}}x^{^{n}}+a_{_{n-1}}x^{^{n-1}}+a_{_{n-2}}x^{^{n-2}}+...+a_{_{2}}x^{^{2}}+a_{_{1}}x+a_{_{0}}.\\ &\textrm{Tunjukkan bahwa}\: \: \underset{x\rightarrow a}{\textrm{lim}}\: f(x)=f(a) \end{aligned}\\\\\\ \begin{aligned}\textbf{Bukti}:\, \, \, &\\ &\\ \underset{x\rightarrow a}{\textrm{lim}}\: f(x)&=\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{n}}x^{^{n}}+a_{_{n-1}}x^{^{n-1}}+a_{_{n-2}}x^{^{n-2}}+...+a_{_{2}}x^{^{2}}+a_{_{1}}x^{1}+a_{_{0}}\\ &=\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{n}}x^{^{n}}+\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{n-1}}x^{^{n-1}}+\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{n-2}}x^{^{n-2}}+...+\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{2}}x^{^{2}}+\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{1}}x^{^{1}}+\underset{x\rightarrow a}{\textrm{lim}}\: a_{_{0}}\\ &=a_{n}\underset{x\rightarrow a}{\textrm{lim}}\: x^{n}+a_{n-1}\underset{x\rightarrow a}{\textrm{lim}}\: x^{n-1}+a_{n-2}\underset{x\rightarrow a}{\textrm{lim}}\: x^{n-2}+...+ a_{2}\underset{x\rightarrow a}{\textrm{lim}}\: x^{2}+a_{1}\underset{x\rightarrow a}{\textrm{lim}}\: x^{1}+a_{0}\\ &=a_{n}\left ( \underset{x\rightarrow a}{\textrm{lim}}\: x \right )^{n}+a_{n-1}\left ( \underset{x\rightarrow a}{\textrm{lim}}\: x \right )^{n-1}+a_{n-2}\left ( \underset{x\rightarrow a}{\textrm{lim}}\: x \right )^{n-2}+...+a_{2}\left ( \underset{x\rightarrow a}{\textrm{lim}}\: x \right )^{2}+a_{1}\underset{x\rightarrow a}{\textrm{lim}}\: x+a_{0}\\ &=a_{n}a^{n}+a_{n-1}a^{n-1}+a_{n-2}a^{n-2}+...+a_{2}a^{2}+a_{1}a+a_{0}\\ &=f(a)\qquad \blacksquare \end{aligned}.

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Pemfaktoran}}\\\hline \begin{aligned}\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{2}-4}{x-2}&=\displaystyle \frac{4-4}{2-2}=\frac{0}{0}\\ &\textrm{hasil seperti ini}\\ &\textrm{harus dihindari, maka}\\ \underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{2}-4}{x-2}&=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{(x-2)(x+2)}{(x-2)}\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: (x+2)\\ &=2+2\\ &=4\\ & \end{aligned}&\begin{aligned}\underset{t\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{t^{3}-8}{t^{2}+t-6}&=\displaystyle \frac{8-8}{4+2-6}=\frac{0}{0}\\ &\textrm{hasil seperti ini lagi-lagi}\\ &\textrm{harus dihindari, maka}\\ \underset{t\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{t^{3}-8}{t^{2}+t-6}&=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{(t-2)(t^{2}+2t+4)}{(t-2)(t+3)}\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{t^{2}+2t+4}{t+3}\\ &=\displaystyle \frac{2^{2}+2.2+4}{2+3}\\ &=\displaystyle \frac{12}{5} \end{aligned} \\\hline \end{array}.

\begin{array}{|l|l|}\hline \multicolumn{2}{|c|}{\textrm{Mengalikan dengan faktor sekawan}}\\\hline \begin{aligned}\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{x-4}{\sqrt{x}-2}&=\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{x-4}{\sqrt{x}-2}\times \displaystyle \frac{\sqrt{x}+2}{\sqrt{x}+2}\\ &=\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{(x-4)\left ( \sqrt{x}+2 \right )}{x-4}\\ &=\underset{x\rightarrow 0}{\textrm{lim}}\: \left (\sqrt{x}+2 \right )\\ &=\sqrt{0}+2\\ &=2\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}&\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{\sqrt{x+2}-\sqrt{6-x}}{x-2}\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{\sqrt{x+2}-\sqrt{6-x}}{x-2}\times \left ( \displaystyle \frac{\sqrt{x+2}+\sqrt{6-x}}{\sqrt{x+2}+\sqrt{6-x}} \right )\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{(x+2)-(6-x)}{(x-2)\left ( \sqrt{x+2}+.... \right )}\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{2(x-2)}{(x-2)\left ( \sqrt{x+2}+.... \right )}\\ &=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{....}{\sqrt{x+2}+....}\\ &=\displaystyle \frac{....}{\sqrt{4}+\sqrt{....}}\\ &=\displaystyle \frac{....}{2+....}\\ &=\displaystyle \frac{....}{....}\\ & \end{aligned} \\\hline \end{array}.

Berikut soal aplikasi untuk model pemfaktoran, yaitu:

\begin{aligned}&\textrm{Tentukanlah nilai}\: \: a\: \: \textrm{dan}\: \: b\: \: \textrm{jika}\: \: \underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{ax+b}{x^{4}-81}=5\\ &\textrm{Jawab}: \end{aligned}\\\\\\ \begin{array}{|l|l|}\hline \begin{aligned}\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{ax+b}{x^{4}-81}&=5\\ \bullet \quad \displaystyle \frac{f(3)}{g(3)}&=\frac{0}{0}\Leftrightarrow \displaystyle \frac{3a+b}{3^{4}-81}=\frac{0}{0}\\ \Rightarrow b&=-3a\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned}&\begin{aligned}\textrm{Dengan bantuan}&\: \textrm{limit \textsl{kanan} kita mendapatkan}\\ \textrm{dengan}\: \: \: \: x=3+&h\: \: \Rightarrow \: \: h\rightarrow 0,\: \textrm{maka}\\ \bullet \quad \underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{ax+b}{x^{4}-81}&=\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{a(3+h)+b}{(3+h)^{4}-81}=5\\ 5&=\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{3a+ah+b}{h^{4}+12h^{3}+54h^{2}+108h+81-81}\\ 5&=\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{3a+ah+(-3a)}{h^{4}+12h^{3}+54h^{2}+108h}\\ 5&=\underset{h\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{a}{h^{3}+12h^{2}+54h+108}\\ 5&=\displaystyle \frac{a}{0+0+0+108}\\ 108\times 5&=a\\ a&=540\\ \textrm{sehingga},\quad &\\ b&=-3a\\ &=-3(540)\\ &=-1620 \end{aligned}\\\hline \end{array}.

Sebagai pembanding untuk mengecek jawaban di atas adalah sebagai berikut:

\begin{aligned}\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{ax+b}{x^{4}-81}&=\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{540x-1620}{x^{4}-81}\: \: ,\qquad\quad \textrm{ingat}\: \: \begin{cases} a & =540 \\ \textrm{dan}&\\ b & =-1620 \end{cases}\\ &=\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{540(x-3)}{(x^{2}+9)(x^{2}-9)}\\ &=\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{540(x-3)}{(x^{2}+9)(x+3)(x-3)}\\ &=\underset{x\rightarrow 3}{\textrm{lim}}\: \displaystyle \frac{540}{(x^{2}+9)(x+3)}\\ &=\displaystyle \frac{540}{(3^{2}+9)(3+3)}\\ &=\displaystyle \frac{540}{(18)(6)}\\ &=5 \end{aligned}.

\LARGE\fbox{\LARGE\fbox{LATIHAN SOAL}}.

Tentukanlah limit-limit berikut!

\begin{array}{lll}\\ 1.&\textrm{a}.&\underset{x\rightarrow 1}{\textrm{lim}}\: (x+1)(x-8)\\ &\textrm{b}.&\underset{x\rightarrow 2}{\textrm{lim}}\: (3x+5)^{2}\\ &\textrm{c}.&\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{\sqrt{x^{2}+5}}{x+2}\\ &\textrm{d}.&\underset{x\rightarrow 1}{\textrm{lim}}\: \left ( \displaystyle \frac{\sqrt{3x^{2}+2x}}{\sqrt{5x-6}}-\frac{x}{2} \right ) \end{array}.

\begin{array}{lll}\\ 2.&\textrm{a}.&\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{3}-8}{x-2}\\ &\textrm{b}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{2x+\sqrt{x}}{\sqrt{x}}\\ &\textrm{c}.&\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{3}+3x^{2}+4}{x^{3}-2x^{2}-4x+8}\\ &\textrm{d}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{2x^{9}-9x^{2}}{5x^{7}-x^{2}} \end{array}.

\begin{array}{lll}\\ 3.&\textrm{a}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sqrt{9+x}-\sqrt{9-x}}{x}\\ &\textrm{b}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sqrt{x^{2}+x}+\sqrt{x}}{x\sqrt{x}}\\ &\textrm{c}.&\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{\sqrt{2x^{2}+5}-\sqrt{x}+7}{\sqrt{3x-4}-\sqrt{x}}\\ &\textrm{d}.&\underset{x\rightarrow 1}{\textrm{lim}}\: \displaystyle \frac{\sqrt[3]{2x+6}-2}{x-1}\\ &\textrm{e}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sqrt{a-x}-\sqrt{a+x}}{\sqrt{b-x}-\sqrt{b+x}} \end{array}.

\begin{array}{lll}\\ 4.&\textrm{a}.&\underset{x\rightarrow -3}{\textrm{lim}}\: \left (\displaystyle \frac{x+6}{x^{2}+5x+6}-\displaystyle \frac{x}{x+3} \right )\\ &\textrm{b}.&\underset{x\rightarrow 2}{\textrm{lim}}\: \left (\displaystyle \frac{x-12}{x^{2}+2x-8}-\displaystyle \frac{x^{2}+1}{3x^{2}-15x+18} \right )\\ \end{array}.

Tentukanlah nilai  a  dan  b berikut!

\begin{array}{lll}\\ 5.&\textrm{a}.&\underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sqrt{ax+b}}{x}=\displaystyle 1\\ &\textrm{b}.&\underset{x\rightarrow 4}{\textrm{lim}}\: \displaystyle \frac{ax+b-\sqrt{x}}{x-4}=\displaystyle \frac{3}{4}\\ &\textrm{b}.&\underset{x\rightarrow 4}{\textrm{lim}}\: \displaystyle \frac{ax+b-\sqrt{x}}{x-4}=\displaystyle \frac{1}{2}\\ &\textrm{c}.&\underset{x\rightarrow 4}{\textrm{lim}}\: \displaystyle \frac{ax^{2}+bx-\sqrt{x}}{x^{2}-16}=\displaystyle \frac{1}{2} \end{array}.

Sumber Referensi

  1. Djumanta, W. dan Sudrajat, R. (2008). Mahir Mengembangkan Kemampuan Matematika untuk Sekolah Menengah Atas/Madrasah Aliyah Kelas XI Program Ilmu Pengetahuan Alam. Jakarta: Pusat Perbukuan, Departemen Pendidikan Nasional.
  2. Kartini, Suprapto, Subandi, dan Setiyadi, U. (2005). Matematika Program Studi Ilmu Alam Kelas XI untuk SMA dan MA. Klaten: Intan Pariwara.

 

 

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Limit Fungsi

A. Pengertian Limit

Adalah nilai hampiran melalui pendekatan intuitif suatu variabel pada suatu bilangan real. Notasi  limit adalah:

\LARGE\boxed{\lim_{x \to a}\: f(x)=L}.

  • Notasi di atas dapat dengan : ” limit suatu fungsi f(x) pada saat x mendekati a adalah sama dengan L
  • Limit fungsi f(x) ada jika limit(harga batas) tersebut memiliki limit kiri dan limit kanan dengan harga yang sama.
  • Limit kiri adalah pendekatan secara intuitif nilai fungsi real dari arah sebelah kiri dan dinotasikan dengan  \LARGE\boxed{\lim_{x \to a^{-}}\: f(x)=L}
    .
  • Limit kanan adalah pendekatan secara intuitif nilai fungsi real dari arah sebelah kanan dan dinotasikan dengan  \LARGE\boxed{\lim_{x \to a^{+}}\: f(x)=L}
    .
  • Secara intuitif di sini diartikan dengan mulai menghitung nilai-nilai fungsi di sekitar titik tersebut.

\LARGE\fbox{\LARGE\fbox{CONTOH SOAL}}.

\begin{array}{ll}\\ 1.&\textrm{Berikut nilai limit untuk}\\ &\textrm{a}.\quad \underset{x \to 1}{\textrm{lim}}\: x=1\\ &\qquad \textrm{dibaca}:\: \textrm{limit \textsl{x} sama dengan 1, jika \textsl{x} mendekat 1}\\ &\textrm{b}.\quad \underset{x \to 2}{\textrm{lim}}\: x^{2}=4\\ &\qquad \textrm{dibaca}:\: \textrm{limit} \: \: x^{2}\: \: \textrm{sama dengan 4, jika \textsl{x} mendekati 2}\\ &\textrm{c}.\quad \underset{x \to 3}{\textrm{lim}}\: x^{3}=27\\ &\qquad \textrm{dibaca}:\: \textrm{limit} \: \: x^{3}\: \: \textrm{sama dengan 27, jika \textsl{x} mendekati 3}\\ &\textrm{d}.\quad \underset{x \to 4}{\textrm{lim}}\: 2x+5=13\\ &\qquad \textrm{dibaca}:\: \textrm{limit} \: \: 2x+5\: \: \textrm{sama dengan 13, jika \textsl{x} mendekati 4}\\ &\textrm{e}.\quad \underset{x \to 2}{\textrm{lim}}\: \displaystyle \frac{x^{2}-4}{x-2}=4\\ &\qquad \textrm{dibaca}:\: \textrm{limit} \: \: \displaystyle \frac{x^{2}-4}{x-2}\: \: \textrm{sama dengan 4, jika \textsl{x} mendekati 2}\\ &\textrm{f}.\quad \underset{x \to 3}{\textrm{lim}}\: \displaystyle \frac{x^{2}-9}{x-3}=6\\ &\qquad \textrm{dibaca}:\: \textrm{limit} \: \: \displaystyle \frac{x^{2}-9}{x-3}\: \: \textrm{sama dengan 6, jika \textsl{x} mendekati 3}\\ \end{array}.

\begin{array}{ll}\\ 2.&\textrm{Perhatikanlah CONTOH SOAL pada No.1 e di atas}\: ,\\ &\textrm{Sebagai penjelasannya adalah sebagai berikut}:\\\\ &\qquad \begin{array}{|l|l|r|r|}\hline x\rightarrow 2^{-}&\multicolumn{2}{|c|}{\begin{aligned}&\\ &\displaystyle \frac{x^{2}-4}{x-2}\\ & \end{aligned}}&x\rightarrow 2^{+}\\\hline 1,7&3,7&4,2&2,2\\\hline 1,8&3,8&4,1&2,1\\\hline 1,99&3,99&4,01&2,01\\\hline 1,999&3,999&4,001&2,001\\\hline \: \: \: \: \downarrow &\: \: \: \: \downarrow &\downarrow \: \: \: \: &\downarrow \: \: \: \: \\ \: \: \: \: 2^{-}&\: \: \: \: 4&4\: \: \: \: &2^{+}\: \: \\\hline \multicolumn{4}{|c|}{\underset{x\rightarrow 2^{-}}{\textrm{lim}}\: f(x)=\underset{x\rightarrow 2^{+}}{\textrm{lim}}\: f(x)}\\\hline \end{array} \end{array}.

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned} &\\\textrm{Walaupun nilai limit}&\\ f(x)&=\displaystyle \frac{x^{2}-4}{x-2}\: \: \textrm{saat x mendekati 2 \textbf{ada} dan bernilai 4},\\ \textrm{masih }&\textrm{ada beberapa hal yang perlu dicermati, antara lain}:\\ & \end{aligned}}\\\hline x=2&x\neq 2\\\hline \begin{aligned}f(x)&=\displaystyle \frac{x^{2}-4}{x-2}\\ f(2)&=\displaystyle \frac{0}{0},\quad \textrm{hasil ini dinamakan bentuk taktentu}\\ &\quad\qquad \textrm{dan tidak ada definisi untuk bentuk ini}\\ &\\ &\\ & \end{aligned}&\begin{aligned}f(x)&=\displaystyle \frac{x^{2}-4}{x-2}\\ f(x)&=\displaystyle \frac{(x+2)(x-2)}{(x-2)}\\ f(x)&=x+2\\ &\textrm{adalah berupa garis lurus yang}\\ &\textrm{terputus di titik}\: \: (2,4) \\ &\textrm{sebagaimana ilustrasi gambar berikut} \end{aligned}\\\hline \end{array}.

387

\begin{array}{ll}\\ 3.&\textrm{Diketahui suatu fungsi}\: \: f(x)\: \: \textrm{dirumuskan sebagai}\\\\ &\qquad\qquad\qquad f(x)=\begin{cases} 2x-3 &, \: \: \textrm{untuk}\: \: x\leq 3\\ x-1 &, \: \: \textrm{untuk}\: \: x>3 \end{cases}\\\\ &\textrm{Tentukanlah}\: \: \underset{x\rightarrow 3}{\textrm{lim}}\: f(x)\end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\textrm{Perhatikan bahwa}&\\ &\\ &\begin{array}{|l|l|r|r|}\hline x\rightarrow 3^{-}&\multicolumn{2}{|c|}{\begin{aligned}&\\ &f(x)=\begin{cases} 2x-3 &, \: \: \textrm{untuk}\: \: x\leq 3\\ x-1 &, \: \: \textrm{untuk}\: \: x>3 \end{cases}\\ & \end{aligned}}& x\rightarrow 3^{+}\\\hline 2,7&2,4&2,2&3,2\\\hline 2,8&2,6&2,1&3,1\\\hline 2,99&2.98&2,01&3,01\\\hline 2,999&2,998&2,001&3,001\\\hline \: \: \: \: \downarrow&\: \: \: \: \downarrow&\downarrow\: \: \: \: \: \: &\downarrow\: \: \: \: \\\hline \: \: \: \: 3&\: \: \: \: 3&2\: \: \: \: \: \: &3\: \: \: \: \\\hline \multicolumn{4}{|c|}{\underset{x\rightarrow 3^{-}}{\textrm{lim}}\: f(x)\: \neq \: \underset{x\rightarrow 3^{+}}{\textrm{lim}}\: f(x)}\\\hline \end{array}\\ &\\ \textrm{Sehingga nilainya}\, &\, \textbf{limit ini tidak ada} \end{aligned}.

Selanjutnya berkaitan dengan bentuk taktentu, perlu kita perhatikan juga bentuk tak tentu semisal yang mungkin, yaitu:

\begin{array}{|l|l|l|}\hline \multicolumn{3}{|c|}{\textrm{Bentuk tak tentu yang harus dihindari diantaranya}}\\\hline \displaystyle \frac{0}{0}&\displaystyle \frac{\infty }{\infty }&\infty -\infty \\\hline &\multicolumn{2}{l|}{\textrm{Misalkan saja}}\\\hline &\begin{aligned}\displaystyle \frac{\infty }{\infty }&=\displaystyle \frac{\displaystyle \frac{1}{0}}{\displaystyle \frac{1}{0}}\\ &=\displaystyle \frac{0}{0} \end{aligned}&\begin{aligned}\infty -\infty&=\displaystyle \frac{1}{0}-\frac{1}{0}\\ &=\displaystyle \frac{1-1}{0}\\ &=\displaystyle \frac{0}{0} \end{aligned}\\\hline \end{array}.

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InsyaAllah

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Lanjutan Contoh Soal Fungsi Komposisi dan Fungsi Invers (3)

\begin{array}{ll}\\ \fbox{21}.&\textrm{Jika}\: \: f^{-1}(x)\: \: \: \textrm{adalah invers dari}\: \: f(x)=\displaystyle \frac{2x+5}{3x-4},\: x\neq \displaystyle \frac{4}{3},\: \: \textrm{maka}\: \: f^{-1}(2)=....\\ &\begin{array}{llll}\\ \textrm{a}.\quad 2,75&&\textrm{d}.\quad 3,5\\ \textrm{b}.\quad 3&\textrm{c}.\quad 3,25&\textrm{e}.\quad 3,75 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{aligned}f(x)&=\displaystyle \frac{2x+5}{3x-4}\\ y&=\displaystyle \frac{2x+5}{3x-4}\\ y(3x-4)&=2x+5\\ 3xy-4y&=2x+5\\ 3xy-2x&=4y+5\\ x(3y-2)&=(4y+5)\\ x&=\displaystyle \frac{4y+5}{3y-2}\\ f^{-1}(y)&=\displaystyle \frac{4y+5}{3y-2}\\ f^{-1}(x)&=\displaystyle \frac{4x+5}{3x-2}\\ f^{-1}(2)&=\displaystyle \frac{4.2+5}{3.2-2}\\ &=\displaystyle \frac{13}{4}\\ &=3,25 \end{aligned}.

\begin{array}{ll}\\ \fbox{22}.&\textrm{Jika invers fungsi}\: \: f(x)\: \: \textrm{adalah}\: \: f^{-1}(x)=\displaystyle \frac{2x}{3-x}\: \: \: \textrm{maka}\: \: f(-3)=....\\ &\begin{array}{llll}\\ \textrm{a}.\quad 9&&\textrm{d}.\quad -\displaystyle \frac{3}{7}\\ \textrm{b}.\quad \displaystyle \frac{9}{5}&\textrm{c}.\quad 1&\textrm{e}.\quad -1 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{a}\\\\ \textbf{Alternatif 1}\\\\ \begin{aligned}f^{-1}(x)&=\displaystyle \frac{2x}{3-x}\\ f^{-1}(y)&=\displaystyle \frac{2y}{3-y}\\ x&=\displaystyle \frac{2y}{3-y}\\ x(3-y)&=2y\\ 3x-xy&=2y\\ -xy-2y&=-3x\\ y(x+2)&=3x\\ y&=\displaystyle \frac{3x}{x+2}\\ f(x)&=\displaystyle \frac{3x}{x+2}\\ f(-3)&=\displaystyle \frac{3.(-3)}{(-3)+2}\\ &=\displaystyle \frac{-9}{-1}\\ &=9 \end{aligned}.

\textbf{Alternatif 2}\\\\ \begin{aligned}f(x)&=\displaystyle \frac{ax+b}{cx+d}&\Rightarrow f^{-1}(x)=\displaystyle \frac{-dx+b}{cx-a}\\ f^{-1}(x)&=\displaystyle \frac{2x}{3-x}=\displaystyle \frac{2x+0}{-x+3}&\Rightarrow \left (f^{-1} \right )^{-1}(x)=f(x)=\displaystyle \frac{-3x}{-x-2}\\ &&f(-3)=\displaystyle \frac{-3.(-3)}{-(-3)-2}\\ &&f(-3)=\displaystyle \frac{9}{1}\\ &&f(-3)=9 \end{aligned}.

\begin{array}{ll}\\ \fbox{23}.&\textrm{Diketahui fungsi}\: \: f(x)=\displaystyle \frac{9}{x+3}\: \: \textrm{dan}\: \: g(x)=x^{2}\:.\: \textrm{Nilai untuk}\: \: (g\circ f)^{-1}(4)\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 3&&\textrm{d}.\quad \displaystyle \frac{2}{3}\\ \textrm{b}.\quad 2&\textrm{c}.\quad \displaystyle \frac{3}{2} &\textrm{e}.\quad \displaystyle \frac{1}{2}\end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{aligned}(g\circ f)(x)&=g(f(x))\\ y&=\left ( \displaystyle \frac{9}{x+3} \right )^{2}\\ \sqrt{y}&=\displaystyle \frac{9}{x+3}\\ x+3&=\displaystyle \frac{9}{\sqrt{y}}\\ x&=\displaystyle \frac{9}{\sqrt{y}}-3\\ (g\circ f)^{-1}(y)&=\displaystyle \frac{9}{\sqrt{y}}-3\\ (g\circ f)^{-1}(x)&=\displaystyle \frac{9}{\sqrt{x}}-3\\ (g\circ f)^{-1}(4)&=\displaystyle \frac{9}{\sqrt{4}}-3\\ &=\displaystyle \frac{9}{2}-3\\ &=\displaystyle \frac{3}{2} \end{aligned}.

\begin{array}{ll}\\ \fbox{24}.&\textrm{Diketahui fungsi}\: \: f\: \: \textrm{dan}\: \: g\: \: \textrm{adalah}\: \: f:x\rightarrow 2x^{\frac{2}{3}}\: \: \textrm{dan} \: \: g:x\rightarrow x^{\frac{2}{3}}.\: \: \textrm{Nilai}\: \: (g\circ f)^{-1}\left ( \sqrt{2} \right )=....\\ \end{array}\\\\\\ \textrm{Jawab}:\\\\ \begin{aligned}(g\circ f)(x)&=\left ( 2x^{\frac{2}{3}} \right )^{\frac{2}{3}}=2^{\frac{2}{3}}.x^{\frac{4}{9}}\\ y&=2^{\frac{2}{3}}.x^{\frac{4}{9}}\\ 2^{\frac{2}{3}}.x^{\frac{4}{9}}&=y\\ x^{\frac{4}{9}}&=\displaystyle \frac{y}{2^{\frac{2}{3}}}\\ x&=\left ( \displaystyle \frac{y}{2^{\frac{2}{3}}} \right )^{\frac{9}{4}}\\ (g\circ f)^{-1}(y)&=\left ( \displaystyle \frac{y}{2^{\frac{2}{3}}} \right )^{\frac{9}{4}}\\ (g\circ f)^{-1}(x)&=\left ( \displaystyle \frac{x}{2^{\frac{2}{3}}} \right )^{\frac{9}{4}}\\ (g\circ f)^{-1}\left ( \sqrt{2} \right )&=\left ( \displaystyle \frac{\sqrt{2}}{2^{\frac{2}{3}}} \right )^{\frac{9}{4}}\\ &=\left ( \displaystyle \frac{2^{\frac{1}{2}}}{2^{\frac{2}{3}}} \right )^{\frac{9}{4}}\\ &=\left ( 2^{\frac{1}{2}-\frac{2}{3}} \right )^{\frac{9}{4}}\\ &=\left ( 2^{-\frac{1}{6}} \right )^{\frac{9}{4}}\\ &=2^{-\frac{3}{8}}\\ &=\displaystyle \frac{1}{2^{\frac{3}{8}}} \end{aligned}.

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Lanjutan Contoh Soal Fungsi Komposisi dan Fungsi Invers (2)

\begin{array}{ll}\\ \fbox{11}.&\textrm{Fungsi}\: \: f(x)=\sqrt{\displaystyle \frac{x^{2}-5x}{1-x}}\: \: \textrm{terdefinisi untuk daerah}....\\\\ &\textrm{a}.\quad x\leq 0\: \: \textrm{atau}\: \: 1\leq x\leq 5\\ &\textrm{b}.\quad x< 0\: \: \textrm{atau}\: \: 1< x< 5\\ &\textrm{c}.\quad x\leq 0\: \: \textrm{atau}\: \: 1< x\leq 5\\ &\textrm{d}.\quad 0\leq x\leq 1\: \: \textrm{atau}\: \: x\geq 5\\ &\textrm{e}.\quad 0< x< 1\: \: \textrm{atau}\: \: x< 5\\ \end{array}\\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\begin{aligned}&\\ &f(x)=\sqrt{\displaystyle \frac{x^{2}-5x}{1-x}}\\ & \end{aligned}}\\\hline \textrm{Domain}&\textrm{Keterangan}\\\hline \begin{aligned}&\textrm{Supaya}\: \textrm{terdefinisi, maka}\\ &\frac{x^{2}-5x}{1-x}\geq 0\\ &\frac{x(x-5)}{(1-x)}\geq 0\\ &\frac{x(x-5)}{(x-1)}\leq 0\\ &\textrm{dengan}\: \textrm{garis bilangan}\\ &\begin{array}{lllllllllll}\\ -&-&+&+&+&-&-&-&+&+&+\\ &&&&&&&&&&\\\hline &&&&&&&&&&\\ &&0&&1&&&&5&& \end{array}\\ &\textrm{Jadi},\: \: D_{_{f}}=\left \{ x|x\leq 0\: \: \textrm{atau}\: \: 1< x\leq 5,\: \: x\in \mathbb{R} \right \} \end{aligned}&\begin{aligned}&\textrm{Karena}\: \: x=1\\ &\textrm{sebagai penyebut},\\ &\textrm{maka tidak boleh}\\ &\textrm{sama dengan}\\ &\textrm{nol}\\ &\\ &\\ &\\ &\\ &\\ &\\ &\\ & \end{aligned} \\\hline \end{array}.

\begin{array}{ll}\\ \fbox{12}.&\textrm{Jika}\: \: f(x)=2x^{2}-5x,\: \: \textrm{maka}\: \: f(x+1)\\ &\textrm{a}.\quad 2x^{2}+x-3\\ &\textrm{b}.\quad 2x^{2}-x-3\\ &\textrm{c}.\quad 2x^{2}-x+3\\ &\textrm{d}.\quad 2x^{2}-4x+1\\ &\textrm{e}.\quad 2x^{2}+4x+1\\ \end{array}\\\\\\ \textrm{Jawab}:\textbf{b}\\\\ \begin{aligned}f(x)&=2x^{2}-5x\\ f(x+1)&=2(x+1)^{2}-5(x+1)\\ &=2(x^{2}+2x+1)-5x-5\\ &=2x^{2}-x-3 \end{aligned}.

\begin{array}{ll}\\ \fbox{13}.&\textrm{Diketahui}\: \: f:\mathbb{R}\rightarrow \mathbb{R},\: \: \textrm{dan}\: \: g:\mathbb{R}\rightarrow \mathbb{R},\: \: \textrm{didefinisikan dengan}\: \: f(x)=x^{3}+4\\ &\textrm{dan}\: \: g(x)=2\sin x,\: \: \textrm{maka nilai}\: \: (f\circ g)\left ( -\displaystyle \frac{1}{2}\pi \right )\: \: \textrm{adalah}....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -4&&\textrm{d}.\quad 6\\ \textrm{b}.\quad 2&\textrm{c}.\quad 3&\textrm{e}.\quad 12 \end{array} \end{array}\\\\\ \textrm{Jawab}:\textbf{a}\\\\ \begin{aligned}(f\circ g)(x)&=f(g(x))\\ &=\left ( 2\sin x \right )^{3}+4\\ (f\circ g)\left ( -\displaystyle \frac{1}{2}\pi \right )&=\left ( 2\sin \left ( -\displaystyle \frac{1}{2}\pi \right ) \right )^{3}+4\\ &=\left ( -2\sin \left ( \displaystyle \frac{1}{2}\pi \right ) \right )^{3}+4\\ &=\left ( -2\sin \left ( 90^{\circ} \right ) \right )^{3}+4\\ &=(-2.1)^{3}+4\\ &=-8+4\\ &=-4 \end{aligned}.

\begin{array}{ll}\\ \fbox{14}.&\textrm{Diketahui}\: \: (g\circ f)(1)=10,\: \: \textrm{dan}\: \: f(1)=5.\: \: \textrm{Jika}\: \: g(x)=2x-p\\ &\textrm{maka nilai}\: \: p=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad -5&&\textrm{d}.\quad 2\\ \textrm{b}.\quad -2&\textrm{c}.\quad 0&\textrm{e}.\quad 5 \end{array} \end{array}\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{aligned}(g\circ f)(x)&=g(f(x))\\ (g\circ f)(x)&=2.f(x)-p\\ (g\circ f)(1)&=2.f(1)-p=10\\ &\: \: \: \: \: \: 2.5-p=10\\ &\quad\quad\quad \: \: \: \: p=10-10\\ &\quad\quad\quad \: \: \: \: p=0 \end{aligned}.

\begin{array}{ll}\\ \fbox{15}.&\textrm{Jika}\: \: g(x)=2x-5\: \: \textrm{dan}\: \: (f\circ g)(x)=6x-13, \textrm{maka}\: \: f(3)=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad 11&&\textrm{d}.\quad 14\\ \textrm{b}.\quad 12&\textrm{c}.\quad 13&\textrm{e}.\quad 15 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{a}\\\\ \begin{aligned}(f\circ g)(x)&=6x-13\\ f(g(x))&=6x-13,\: \: \: \: g(x)\: \: \textrm{dimisalkan dengan sesuatu, misalkan}\: \: a\: \: \textrm{saja}\\ \textrm{sehing}&\textrm{ga}\\ f(a)&=6\left ( \displaystyle \frac{a+5}{2} \right )-13,\: \: \textrm{ingat}\: \: a=2x-5\rightarrow x=\displaystyle \frac{a+5}{2}\\ f(3)&=6\left ( \displaystyle \frac{3+5}{2} \right )-13\\ &=24-13\\ &=11 \end{aligned}.

\begin{array}{ll}\\ \fbox{16}.&\textrm{Diketahui}\: \: f(x)=2x+5\: \: \textrm{dan}\: \: g(x)=\displaystyle \frac{x-1}{x-4}. \textrm{Jika}\: \: (f\circ g)(a)=5,\: \: \textrm{maka}\: \: a=.... \\ &\begin{array}{lll}\\ \textrm{a}.\quad -2&&\textrm{d}.\quad 1\\ \textrm{b}.\quad -1&\textrm{c}.\quad 0&\textrm{e}.\quad 2 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{d}\\\\ \begin{aligned}(f\circ g)(x)&=f(g(x))\\ (f\circ g)(x)&=2\left ( \displaystyle \frac{x-1}{x-4} \right )+5\\ (f\circ g)(a)&=2\left ( \displaystyle \frac{a-1}{a-4} \right )+5\\ 5&=2\left ( \displaystyle \frac{a-1}{a-4} \right )+5\\ 0&=\left ( \displaystyle \frac{a-1}{a-4} \right )\\ 0&=a-1\\ 1&=a \end{aligned}.

\begin{array}{ll}\\ \fbox{17}.&\textrm{Jika}\: \: f(x)=\sqrt{x+1}\: \: \textrm{dan}\: \: (f\circ g)(x)=2\sqrt{x-1},\: \: \textrm{maka}\: \: g(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad 2x-1&&\textrm{d}.\quad 4x-3\\ \textrm{b}.\quad 2x-3&\textrm{c}.\quad 4x-5&\textrm{e}.\quad 5x-4 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{aligned}(f\circ g)(x)&=2\sqrt{x-1}\\ f(g(x))&=2\sqrt{x-1}\\ \sqrt{g(x)+1}&=2\sqrt{x-1}\\ \sqrt{g(x)+1}&=\sqrt{4(x-1)}\\ \sqrt{g(x)+1}&=\sqrt{4x-4}\\ \sqrt{g(x)+1}&=\sqrt{(4x-5)+1}\\ \textrm{sehingga}&\\ g(x)&=4x-5 \end{aligned}.

\begin{array}{ll}\\ \fbox{18}.&\textrm{Jika}\: \: f(x)=2x-4\: \: \textrm{dan}\: \: (g\circ f)(x)=4x^{2}-24x+32,\: \: \textrm{maka}\: \: g(x)=....\\ &\begin{array}{lll}\\ \textrm{a}.\quad x^{2}-4x+8&&\textrm{d}.\quad x^{2}+4x\\ \textrm{b}.\quad x^{2}-4x-8&\textrm{c}.\quad x^{2}+4x+8&\textrm{e}.\quad x^{2}-4x \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{e}\\\\ \begin{aligned}(g\circ f)(x)&=4x^{2}-24x+32\\ g(f(x))&=4x^{2}-24x+32\\ g(a)&=4\left ( \displaystyle \frac{a+4}{2} \right )^{2}-24\left ( \displaystyle \frac{a+4}{2} \right )+32,\: \: \textrm{dengan}\: \: a=2x-4\rightarrow x=\displaystyle \frac{a+4}{2}\\ g(a)&=(a+4)^{2}-12(a+4)+32\\ g(a)&=a^{2}+8a+16-12a-48+32\\ g(a)&=a^{2}-4a\\ \textrm{maka},&\\ g(x)&=x^{2}-4x \end{aligned}.

\begin{array}{ll}\\ \fbox{19}.&\textrm{(UMPTN 1997)Fungsi}\: \: f:\mathbb{R}\rightarrow \mathbb{R}\: \: \textrm{dan}\: \: g:\mathbb{R}\rightarrow \mathbb{R},\: \: \textrm{ditentukan oleh}\: \: g(x)=x^{2}-3x+1\: \: \textrm{dan}\: \: (f\circ g)(x)=2x^{2}-6x-1,\\ &\textrm{maka}\: \: f(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.\quad 2x+3&&\textrm{d}.\quad 2x-2\\ \textrm{b}.\quad 2x+2&\textrm{c}.\quad 2x-1&\textrm{e}.\quad 2x-3 \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{e}.

\begin{array}{|c|c|}\hline \textrm{Pembahasan}&\textrm{Proses pemisalan}\\\hline \begin{aligned}(f\circ g)(x)&=2x^{2}-6x-1\\ f(g(x))&=2x^{2}-6x-1\\ f(a)&=2\left ( \displaystyle \frac{3}{2}\pm \sqrt{a+\displaystyle \frac{5}{4}}\right )^{2}-6\left ( \displaystyle \frac{3}{2}\pm \sqrt{a+\displaystyle \frac{5}{4}} \right )-1\\ &=2\left ( \displaystyle \frac{9}{4}\pm 3\sqrt{a+\displaystyle \frac{5}{4}}+a+\displaystyle \frac{5}{4} \right )-9\pm 6\sqrt{a+\displaystyle \frac{5}{4}}-1\\ &=2\left ( a+\displaystyle \frac{7}{2}\pm 3\sqrt{a+\displaystyle \frac{5}{4}} \right )-10\pm 6\sqrt{a+\displaystyle \frac{5}{4}}\\ &=2a+7-10\\ &=2a-3\\ f(x)&=2x-3 \end{aligned}&\begin{aligned}a&=x^{2}-3x+1\\ a-1&=x^{2}-3x\\ x^{2}-3x&=a-1\\ x^{2}-3x+\left (\displaystyle \frac{3}{2} \right )^{2}&=a-1+\left (\displaystyle \frac{3}{2} \right )^{2},\: \: \textrm{dengan kuadrat sempurna}\\ \left ( x-\displaystyle \frac{3}{2} \right )^{2}&=a+\displaystyle \frac{5}{4}\\ x&=\displaystyle \frac{3}{2}\pm \sqrt{a+\displaystyle \frac{5}{4}}\\ &\\ &\\ & \end{aligned}\\\hline \end{array}.

atau sebagai alternatif malahan agak pendek, yaitu :

\begin{aligned}(f\circ g)(x)&=2x^{2}-6x-1\\ f(g(x))&=2x^{2}-6x-1\\ f(x^{2}-3x+1)&=2x^{2}-6x-1,\: \: \textrm{dengan sedikit manipulasi aljabar, akan menjadi bentuk}\\ f(x^{2}-3x+1)&=2(x^{2}-3x+1)-3\\ \textrm{maka},\qquad &\\ f(x)&=2x-3 \end{aligned}.

\begin{array}{ll}\\ \fbox{20}.&\textrm{Fungsi}\: \: f:\mathbb{R}\rightarrow \mathbb{R}\: \: \: \textrm{dirumuskan dengan}\: \: f(x)=\displaystyle \frac{2x-1}{3x+4},\: x\neq \displaystyle \frac{-4}{3}.\: \: \textrm{Invers fungsi}\: \: f\: \: \textrm{adalah}\: \: f^{-1}(x)=....\\ &\begin{array}{llll}\\ \textrm{a}.\quad \displaystyle \frac{4x-1}{3x+2},\: x\neq \displaystyle \frac{-2}{3}&&\textrm{d}.\quad \displaystyle \frac{4x+1}{3x-2},\: x\neq \displaystyle \frac{2}{3}\\\\ \textrm{b}.\quad \displaystyle \frac{4x-1}{3x-2},\: x\neq \displaystyle \frac{2}{3}&\textrm{c}.\quad \displaystyle \frac{4x+1}{2-3x},\: x\neq \displaystyle \frac{2}{3}&\textrm{e}.\quad \displaystyle \frac{4x+1}{3x+2},\: x\neq \displaystyle \frac{-2}{3} \end{array} \end{array}\\\\\\ \textrm{Jawab}:\textbf{c}\\\\ \begin{aligned}f(x)&=\displaystyle \frac{2x-1}{3x+4}\\ y&=\displaystyle \frac{2x-1}{3x+4}\\ (3x+4)y&=2x-1\\ 3xy+4y&=2x-1\\ 3xy-2x&=-4y-1\\ x(3y-2)&=-4y-1\\ x(2-3y)&=4y+1\\ x&=\displaystyle \frac{4y+1}{2-3y}\\ f^{-1}(y)&=\displaystyle \frac{4y+1}{2-3y}\\ f^{-1}(x)&=\displaystyle \frac{4x+1}{2-3x},\: \: x\neq \frac{2}{3} \end{aligned}.

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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}.

386

\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}.

 

 

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