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Limit Fungsi Trigonometri

Perhatikanlah ilustrsi gambar berikut ini

serta ,

\begin{array}{|c|c|c|}\hline \multicolumn{3}{|c|}{\textrm{Luas Daerah}}\\\hline \triangle \textrm{AOB}&\textrm{Juring AOB}&\triangle \textrm{AOC}\\\hline \begin{aligned}\textrm{L}_{_{\triangle \textrm{AOB}}}&=\displaystyle \frac{1}{2}.OA.BD\\ &=\displaystyle \frac{1}{2}.OA.OB.\sin x\\ &=\displaystyle \frac{1}{2}.1.1.\sin x\\ &=\displaystyle \frac{1}{2}\sin x \end{aligned}&\begin{aligned}\textrm{L}_{_{\textrm{Juring AOB}}}&=\displaystyle \frac{x}{2\pi }.\pi r^{2}\\ &=\displaystyle \frac{x}{2\pi }.\pi .1^{2}\\ &=\displaystyle \frac{1}{2}x\\ &\\ & \end{aligned}&\begin{aligned}\textrm{L}_{_{\triangle \textrm{AOB}}}&=\displaystyle \frac{1}{2}.OA.AC\\ &=\displaystyle \frac{1}{2}.OA.OA.\tan x\\ &=\displaystyle \frac{1}{2}.1.1.\tan x\\ &=\displaystyle \frac{1}{2}\tan x \end{aligned}\\\hline \end{array}.

Untuk prinsip apit

\begin{array}{|c|c|}\hline \multicolumn{2}{|c|}{\textrm{L}_{_{\triangle \textrm{AOB}}}\leq \textrm{L}_{_{\textrm{Juring AOB}}}\leq \textrm{L}_{_{\triangle \textrm{AOB}}}}\\\hline \begin{array}{rlclll}\\ \displaystyle \frac{1}{2}\sin x&\leq &\displaystyle \frac{1}{2}x&\leq &\displaystyle \frac{1}{2}\tan x\\\\ \sin x&\leq &x&\leq &\displaystyle \frac{\sin x}{\cos x}\\\\ 1&\leq &\displaystyle \frac{x}{\sin x}&\leq &\displaystyle \frac{1}{\cos x}\\\\ 1&\geq &\displaystyle \frac{\sin x}{x}&\geq &\cos x\\\\ \cos x&\leq &\displaystyle \frac{\sin x}{x}&\leq &1\\\\ &&&& \end{array}&\begin{array}{rlclll}\\ \displaystyle \frac{1}{2}\sin x&\leq &\displaystyle \frac{1}{2}x&\leq &\displaystyle \frac{1}{2}\tan x\\\\ \displaystyle \sin x&\leq &x&\leq &\displaystyle \tan x\\\\ \displaystyle \frac{\sin x}{\tan x}&\leq &\displaystyle \frac{x}{\tan x}&\leq &1\\\\ \cos x&\leq &\displaystyle \frac{ x}{\tan x}&\leq &1\\\\ \displaystyle \frac{1}{\cos x}&\geq &\displaystyle \frac{\tan x}{x}&\geq &1\\ 1&\leq &\displaystyle \frac{\tan x}{x}&\leq &\displaystyle \frac{1}{\cos x} \end{array} \\\hline \end{array}.

\begin{aligned}&\textrm{Jika}\: \: x\rightarrow 0,\: \textrm{maka}\: \: \underset{x\rightarrow 0}{\textrm{lim}}\: \cos x=1,\: \textrm{akan diperoleh}\: \:\begin{cases} \bullet & \underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{x}{\sin x}=1\\\\ \bullet & \underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\sin x}{x}=1 \\\\ \bullet & \underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{x}{\tan x}=1 \\\\ \bullet & \underset{x\rightarrow 0}{\textrm{lim}}\: \displaystyle \frac{\tan x}{x}=1 \end{cases} \end{aligned}.

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

\begin{aligned}1.\quad \underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan 4x}{\sin 7x}&=\underset{x\rightarrow 0 }{\textrm{lim}}\: \displaystyle \frac{\tan 4x}{\sin 7x} \\ &=\underset{x\rightarrow 0 }{\textrm{lim}}\: \left ( \displaystyle \frac{\tan 4x}{4x} \right )\left ( \displaystyle \frac{7x}{\sin 7x} \right )\left ( \displaystyle \frac{4x}{7x} \right )\\ &=1.1.\displaystyle \frac{4}{7}\\ &=\displaystyle \frac{4}{7}\\ \end{aligned}.

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Limit tak Hingga dan tak Hingga

  • Istilah “banyak sekali, tak terhingga, dan tidak terbatas” tidak dapat ditentukan besarnya
  • Dalam matematika istilah ini dilambangkan dengan “∞”
  • Perhatikanlah gambar garfik fungsi  f(x)=\displaystyle \frac{1}{x}
    berikut!

  • bagaimana nilai limitnya?

\begin{array}{|l|l|r|r|}\hline \quad x&\multicolumn{2}{|c|}{f(x)=\displaystyle \frac{1}{x}}&x\quad \\\hline -1&-1&1&1\\\hline -2&-0,5&0,5&2\\\hline -3&-0,33&0,33&3\\\hline -4&-0,25&0,25&4\\\hline \quad \vdots &\: \: \: \quad \vdots &\vdots \quad &\vdots \\\hline -10&-0,1&0,1&10\\\hline -100&-0,01&0,01&100\\\hline \: \quad \vdots &\qquad \vdots &\vdots \: \: \quad &\vdots \: \: \\\hline -10^{6}&-0,000001&0,000001&10^{6}\\\hline \: \quad \vdots &\qquad \vdots &\vdots \: \: \quad &\vdots \: \: \\\hline \cdots \rightarrow -\infty &\quad \rightarrow 0&\rightarrow 0\quad &\cdots \rightarrow \infty \\\hline \end{array}.

  • Sehingga, untuk   x\rightarrow \infty ,\quad \textrm{nilai}\: \: \displaystyle \frac{1}{x}\: \: \textrm{mendekati}\: \: 0.
  • Kondisi di atas dapat dituliskan dengan  \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{1}{x}=0.
  • Dalam menyelesaikan harga limitnya hindari bentuk – bentu tak tentu berikut:

\begin{array}{|c|c|c|c|}\hline \multicolumn{4}{|c|}{\textrm{Bentuk}}\\\hline \multicolumn{2}{|c|}{\underset{x\rightarrow \infty }{\textrm{lim}}\: f(x)=\displaystyle \frac{\infty }{\infty }}&\multicolumn{2}{|c|}{\underset{x\rightarrow \infty }{\textrm{lim}}\: f(x)=\infty -\infty }\\\hline \multicolumn{2}{|l|}{\begin{aligned}&\textrm{maka bagilah}\\ &\textrm{pembilang dan penyebut}\\ &\textrm{dengan}\: \: x\: \: \textrm{berderajat tertinggi} \end{aligned}}&\multicolumn{2}{|l|}{\begin{aligned}&\textrm{maka arahkan jawaban}\\ &\textrm{ke bentuk}\: \: \displaystyle \frac{\infty }{\infty }\: \: \textrm{dengan}\\ &\textrm{cara perkalian sekawan} \end{aligned}}\\\hline \end{array}.

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

\begin{aligned}1.\quad \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2x^{2}-5x+7}{3x^{2}+4x+8}&=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{2x^{2}}{x^{2}}-\displaystyle \frac{5x}{x^{2}}+\displaystyle \frac{7}{x^{2}}}{\displaystyle \frac{3x^{2}}{x^{2}}+\displaystyle \frac{4x}{x^{2}}+\frac{8}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2-\displaystyle \frac{5}{x}+\frac{7}{x^{2}}}{3+\displaystyle \frac{4}{x}+\displaystyle \frac{8}{x^{2}}}\\ &=\displaystyle \frac{2-0+0}{3+0+0}\\ &=\displaystyle \frac{2}{3} \end{aligned}.

\begin{aligned}2.\quad \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2x^{2}+9}{5x^{2}-4x+2}&=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\displaystyle \frac{2x^{2}}{x^{2}}+\displaystyle \frac{9}{x^{2}}}{\displaystyle \frac{5x^{2}}{x^{2}}-\displaystyle \frac{4x}{x^{2}}+\frac{2}{x^{2}}}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2+\displaystyle \frac{9}{x^{2}}}{5-\displaystyle \frac{4}{x}+\displaystyle \frac{2}{x^{2}}}\\ &=\displaystyle \frac{2+0}{5-0+0}\\ &=\displaystyle \frac{2}{5} \end{aligned}.

\begin{aligned}3.\quad \underset{x\rightarrow \infty }{\textrm{lim}}\: \left ( \sqrt{x^{2}+3x}-\sqrt{x^{2}-5x} \right )&=\underset{x\rightarrow \infty }{\textrm{lim}}\: \left ( \sqrt{x^{2}+3x}-\sqrt{x^{2}-5x} \right )\times \displaystyle \frac{\left ( \sqrt{x^{2}+3x}+\sqrt{x^{2}-5x} \right )}{\left ( \sqrt{x^{2}+3x}+\sqrt{x^{2}-5x} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\left ( x^{2}+3x \right )-\left ( x^{2}-5x \right )}{\left ( \sqrt{x^{2}+3x}+\sqrt{x^{2}-5x} \right )}\\ &=\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8x}{\left ( \sqrt{x^{2}+3x}-\sqrt{x^{2}-5x} \right )}\\ &= \underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{8}{\left ( \sqrt{1+\displaystyle \frac{3}{x}}+\sqrt{1-\displaystyle \frac{5}{x}} \right )}\\ &=\displaystyle \frac{8}{\sqrt{1+0}+\sqrt{1-0}}\\ &=\displaystyle \frac{8}{\sqrt{1}+\sqrt{1}}\\ &=\displaystyle \frac{8}{1+1}\\ &=4 \end{aligned}.

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

\begin{array}{lll}\\ 1.&\textrm{a}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{5x+6x^{2}}{7-2x^{2}-4x^{3}}\\ &\textrm{b}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\left ( 3-2x \right )^{2}}{\left ( 1+3x \right )^{2}}\\ &\textrm{c}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\left ( 2x-1 \right )^{3}}{\left ( 3x-1 \right )\left ( 5x^{2}+x \right )}\\ &\textrm{d}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\left ( 3x^{2}-4 \right )^{2}}{6x^{4}+3x^{2}-5x+1}\\ \end{array}.

\begin{array}{lll}\\ 2.&\textrm{a}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{2x}{\sqrt{3x^{2}+4}}\\ &\textrm{b}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{\left ( x-5 \right )}{x+\sqrt{x^{2}+3x}}\\ &\textrm{c}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \frac{6x}{\sqrt{x^{2}+1}+\sqrt{9x^{2}-2x}}\\ \end{array}.

\begin{array}{lll}\\ 3.&\textrm{a}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \left ( \sqrt{x-4}-\sqrt{x+2} \right )\\ &\textrm{b}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \left ( \sqrt{x^{2}-3x}-\sqrt{x^{2}+5x} \right )\\ &\textrm{c}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \left ( \sqrt{4x^{2}-5x-3}-2x \right )\\ &\textrm{d}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \sqrt{4x^{2}-6x+7}-2x-1\\ \end{array}.

\begin{array}{lll}\\ 4.&\textrm{a}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \left ( \sqrt{x-4}+\sqrt{x+2} \right )\\ &\textrm{b}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \left ( \sqrt{x^{2}-3x}+\sqrt{x^{2}+5x} \right )\\ &\textrm{c}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \left ( \sqrt{4x^{2}-5x-3}+2x \right )\\ &\textrm{d}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \displaystyle \sqrt{4x^{2}-6x+7}+2x-1\\ \end{array}.

\begin{array}{lll}\\ 5.&\textrm{a}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \left ( \displaystyle \frac{1}{1.2}+\displaystyle \frac{1}{2.3}+\displaystyle \frac{1}{3.4}+\displaystyle \frac{1}{4.5}+\cdots +\displaystyle \frac{1}{x\left ( x+1 \right )} \right )=...\\ &\textrm{b}.&\underset{x\rightarrow \infty }{\textrm{lim}}\: \left ( \displaystyle \frac{1}{1.3}+\displaystyle \frac{1}{3.5}+\displaystyle \frac{1}{5.7}+\displaystyle \frac{1}{7.9}+\cdots +\displaystyle \frac{1}{\left ( 2x+1 \right )\left ( 2x+3 \right )} \right )=...\\ \end{array}.

Sumber Referensi

  1. 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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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}.

soal aplikasi lainnya untuk model pemfaktoran:

\begin{aligned}&\textrm{Tentukanlah harga}\: \: n\in \mathbb{N}\: ,\: \textrm{jika}\: \: \: \underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{n}-2^{n}}{x-2}=192\end{aligned}\\\\ \textrm{Jawab}:\\\\ \begin{aligned}\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{x^{n}-2^{n}}{x-2}&=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \frac{(x-2)\left ( x^{n-1}+x^{n-2}.2+x^{n-3}.2^{2}+...+x^{3}.2^{n-4}+x^{2}.2^{n-3}+x.2^{n-2}+2^{n-1} \right )}{(x-2)}\\\\ 192&=\underset{x\rightarrow 2}{\textrm{lim}}\: \displaystyle \left (x^{n-1}+x^{n-2}.2+x^{n-3}.2^{2}+...+x^{3}.2^{n-4}+x^{2}.2^{n-3}+x.2^{n-2}+2^{n-1} \right )\\ 192&=2^{n-1}+2^{n-2}.2+2^{n-3}.2^{2}+...+2^{3}.2^{n-4}+2^{2}.2^{n-3}+2.2^{n-2}+2^{n-1}\\ 192&=\underset{\textrm{sebanyak n faktor}}{\underbrace{2^{n-1}+2^{n-1}+2^{n-1}+...+2^{n-1}+2^{n-1}+2^{n-1}+2^{n-1}}}\\ 192&=n.2^{n-1}\\ 6.32&=n.2^{n-1}\\ 6.2^{6-1}&=n.2^{n-1} \end{aligned}\\\\ \textrm{Jadi},\: n=6.

\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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