PAS MATEMATIKA BLOG: composition function and inverse function

Tampilkan postingan dengan label composition function and inverse function. Tampilkan semua postingan

Contoh Soal 4 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{ll} 16. & Jika f dan g adalah fungsi yang \\ mempunyai invers dan memenuhi \\ f (2 x) = g (x - 3), maka f^{- 1} (x) \\ adalah . . . . \\ \begin{array}{llllll} a . & g^{- 1} (\frac{x}{2} - \frac{2}{3}) \\ b . & g^{- 1} (\frac{x}{2}) - \frac{2}{3} \\ c . & g^{- 1} (2 x + 6) \\ d . & 2 g^{- 1} (x) - 6 \\ e . & 2 g^{- 1} (x) + 6 \end{array} \\ (SBMPTN 2016 Mat Das) \\ Jawab : \\ \begin{aligned} Misalkan bahwa \\ f (2 x) = g (x - 3) = x, \\ maka \\ {\begin{cases} f^{- 1} (x) & = 2 x \\ g^{- 1} (x) & = x - 3 \end{cases} \end{aligned} \\ \begin{array}{cc} Sintak & Hasil \\ \begin{aligned} g^{- 1} (x) & = x - 3 \\ x & = g^{- 1} (x) + 3 \end{aligned} & \begin{aligned} f^{- 1} (x) & = 2 x \\ = 2 (g^{- 1} (x) + 3) \\ = 2 g^{- 1} (x) + 6 \end{aligned} \end{array} \end{array}$ .

$\begin{array}{ll} 17. & Jika f^{- 1} (x) = \frac{x - 1}{5} dan g^{- 1} (x) = \frac{3 - x}{2}, \\ maka {(f \circ g)}^{- 1} (6) = . . . . \\ \begin{array}{llllll} a . & - 1 & d . & 2 \\ b . & 0 & c . & 1 & e . & 3 \end{array} \\ (UMPTN 1995) \\ Jawab : \\ \begin{aligned} Diketahui b & ahwa : \\ {\begin{cases} f^{- 1} (x) & = \frac{x - 1}{5} \\ g^{- 1} (x) & = \frac{3 - x}{2} \end{cases} \\ {(f \circ g)}^{- 1} (x) & = (g^{- 1} \circ f^{- 1}) (x) \\ = \frac{3 - (\frac{x - 1}{5})}{2} \\ {(f \circ g)}^{- 1} (6) & = \frac{3 - (\frac{6 - 1}{5})}{2} \\ = \frac{3 - 1}{2} = \frac{2}{2} \\ = 1 \end{aligned} \end{array}$ .

$\begin{array}{ll} 18. & Invers dari f (x) = 125^{x} adalah f^{- 1} (x) \\ Nilai dari f^{- 1} (5 \sqrt{5}) = . . . . \\ \begin{array}{llllll} a . & 1 & d . & \frac{3}{5} \\ b . & \frac{1}{2} & c . & \frac{1}{6} & e . & - \frac{1}{2} \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa f (x) = 125^{x}, maka \\ f (x) = y = 125^{x} \\ Kedua ruas dilogkan masing-masing \\ \log y = \log 125^{x} \\ \Leftrightarrow \log y = x \log 125 \\ \Leftrightarrow x \log 125 = \log y \\ \Leftrightarrow x = \frac{\log y}{\log 125} \\ \Leftrightarrow x =^{125} \log y \\ \Leftrightarrow f^{- 1} (x) =^{125} \log x \\ Selanjutnya \\ f^{- 1} (x) =^{125} \log x \\ \Leftrightarrow f^{- 1} (5 \sqrt{5}) =^{125} \log (5 \sqrt{5}) \\ \Leftrightarrow f^{- 1} (5 \sqrt{5}) =^{5^{3}} \log 5^{.^{\frac{3}{2}}} \\ \Leftrightarrow f^{- 1} (5 \sqrt{5}) = \frac{\frac{3}{2}}{3}^{5} \log 5 \\ \Leftrightarrow f^{- 1} (5 \sqrt{5}) = \frac{1}{2} .1 \\ \Leftrightarrow f^{- 1} (5 \sqrt{5}) = \frac{1}{2} \end{aligned} \end{array}$

Contoh Soal 3 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{ll} 11. & Diketahui beberapa fungsi memiliki \\ sifat-sifat sebagaimana berikut ini : \\ (i) Φ (- x) = - Φ (x) untuk setiap x \\ (i i) Φ (- x) = Φ (x) untuk setiap x \\ Jika diketahui fungsi f dan g \\ memiliki sifat (i) dan fungsi h dan k \\ memiliki sifat (i i), maka pernyataan \\ berikut yang salah adalah . . . . \\ (1) (f + g) (- x) = - (f + g) (x) \\ (2) (f . k) (- x) = - (f . k) (x) \\ (3) (h - k) (- x) = (h - k) (x) \\ (4) (h - g) (- x) = (h - g) (x) \\ \begin{array}{llllll} a . & (1), (2) dan (3) \\ b . & (1) dan (3) \\ c . & (2) dan (4) \\ d . & (4) saja \\ e . & semuanya benar \end{array} \\ (SIMAK UI 2014 Mat Das) \\ Jawab : \\ \begin{aligned} Diketahui bahwa : \\ {\begin{cases} Φ (- x) = - Φ (x) \\ (fungsi ganjil) {\begin{cases} f & misal f (x) = x \\ g & misal g (x) = 2 x \end{cases} \\ Φ (- x) = Φ (x) \\ (fungsi genap) {\begin{cases} h & misal h (x) = x^{2} \\ k & misal k (x) = 2 x^{2} \end{cases} \end{cases} \end{aligned} \\ \begin{array}{c} \begin{aligned} (1) (f + g) (- x) & = - (f + g) (x) \\ benar \\ (2) (f . k) (- x) & = - (f . k) (x) \\ benar \end{aligned} \\ \begin{aligned} (3) (h - k) (- x) & = (h - k) (x) \\ benar \\ (4) (h - g) (- x) & = (h - g) (x) \\ salah \end{aligned} \end{array} \end{array}$ .

$\begin{array}{ll} 12. & Jika f (x) = \frac{1}{2 x - 1} dan (f \circ g) (x) = \frac{x}{3 x - 2} \\ maka g (x) = . . . . \\ (UMPTN 1998) \\ \begin{array}{llllll} a . & x + \frac{1}{2} & d . & 1 - \frac{2}{x} \\ b . & x - \frac{1}{2} & c . & 2 - \frac{1}{x} & e . & 2 - \frac{1}{2 x} \end{array} \\ Jawab : \\ \begin{aligned} Alternatif 1 \\ Diketahui bahwa f (x) = \frac{1}{2 x - 1}, dan \\ (f \circ g) (x) = \frac{x}{3 x - 2}, maka \\ \Leftrightarrow (f \circ g) (x) = f (g (x)) = \frac{x}{3 x - 2} \\ \Leftrightarrow \frac{1}{2 g (x) - 1} = \frac{x}{3 x - 2} \\ \Leftrightarrow \frac{1}{2 g (x) - 1} = \frac{1}{(\frac{3 x - 2}{x})} \\ Dari bentuk di atas didapatkan \\ 2 g (x) - 1 = \frac{3 x - 2}{x} \\ \Leftrightarrow 2 g (x) = 1 + (3 - \frac{2}{x}) \\ \Leftrightarrow 2 g (x) = 4 - \frac{2}{x} \\ \Leftrightarrow g (x) = 2 - \frac{1}{x} \end{aligned} \\ \begin{aligned} Alternatif 2 \\ Diketahui bahwa f (x) = \frac{1}{2 x - 1}, dengan \\ f^{- 1} (x) = \frac{x + 1}{2 x} . . . . . . . . . (tunjukkan sendiri) \\ serta (f \circ g) (x) = \frac{x}{3 x - 2}, maka \\ g (x) = (f^{- 1} \circ f \circ g) (x) = (I \circ g) (x) = g (x) \\ \Leftrightarrow g (x) = \frac{f (g (x)) + 1}{2 (f (g (x)))} \\ \Leftrightarrow g (x) = \frac{(\frac{x}{3 x - 2}) + 1}{2 (\frac{x}{3 x - 2})} \\ \Leftrightarrow g (x) = \frac{\frac{4 x - 2}{3 x - 2}}{\frac{2 x}{3 x - 2}} \\ \Leftrightarrow g (x) = \frac{4 x - 2}{2 x} \\ \Leftrightarrow g (x) = 2 - \frac{1}{x} \end{aligned} \end{array}$ .

$\begin{array}{ll} 13. & Jika f (x) = x^{2} - 9 dan (f \circ g) (x) = x (x - 6) \\ rumus fungsi g (x) = . . . . \\ \begin{array}{llllll} a . & x + 3 & d . & 3 x + 1 \\ b . & x - 3 & c . & - x & e . & x \end{array} \\ Jawab : \\ \begin{aligned} Alternatif 1 \\ Diketahui bahwa f (x) = x^{2} - 9, dan \\ (f \circ g) (x) = x (x - 6) = x^{2} - 6 x, maka \\ \Leftrightarrow (f \circ g) (x) = f (g (x)) = x^{2} - 6 x \\ \Leftrightarrow {(g (x))}^{2} - 9 = x^{2} - 6 x \\ \Leftrightarrow {(g (x))}^{2} - 9 = x^{2} - 6 x + 9 - 9 \\ \Leftrightarrow {(g (x))}^{2} - 9 = (x - 3)^{2} - 9 \\ Dari bentuk di atas didapatkan \\ g (x) = x - 3 \end{aligned} \\ \begin{aligned} Alternatif 2 \\ Diketahui bahwa f (x) = x^{2} - 9, dengan \\ f^{- 1} (x) = \sqrt{x + 9} . . . . . . . (tunjukkan sendiri) \\ serta (f \circ g) (x) = x^{2} - 6 x, maka \\ g (x) = (f^{- 1} \circ f \circ g) (x) = (I \circ g) (x) = g (x) \\ \Leftrightarrow g (x) = \sqrt{(f (g (x))) + 9} \\ \Leftrightarrow g (x) = \sqrt{x^{2} - 6 x + 9} \\ \Leftrightarrow g (x) = \sqrt{(x - 3)^{2}} \\ \Leftrightarrow g (x) = x - 3 \end{aligned} \end{array}$ .

$\begin{array}{ll} 14. & Jika f (x) = \frac{1}{\sqrt{x^{2} - 2}} dan \\ (f \circ g) (x) = \frac{1}{\sqrt{x^{2} + 6 x + 7}}, \\ maka g (x + 2) = . . . . \\ \begin{array}{llllll} a . & \frac{1}{x + 3} & d . & x + 3 \\ b . & \frac{1}{x - 2} & c . & x - 2 & e . & x + 5 \end{array} \\ (UM UGM 2010 Mat Das) \\ Jawab : \\ Alternatif 1 \\ \begin{aligned} (f \circ g) (x) & = \frac{1}{\sqrt{x^{2} + 6 x + 7}} \\ f (g (x)) & = \frac{1}{\sqrt{x^{2} + 6 x + 7}} \\ \frac{1}{\sqrt{{(g (x))}^{2} - 2}} & = \frac{1}{\sqrt{x^{2} + 6 x + 7}} \\ {(g (x))}^{2} - 2 & = x^{2} + 6 x + 7 \\ {(g (x))}^{2} & = x^{2} + 6 x + 9 \\ g (x) & = \sqrt{x^{2} + 6 x + 9} = \sqrt{(x + 3)^{2}} \\ g (x) & = x + 3 \\ g (x + 2) & = (x + 2) + 3 \\ = x + 5 \end{aligned} \\ \begin{aligned} Alternatif 2 \\ Diketahui bahwa f (x) = \frac{1}{\sqrt{x^{2} - 2}}, dengan \\ f^{- 1} (x) = \sqrt{\frac{1}{x^{2}} + 2} . . . . . . . (akan ditunjukkan) \\ serta (f \circ g) (x) = \frac{1}{\sqrt{x^{2} + 6 x + 7}}, maka \\ \begin{array}{cc} \begin{aligned} f (x) = y & = \frac{1}{\sqrt{x^{2} - 2}} \\ y^{2} & = \frac{1}{x^{2} - 2} \\ x^{2} - 2 & = \frac{1}{y^{2}} \\ x^{2} & = \frac{1}{y^{2}} + 2 \\ x & = \sqrt{\frac{1}{y^{2}} + 2} \\ f^{- 1} (y) & = \sqrt{\frac{1}{y^{2}} + 2} \\ f^{- 1} (x) & = \sqrt{\frac{1}{x^{2}} + 2} \end{aligned} & \begin{aligned} g (x) & = (f^{- 1} \circ f \circ g) (x) \\ = \sqrt{\frac{1}{{(\frac{1}{\sqrt{x^{2} + 6 x + 7}})}^{2}} + 2} \\ = \sqrt{(x^{2} + 6 x + 7) + 2} \\ = \sqrt{(x^{2} + 6 x + 9)} \\ = \sqrt{(x + 3)^{2}} \\ = x + 3 \\ g (x + 2) & = (x + 2) + 3 \\ = x + 5 \end{aligned} \end{array} \end{aligned} \end{array}$ .

$\begin{array}{ll} 15. & Jika g (x) = 2 x + 4 dan (f \circ g) (x) = 4 x^{2} + 8 x - 3, \\ maka f^{- 1} (x) = . . . . \\ \begin{array}{llllll} a . & x + 9 \\ b . & \sqrt{x} + 2 \\ c . & x^{2} - 4 x - 3 \\ d . & \sqrt{x + 1} + 2 \\ e . & \sqrt{x + 7} + 2 \end{array} \\ Jawab : \\ \begin{array}{ccc} Sintak 1 & Sintak 2 & Hasil Invers \\ \begin{aligned} g (x) = y & = 2 x + 4 \\ y - 4 & = 2 x \\ x & = \frac{y - 4}{2} \\ f^{- 1} (y) & = \frac{y - 4}{2} \\ f^{- 1} (x) & = \frac{x - 4}{2} \end{aligned} & \begin{aligned} f (x) & = (f \circ g \circ g^{- 1}) (x) \\ = 4 {(g^{^{- 1}} (x))}^{2} + 8 (g^{- 1} (x)) - 3 \\ = 4 {(\frac{x - 4}{2})}^{2} + 8 (\frac{x - 4}{2}) - 3 \\ = (x^{2} - 8 x + 16) + 4 x - 16 - 3 \\ = x^{2} - 4 x - 3 \\ = x^{2} - 4 x + 4 - 7 \\ = (x - 2)^{2} - 7 \end{aligned} & \begin{aligned} f (x) = y & = (x - 2)^{2} - 7 \\ y + 7 & = (x - 2)^{2} \\ \sqrt{y + 7} & = (x - 2) \\ (x - 2) & = \sqrt{y + 7} \\ x & = \sqrt{y + 7} + 2 \\ f^{- 1} (y) & = \sqrt{y + 7} + 2 \\ f^{- 1} (x) & = \sqrt{x + 7} + 2 \end{aligned} \end{array} \end{array}$

Contoh Soal 2 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{ll} 6. & Fungsi g : R \to R ditentukan oleh \\ g (x) = x^{2} - x + 3 dan f : R \to R \\ sehingga (f \circ g) (x) = 3 x^{2} - 3 x + 4, \\ maka fungsi f (x - 2) = . . . . \\ \begin{array}{llllll} a . & 2 x - 11 & d . & 3 x - 7 \\ b . & 2 x - 7 & c . & 3 x + 1 & e . & 3 x - 11 \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa : \\ {\begin{cases} f (x) = f (x) \\ g (x) = x^{2} - x + 3 \end{cases} \\ (f \circ g) (x) = 3 x^{2} - 3 x + 4 \\ \begin{aligned} f (g (x)) & = 3 x^{2} - 3 x + 4 \\ f (x^{2} - & x + 3) = 3 (x^{2} - x + 3) - 5 \\ Sehing & ga \\ f (x) & = 3 x - 5 \\ f (x - 2) & = 3 (x - 2) - 5 \\ = 3 x - 11 \end{aligned} \end{aligned} \end{array}$

$\begin{array}{ll} 7. & Fungsi f : R \to R dan g : R \to R \\ dengan f (x) = x - 3 dan \\ g (x) = x^{2} + 5. Jika (f \circ g) (x) = (g \circ f) (x) \\ maka nilai x adalah . . . . \\ \begin{array}{llllll} a . & 1 & d . & 4 \\ b . & 2 & c . & 3 & e . & 5 \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa : \\ {\begin{cases} f (x) = x - 3 \\ g (x) = x^{2} + 5 \end{cases} \\ (f \circ g) (x) = (g \circ f) (x) \\ \begin{aligned} f (g (x)) & = g (f (x)) \\ (x^{2} + 5) - 3 & = (x - 3)^{2} + 5 \\ x^{2} + 2 & = x^{2} - 6 x + 14 \\ 6 x & = 14 - 2 \\ x & = \frac{12}{6} \\ x & = 2 \end{aligned} \end{aligned} \end{array}$ .

$\begin{array}{ll} 8. & Fungsi f : R \to R dan g : R \to R \\ dengan f (x) = 3 x - 10 dan \\ g (x) = 4 x + n . Jika \\ (g \circ f) (x) - (f \circ g) (x) = 0 \\ maka nilai n adalah . . . . \\ \begin{array}{llllll} a . & 15 & d . & - 10 \\ b . & 10 & c . & 5 & e . & - 15 \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa : \\ {\begin{cases} f (x) = 3 x - 10 \\ g (x) = 4 x + n \end{cases} \\ (g \circ f) (x) - (f \circ g) (x) = 0 \\ \begin{aligned} g (f (x)) & = f (g (x)) \\ 3 (4 x + n) - 10 & = 4 (3 x - 10) + n \\ 12 x + 3 n - 10 & = 12 x - 40 + n \\ 2 n & = - 30 \\ x & = \frac{- 30}{2} \\ x & = - 15 \end{aligned} \end{aligned} \end{array}$ .

$\begin{array}{ll} 9. & Jika f (x) = \sqrt{2 x + 3} \\ dan g (x) = x^{2} + 1, maka (f \circ g) (2) = . . . . \\ \begin{array}{llllll} a . & 2, 24 & d . & 6 \\ b . & 3 & c . & 3, 61 & e . & 6, 16 \end{array} \\ (SAT Subject Test) \\ Jawab : c \\ \begin{aligned} (f \circ g) (x) & = f (g (x)) \\ = \sqrt{2 g (x) + 3} \\ = \sqrt{2 (x^{2} + 1) + 3} \\ (f \circ g) (2) & = \sqrt{2 (2^{2} + 1) + 3} \\ = \sqrt{2 (5) + 3} \\ = \sqrt{13} \\ \approx 3, 61 \end{aligned} \end{array}$

$\begin{array}{ll} 10. & Misalkan f (x) = x^{2}, g (x) = 2 x \\ dan h (x) = 1 - x . \\ Fungsi (f \circ g \circ h) (x) = . . . . \\ \begin{array}{llllll} a . & 4 x^{2} - 8 x + 4 \\ b . & 4 x^{2} + 8 x - 4 \\ c . & 2 x^{2} - 4 x + 1 \\ d . & x^{2} - 2 x + 1 \\ e . & 4 - 2 x + x^{2} \end{array} \\ Jawab : \\ \begin{aligned} Dike & tahui bahwa \\ ∙ & f (x) = x^{2} \\ ∙ & g (x) = 2 x \\ ∙ & h (x) = 1 - x \\ mak & a \\ (f \circ g \circ h) = f (g (h (x))) \\ = {(2 (1 - x))}^{2} \\ = (2 - 2 x)^{2} \\ = 4 x^{2} - 8 x + 4 \end{aligned} \end{array}$

Contoh Soal 1 Fungsi Komposisi dan Fungsi Invers

$\begin{array}{ll} 1. & Diketahui fungsi f (2 x) = 8 x - 9 \\ dan g (3 x + 1) = 6 x + 3. \\ Rumus untuk (f + g) (x) = . . . . \\ \begin{array}{llllll} a . & 6 x + 8 & d . & 14 x - 6 \\ b . & 6 x - 8 & c . & 14 x + 6 & e . & 6 x - 6 \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa : \\ {\begin{cases} f (2 x) = 8 x - 9 \\ \Rightarrow f (x) = f (2 (\frac{x}{2})) \\ = 8 (\frac{x}{2}) - 9 = 4 x - 9 \\ g (3 x + 1) = 6 x + 3 \\ \Rightarrow g (x) = g (3 (\frac{x - 1}{3}) + 1) \\ = 6 (\frac{x - 1}{3}) + 3 \\ = 2 x + 1 \end{cases} \\ (f + g) (x) = (4 x - 9) + (2 x + 1) \\ = 6 x - 8 \end{aligned} \end{array}$ .

$\begin{array}{ll} 2. & Diketahui fungsi f (x) = 2 x - 1 \\ dan g (x) = x^{2} . Fungsi (f + g) (x^{2}) = . . . . \\ \begin{array}{llllll} a . & x^{2} + 2 x - 1 \\ b . & x^{4} + 2 x^{2} - 1 \\ c . & x^{4} + 2 x - 1 \\ d . & x^{4} + (2 x - 1)^{2} \\ e . & x^{4} + 2 x \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa : \\ {\begin{cases} f (x) = 2 x - 1 \\ g (x) = x^{2} \end{cases} \\ (f + g) (x) = (2 x - 1) + (x^{2}) \\ = x^{2} + 2 x - 1 \\ maka \\ (f + g) (x^{2}) = (x^{2})^{2} + 2 (x^{2}) - 1 \\ = x^{4} + 2 x^{2} - 1 \end{aligned} \end{array}$ .

$\begin{array}{ll} 3. & Jika f (x) = 3 - x, maka \\ f (x^{2}) + {(f (x))}^{2} - 2 f (x) = . . . . \\ \begin{array}{llllll} a . & 2 x^{2} - 6 x + 4 \\ b . & 2 x^{2} + 4 x + 6 \\ c . & 2 x^{2} - 4 x - 6 \\ d . & 6 x + 4 \\ e . & - 4 x + 6 \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa \\ f (x) = 3 - x, sehingga \\ f (x^{2}) + {(f (x))}^{2} - 2 f (x) \\ = (3 - x^{2}) + {(3 - x)}^{2} - 2 (3 - x) \\ = (3 - x^{2}) + (9 - 6 x + x^{2}) - (6 - 2 x) \\ = - x^{2} + x^{2} - 6 x + 2 x + 3 + 9 - 6 \\ = - 4 x + 6 \end{aligned} \end{array}$ .

$\begin{array}{ll} 4. & Diketahui fungsi f : R \to R dan \\ g : R \to R dirumuskan dengan \\ f (x) = x - 1 dan g (x) = x^{2} + 2 x - 3. \\ Fungsi komposisi g atas f \\ dinotasikan dengan \\ \begin{array}{ll} a . & (g \circ f) (x) = x^{2} - 4 \\ b . & (g \circ f) (x) = x^{2} - 5 \\ c . & (g \circ f) (x) = x^{2} - 6 \\ d . & (g \circ f) (x) = x^{2} - 4 x - 4 \\ e . & (g \circ f) (x) = x^{2} - 4 x - 5 \end{array} \\ (UN 2016) \\ Jawab : \\ \begin{aligned} Diketahui & bahwa : \\ {\begin{cases} f (x) & = x - 1 \\ g (x) & = x^{2} + 2 x - 3 \end{cases} \\ (g \circ f) & = g (f (x)) \\ = {(f (x))}^{2} + 2 (f (x)) - 3 \\ = {(x - 1)}^{2} + 2 (x - 1) - 3 \\ = (x^{2} - 2 x + 1) + (2 x - 2) - 3 \\ = x^{2} - 4 \end{aligned} \end{array}$

$\begin{array}{ll} 5. & Diketahui fungsi f (x) = 6 x - 3 \\ dan g (x) = 5 x + 4 dan (f \circ g) (a) = 81 \\ Nilai a adalah . . . . . . . . (Ebtanas 2001) \\ \begin{array}{llllll} a . & - 2 & d . & 2 \\ b . & - 1 & c . & 1 & e . & 27 \end{array} \\ Jawab : \\ \begin{aligned} Diketahui bahwa : \\ {\begin{cases} f (x) = 6 x - 3 \\ g (x) = 5 x + 4 \end{cases} \\ (f \circ g) (a) = 81 \\ maka \\ (f \circ g) (x) = f (g (x)) \\ = 6 (g (x)) - 3 \\ = 6 (5 x + 4) - 3 \\ = 30 x + 24 - 3 \\ = 30 x + 21 = 81 \\ 30 x + 21 = 81 \\ 30 x = 81 - 21 = 60 \\ x = \frac{60}{30} = 2 \end{aligned} \end{array}$ .

Fungsi Komposisi dan Fungsi Invers

$A. Fungsi Komposisi$

Perhatikanlah ilustrasi gambar berikut

$\begin{array}{cc} Syarat & Sifat-sifat \\ \begin{aligned} R_{_{f}} \cap D_{_{g}} \neq {} \end{aligned} & \begin{aligned} 1. & Tidak komutatif \\ (f \circ g) (x) \neq (g \circ f) (x) \\ 2. & Bersifat asosiatif \\ f \circ (g \circ h) (x) = (f \circ g) \circ h (x) \\ 3. & Adanya unsur dentitas \\ (f \circ I) (x) = (I \circ f) (x) = f (x) \end{aligned} \end{array}$ .

$B. Fungsi Invers$

$\begin{aligned} ∙ Suatu fungsi f : A \to B \\ memiliki fungsi invers g : B \to A \\ jika dan hanya jika f \\ merupakan fungsi bijektif \\ ∙ Jika fungsi g ada, maka \\ g dinyatakan dengan f^{- 1} (dibaca : f invers) \end{aligned}$ .

$\begin{array}{ll} Catatan \\ Perlu diingat bahwa pada invers \\ fungsi komposisi berlaku \\ ketentuan sebagai berikut \\ ⧫ & {(g \circ f)}^{- 1} (x) = (f^{- 1} \circ g^{- 1}) (x) \\ ⧫ & {(f \circ g)}^{- 1} (x) = (g^{- 1} \circ f^{- 1}) (x) \\ ⧫ & f (x) = ({(f^{- 1})}^{- 1} (x)) \\ ⧫ & x = f^{- 1} (f (x)) = (f^{- 1} \circ f) (x) \\ = (f \circ f^{- 1}) (x) = f (f^{- 1} (x)) \end{array}$ .

$CONTOH SOAL$

$\begin{array}{ll} 1. & Tentukanlah (f \circ g) (x) dan (g \circ f) (x) Jika : \\ a . f (x) = 2 - x dah g (x) = 5 x + 3 \\ b . f (x) = 2 x + 1 dah g (x) = x^{2} - 4 \\ c . f (x) = \frac{5}{x - 4} dah g (x) = 3 x^{2} \\ d . f (x) = \sqrt{4 - x} dah g (x) = x^{2} + x \\ e . f (x) = x^{3} + 1 dah g (x) = \frac{x}{x - 1} \\ f . f (x) = \frac{3}{x - 2} dah g (x) = \sqrt{x - 4} \end{array}$ .

Jawab:

hanya no. 1 a saja yang dibahas

$\begin{aligned} 1. a . (f \circ g) (x) & = f (g (x)) \\ = f (5 x + 3) \\ = 2 - (5 x + 3) \\ = - 5 x - 1 dan \\ (g \circ f) (x) & = g (f (x)) \\ = g (2 - x) \\ = 5 (2 - x) + 3 \\ = 10 - 5 x + 3 \\ = 13 - 10 x \end{aligned}$ .

$\begin{array}{ll} 2. & Diketahui bahwa g (x) = 3 x + 2 \\ dan (g \circ f) (x) = 4 x - 5. Tentukanlah f (x) \\ Jawab : \\ \begin{aligned} (g \circ f) (x) & = 4 x - 5 \\ g (f (x)) & = 4 x - 5 \\ 3. f (x) + 2 & = 4 x - 5 \\ 3. f (x) & = 4 x - 7 \\ f (x) & = \frac{4 x - 7}{3} \end{aligned} \end{array}$ .

$\begin{array}{ll} 3. & Diketahui bahwa g (x) = x + 4 dan \\ (f \circ g) (x) = 2 x^{2} + 3. Tentukanlah f (x) \\ Jawab : \\ \begin{aligned} (f \circ g) (x) & = 2 x^{2} + 3 \\ f (g (x)) & = 2 x^{2} + 3 \\ f (x + 4) & = 2 x^{2} + 3, \\ misalkan x + 4 = a \Rightarrow x = a - 4, \\ sehingga & , \\ f (a) & = 2 (a - 4)^{2} + 3 \\ f (a) & = 2 (a^{2} - 8 a + 16) + 3 \\ = 2 a^{2} - 16 a + 35 \\ f (x) & = 2 x^{2} - 16 x + 35 \end{aligned} \end{array}$ .

$\begin{array}{ll} 4. & Diketahui f (x) = 3 x dan g (x) = 3^{x} . \\ Tentukanlah rumus untuk^{^{27}} \log (g \circ f) (x) \\ Jawab : \\ \begin{aligned} ^{^{27}} \log (g \circ f) (x) & =^{^{27}} \log g (f (x)) \\ =^{^{27}} \log 3^{3 x} \\ =^{^{3^{3}}} \log 3^{3 x} \\ =^{^{^{(3^{3})}}} \log {(3^{3})}^{x} \\ = x \end{aligned} \end{array}$ .

$\begin{array}{ll} 5. & Tentukanlah invers dari f (x) = 6^{2 x} \\ Jawab : \\ \begin{aligned} f (x) & = 6^{2 x} \\ y & = 6^{2 x} \\ \log y & = \log 6^{2 x} \\ \log y & = 2 x \log 6 \\ \frac{\log y}{\log 6} & = 2 x \\ \frac{\log y}{2 \log 6} & = x \\ \frac{\log y}{\log 6^{2}} & = x \\ \frac{\log y}{\log 36} & = x \\ x & = \frac{\log y}{\log 36} \\ x & =^{^{36}} \log y \\ f^{- 1} (x) & =^{^{36}} \log x \end{aligned} \end{array}$ .

$\begin{array}{ll} 6. & Tentukanlah inver dari \\ f (x) = \frac{2 x + 3}{4 x - 5}, x \neq \frac{5}{4} \\ Jawab : \\ \begin{aligned} f (x) & = \frac{2 x + 3}{4 x - 5} \\ y & = \frac{2 x + 3}{4 x - 5} \\ (4 x - 5) y & = 2 x + 3 \\ 4 x y - 5 y & = 2 x + 3 \\ 4 x y - 2 x & = 5 y + 3 \\ x (4 y - 2) & = 5 y + 3 \\ x & = \frac{5 y + 3}{4 y - 2} \\ f^{- 1} (y) & = \frac{5 y + 3}{4 y - 2} \\ maka, \\ f^{- 1} (x) & = \frac{5 x + 3}{4 x - 2}, x \neq \frac{1}{2} \end{aligned} \end{array}$ .

$\begin{array}{ll} 7. & Jika f (x) = - 2 x - 4 dan g (x) = \frac{20 - 3 x}{2}, \\ maka nilai dari (f \circ g)^{- 1} (2) = . . . . \\ Jawab : \\ Perhatikan bahwa \\ \begin{aligned} (f \circ g)^{- 1} (x) = (g^{- 1} \circ f^{- 1}) (x) \end{aligned} \\ \begin{array}{cc} (f \circ g)^{- 1} (x) & (g^{- 1} \circ f^{- 1}) (x) \\ \begin{aligned} (f \circ g) (x) & = f (g (x)) \\ y & = - 2 (\frac{20 - 3 x}{2}) - 4 \\ y & = 3 x - 20 - 4 \\ y & = 3 x - 24 \\ y + 24 & = 3 x \\ x & = \frac{y + 24}{3} \\ (f \circ g)^{- 1} y & = \frac{y + 24}{3} \\ (f \circ g)^{- 1} (2) & = \frac{2 + 24}{3} \\ = \frac{26}{3} \end{aligned} & \begin{aligned} (g^{- 1} \circ f^{- 1}) (x) & = g^{- 1} (f^{- 1} (x)) \\ = . . . . . . \\ = . . . . \\ = . . . . \\ = . . . . \\ = . . . . \\ = . . . . \\ = . . . . \\ = . . . . \\ = . . . . \\ (g^{- 1} \circ f^{- 1}) (2) & = . . . . \\ = . . . . \end{aligned} \end{array} \end{array}$ .

DAFTAR PUSTAKA

Soedyarto, Nugroho, Maryanto. 2008. Matematika 2 untuk SMA dan MA Kelas XI Program IPA. Jakarta: Pusat Perbukuan Departemen Pendidikan Nasional.
Sunardi, Waluyo, S., Sutrisno, & Subagya. 2005. Matematika 2untuk SMA Kelas 2 IPA. Jakarta: BUMI AKSARA.