PAS MATEMATIKA BLOG: Contoh Soal 1 Fungsi Komposisi dan Fungsi Invers

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

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

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

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

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

PAS MATEMATIKA BLOG

Contoh Soal 1 Fungsi Komposisi dan Fungsi Invers

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