Pembahasan Soal-Soal Fungsi Komposisi Dan Invers Fungsi (Lanjutan)
Untuk postingan yang kemaren saya sudah berusaha untuk membahas kurang lebih 20 nomor soal tentang fungsi komposisi. Nah kesempatan kali ini saya ingin membahas kembali beberapa soal saja kurang lebih 6 nomor dan sisa 6 nomor lagi akan saya posting di pembahasan berikutnya. Pembahasan kali ini tetap sama seperti yang kemarin yakni menggunakan spoiler berbasis CSS. Mari kita coba membahas soalnya.
21. Ditentukan $g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right)$, jika $f\left(x\right)=2x+p$ dan $g\left(x\right)=3x+120$, maka nilai $p$ adalah ....
Jawaban \begin{eqnarray*} g\left(f\left(x\right)\right) & = & 3\left(2x+p\right)+120\\ & = & 6x+3p+120 \end{eqnarray*} \begin{eqnarray*} f\left(g\left(x\right)\right) & = & 2\left(3x+120\right)+p\\ & = & 6x+240+p \end{eqnarray*} Karena $g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right)$ maka \begin{eqnarray*} 6x+3p+120 & = & 6x+240+p\\ 2p & = & 120\\ p & = & 60 \end{eqnarray*} 22. Jika $f\left(3+2x\right)=4-2x+x^{2}$, maka $f\left(1\right)=.....$
Jawaban
Diketahui : $f\left(3+2x\right)=4-2x+x^{2}$.
Misalkan \begin{eqnarray*} y & = & 3+2x\\ x & = & \frac{y-3}{2} \end{eqnarray*} \begin{eqnarray*} f\left(y\right) & = & 4-2\left(\frac{y-3}{2}\right)+\left(\frac{y-3}{2}\right)^{2}\\ & = & 4-y+3+\frac{1}{4}\left(y^{2}-6y+9\right)\\ & = & 7-y+\frac{y^{2}}{4}-\frac{6y}{4}+\frac{9}{4}\\ & = & \frac{28}{4}-\frac{4y}{4}+\frac{y^{2}}{4}-\frac{6y}{4}+\frac{9}{4}\\ & = & \frac{y^{2}}{4}-\frac{10y}{4}+\frac{37}{4}\\ f\left(x\right) & = & \frac{x^{2}}{4}-\frac{10x}{4}+\frac{37}{4}\\ f\left(1\right) & = & \frac{1}{4}-\frac{10}{4}+\frac{37}{4}\\ & = & \frac{28}{4}\\ f\left(1\right) & = & 7 \end{eqnarray*} 23. Dari fungsi $f:\mathbb{R}\to\mathbb{R}$ dan $g:\mathbb{R}\to\mathbb{R}$ diketahui bahwa $f\left(x\right)=x+3$ dan $\left(f\circ g\right)\left(x\right)=x^{2}+6x+7$ maka $g\left(-1\right)=$. ....
Jawaban \begin{eqnarray*} \left(f\circ g\right)\left(x\right) & = & x^{2}+6x+7\\ f\left(g\left(x\right)\right) & = & x^{2}+6x+7\\ g\left(x\right)+3 & = & x^{2}+6x+7\\ g\left(x\right) & = & x^{2}+6x+4\\ g\left(-1\right) & = & 1-6+4\\ g\left(-1\right) & = & -1 \end{eqnarray*} 24. Diberikan fungsi $f:\mathbb{R}\to\mathbb{R}$ dan $g:\mathbb{R}\to\mathbb{R}$ ditentukan oleh $g\left(x\right)=x^{2}-3x+1$. Jika $\left(f\circ g\right)\left(x\right)=2x^{2}-6x-1$ maka $f\left(x\right)=....$
Jawaban \begin{eqnarray*} f\left(g\left(x\right)\right) & = & 2x^{2}-6x-1\\ f\left(x^{2}-3x+1\right) & = & 2x^{2}-6x-1 \end{eqnarray*} Misalkan \begin{eqnarray*} y & = & x^{2}-3x+1\\ y & = & \left(x-1,5\right)\left(x-1,5\right)-1,25\\ y & = & \left(x-1,5\right)^{2}-1,25\\ y+1,25 & = & \left(x-1,5\right)^{2}\\ \sqrt{y+1,25} & = & x-1,5\\ x & = & \sqrt{y+1,25}+1,5 \end{eqnarray*} \begin{eqnarray*} f\left(y\right) & = & 2\left(\sqrt{y+1,25}+1,5\right)^{2}-6\left(\sqrt{y+1,25}+1,5\right)-1\\ & = & 2\left(y+1,25+3\sqrt{y+1,25}+2,25\right)-6\left(\sqrt{y+1,25}+1,5\right)-1\\ & = & 2y+2,5+6\sqrt{y+1,25}+4,5-6\sqrt{y+1,25}-9-1\\ & = & 2y+7-9-1\\ f\left(y\right) & = & 2y-3\\ f\left(x\right) & = & 2x-3 \end{eqnarray*} 25. Suatu pemetaan $f:\mathbb{R}\to\mathbb{R}$ dengan $\left(g\circ f\right)\left(x\right)=2x^{2}+4x+5$dan $g\left(x\right)=2x+3$ maka $f\left(x\right)=$.....
Jawaban \begin{eqnarray*} g\left(f\left(x\right)\right) & = & 2x^{2}+4x+5\\ 2f\left(x\right)+3 & = & 2x^{2}+4x+5\\ 2f\left(x\right) & = & 2x^{2}+4x+2\\ f\left(x\right) & = & x^{2}+2x+1 \end{eqnarray*} 26. Jika fungsi $f$ dan $g$ adalah $f:x\to2x^{\frac{2}{3}}$ dan $g:x\to x^{\frac{2}{3}}$ maka $\left(g\circ f^{-1}\right)\left(\sqrt{2}\right)$ adalah ....
Jawaban \begin{eqnarray*} f\left(x\right) & = & 2x^{\frac{2}{3}}\\ y & = & 2x^{\frac{2}{3}}\\ y^{3} & = & 2x^{2}\\ x^{2} & = & \frac{y^{3}}{2}\\ x & = & \sqrt{\frac{y^{3}}{2}}\\ f^{-1}\left(x\right) & = & \sqrt{\frac{x^{3}}{2}} \end{eqnarray*} \begin{eqnarray*} \left(g\circ f^{-1}\right)\left(x\right) & = & g\left(f^{-1}\left(x\right)\right)\\ & = & \left(\sqrt{\frac{x^{3}}{2}}\right)^{\frac{2}{3}}\\ & = & \left(\left(\frac{x^{3}}{2}\right)^{\frac{1}{2}}\right)^{\frac{2}{3}}\\ & = & \left(\frac{x^{3}}{2}\right)^{\frac{1}{3}}\\ \left(g\circ f^{-1}\right)\left(x\right) & = & \frac{x}{2^{\frac{1}{3}}}\\ \left(g\circ f^{-1}\right)\left(x\right) & = & x\cdot2^{-\frac{1}{3}}\\ \left(g\circ f^{-1}\right)\left(\sqrt{2}\right) & = & 2^{\frac{1}{2}}\cdot2^{-\frac{1}{3}}\\ & = & 2^{\frac{1}{2}-\frac{1}{3}}\\ \left(g\circ f^{-1}\right)\left(\sqrt{2}\right) & = & 2^{\frac{1}{6}} \end{eqnarray*} Sekian dulu yah. Pembahasan selanjutnya saya posting pada postingan mendatang. Selamat belajar
21. Ditentukan $g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right)$, jika $f\left(x\right)=2x+p$ dan $g\left(x\right)=3x+120$, maka nilai $p$ adalah ....
Jawaban \begin{eqnarray*} g\left(f\left(x\right)\right) & = & 3\left(2x+p\right)+120\\ & = & 6x+3p+120 \end{eqnarray*} \begin{eqnarray*} f\left(g\left(x\right)\right) & = & 2\left(3x+120\right)+p\\ & = & 6x+240+p \end{eqnarray*} Karena $g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right)$ maka \begin{eqnarray*} 6x+3p+120 & = & 6x+240+p\\ 2p & = & 120\\ p & = & 60 \end{eqnarray*} 22. Jika $f\left(3+2x\right)=4-2x+x^{2}$, maka $f\left(1\right)=.....$
Jawaban
Diketahui : $f\left(3+2x\right)=4-2x+x^{2}$.
Misalkan \begin{eqnarray*} y & = & 3+2x\\ x & = & \frac{y-3}{2} \end{eqnarray*} \begin{eqnarray*} f\left(y\right) & = & 4-2\left(\frac{y-3}{2}\right)+\left(\frac{y-3}{2}\right)^{2}\\ & = & 4-y+3+\frac{1}{4}\left(y^{2}-6y+9\right)\\ & = & 7-y+\frac{y^{2}}{4}-\frac{6y}{4}+\frac{9}{4}\\ & = & \frac{28}{4}-\frac{4y}{4}+\frac{y^{2}}{4}-\frac{6y}{4}+\frac{9}{4}\\ & = & \frac{y^{2}}{4}-\frac{10y}{4}+\frac{37}{4}\\ f\left(x\right) & = & \frac{x^{2}}{4}-\frac{10x}{4}+\frac{37}{4}\\ f\left(1\right) & = & \frac{1}{4}-\frac{10}{4}+\frac{37}{4}\\ & = & \frac{28}{4}\\ f\left(1\right) & = & 7 \end{eqnarray*} 23. Dari fungsi $f:\mathbb{R}\to\mathbb{R}$ dan $g:\mathbb{R}\to\mathbb{R}$ diketahui bahwa $f\left(x\right)=x+3$ dan $\left(f\circ g\right)\left(x\right)=x^{2}+6x+7$ maka $g\left(-1\right)=$. ....
Jawaban \begin{eqnarray*} \left(f\circ g\right)\left(x\right) & = & x^{2}+6x+7\\ f\left(g\left(x\right)\right) & = & x^{2}+6x+7\\ g\left(x\right)+3 & = & x^{2}+6x+7\\ g\left(x\right) & = & x^{2}+6x+4\\ g\left(-1\right) & = & 1-6+4\\ g\left(-1\right) & = & -1 \end{eqnarray*} 24. Diberikan fungsi $f:\mathbb{R}\to\mathbb{R}$ dan $g:\mathbb{R}\to\mathbb{R}$ ditentukan oleh $g\left(x\right)=x^{2}-3x+1$. Jika $\left(f\circ g\right)\left(x\right)=2x^{2}-6x-1$ maka $f\left(x\right)=....$
Jawaban \begin{eqnarray*} f\left(g\left(x\right)\right) & = & 2x^{2}-6x-1\\ f\left(x^{2}-3x+1\right) & = & 2x^{2}-6x-1 \end{eqnarray*} Misalkan \begin{eqnarray*} y & = & x^{2}-3x+1\\ y & = & \left(x-1,5\right)\left(x-1,5\right)-1,25\\ y & = & \left(x-1,5\right)^{2}-1,25\\ y+1,25 & = & \left(x-1,5\right)^{2}\\ \sqrt{y+1,25} & = & x-1,5\\ x & = & \sqrt{y+1,25}+1,5 \end{eqnarray*} \begin{eqnarray*} f\left(y\right) & = & 2\left(\sqrt{y+1,25}+1,5\right)^{2}-6\left(\sqrt{y+1,25}+1,5\right)-1\\ & = & 2\left(y+1,25+3\sqrt{y+1,25}+2,25\right)-6\left(\sqrt{y+1,25}+1,5\right)-1\\ & = & 2y+2,5+6\sqrt{y+1,25}+4,5-6\sqrt{y+1,25}-9-1\\ & = & 2y+7-9-1\\ f\left(y\right) & = & 2y-3\\ f\left(x\right) & = & 2x-3 \end{eqnarray*} 25. Suatu pemetaan $f:\mathbb{R}\to\mathbb{R}$ dengan $\left(g\circ f\right)\left(x\right)=2x^{2}+4x+5$dan $g\left(x\right)=2x+3$ maka $f\left(x\right)=$.....
Jawaban \begin{eqnarray*} g\left(f\left(x\right)\right) & = & 2x^{2}+4x+5\\ 2f\left(x\right)+3 & = & 2x^{2}+4x+5\\ 2f\left(x\right) & = & 2x^{2}+4x+2\\ f\left(x\right) & = & x^{2}+2x+1 \end{eqnarray*} 26. Jika fungsi $f$ dan $g$ adalah $f:x\to2x^{\frac{2}{3}}$ dan $g:x\to x^{\frac{2}{3}}$ maka $\left(g\circ f^{-1}\right)\left(\sqrt{2}\right)$ adalah ....
Jawaban \begin{eqnarray*} f\left(x\right) & = & 2x^{\frac{2}{3}}\\ y & = & 2x^{\frac{2}{3}}\\ y^{3} & = & 2x^{2}\\ x^{2} & = & \frac{y^{3}}{2}\\ x & = & \sqrt{\frac{y^{3}}{2}}\\ f^{-1}\left(x\right) & = & \sqrt{\frac{x^{3}}{2}} \end{eqnarray*} \begin{eqnarray*} \left(g\circ f^{-1}\right)\left(x\right) & = & g\left(f^{-1}\left(x\right)\right)\\ & = & \left(\sqrt{\frac{x^{3}}{2}}\right)^{\frac{2}{3}}\\ & = & \left(\left(\frac{x^{3}}{2}\right)^{\frac{1}{2}}\right)^{\frac{2}{3}}\\ & = & \left(\frac{x^{3}}{2}\right)^{\frac{1}{3}}\\ \left(g\circ f^{-1}\right)\left(x\right) & = & \frac{x}{2^{\frac{1}{3}}}\\ \left(g\circ f^{-1}\right)\left(x\right) & = & x\cdot2^{-\frac{1}{3}}\\ \left(g\circ f^{-1}\right)\left(\sqrt{2}\right) & = & 2^{\frac{1}{2}}\cdot2^{-\frac{1}{3}}\\ & = & 2^{\frac{1}{2}-\frac{1}{3}}\\ \left(g\circ f^{-1}\right)\left(\sqrt{2}\right) & = & 2^{\frac{1}{6}} \end{eqnarray*} Sekian dulu yah. Pembahasan selanjutnya saya posting pada postingan mendatang. Selamat belajar
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