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 $$\begin{array}{rcl}g\left(f\left(x\right)\right)& =& 3(2x+p)+120\\ & =& 6x+3p+120\end{array}$$ $$\begin{array}{rcl}f\left(g\left(x\right)\right)& =& 2(3x+120)+p\\ & =& 6x+240+p\end{array}$$ Karena $g\left(f\left(x\right)\right)=f\left(g\left(x\right)\right)$ maka $$\begin{array}{rcl}6x+3p+120& =& 6x+240+p\\ 2p& =& 120\\ p& =& 60\end{array}$$ 22. Jika $f(3+2x)=4-2x+x^2$, maka $f\left(1\right)=.....$
Jawaban
Diketahui : $f(3+2x)=4-2x+x^2$.
Misalkan $$\begin{array}{rcl}y& =& 3+2x\\ x& =& \frac{y-3}{2}\end{array}$$ $$\begin{array}{rcl}f\left(y\right)& =& 4-2\left(\frac{y-3}{2}\right)+{\left(\frac{y-3}{2}\right)}^2\\ & =& 4-y+3+\frac{1}{4}(y^2-6y+9)\\ & =& 7-y+\frac{y^2}{4}-\frac{6y}{4}+\frac{9}{4}\\ & =& \frac{28}{4}-\frac{4y}{4}+\frac{y^2}{4}-\frac{6y}{4}+\frac{9}{4}\\ & =& \frac{y^2}{4}-\frac{10y}{4}+\frac{37}{4}\\ f\left(x\right)& =& \frac{x^2}{4}-\frac{10x}{4}+\frac{37}{4}\\ f\left(1\right)& =& \frac{1}{4}-\frac{10}{4}+\frac{37}{4}\\ & =& \frac{28}{4}\\ f\left(1\right)& =& 7\end{array}$$ 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 $(f\circ g)\left(x\right)=x^2+6x+7$ maka $g(-1)=$. ....
Jawaban $$\begin{array}{rcl}(f\circ g)\left(x\right)& =& x^2+6x+7\\ f\left(g\left(x\right)\right)& =& x^2+6x+7\\ g\left(x\right)+3& =& x^2+6x+7\\ g\left(x\right)& =& x^2+6x+4\\ g(-1)& =& 1-6+4\\ g(-1)& =& -1\end{array}$$ 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 $(f\circ g)\left(x\right)=2x^2-6x-1$ maka $f\left(x\right)=....$
Jawaban $$\begin{array}{rcl}f\left(g\left(x\right)\right)& =& 2x^2-6x-1\\ f(x^2-3x+1)& =& 2x^2-6x-1\end{array}$$ Misalkan $$\begin{array}{rcl}y& =& x^2-3x+1\\ y& =& (x-1,5)(x-1,5)-1,25\\ y& =& {(x-1,5)}^2-1,25\\ y+1,25& =& {(x-1,5)}^2\\ \sqrt{y+1,25}& =& x-1,5\\ x& =& \sqrt{y+1,25}+1,5\end{array}$$ $$\begin{array}{rcl}f\left(y\right)& =& 2{(\sqrt{y+1,25}+1,5)}^2-6(\sqrt{y+1,25}+1,5)-1\\ & =& 2(y+1,25+3\sqrt{y+1,25}+2,25)-6(\sqrt{y+1,25}+1,5)-1\\ & =& 2y+2,5+6\sqrt{y+1,25}+4,5-6\sqrt{y+1,25}-9-1\\ & =& 2y+7-9-1\\ f\left(y\right)& =& 2y-3\\ f\left(x\right)& =& 2x-3\end{array}$$ 25. Suatu pemetaan $f:\mathbb{R}\to \mathbb{R}$ dengan $(g\circ f)\left(x\right)=2x^2+4x+5$dan $g\left(x\right)=2x+3$ maka $f\left(x\right)=$.....

Jawaban $$\begin{array}{rcl}g\left(f\left(x\right)\right)& =& 2x^2+4x+5\\ 2f\left(x\right)+3& =& 2x^2+4x+5\\ 2f\left(x\right)& =& 2x^2+4x+2\\ f\left(x\right)& =& x^2+2x+1\end{array}$$ 26. Jika fungsi $f$ dan $g$ adalah $f:x\to 2x^{\frac{2}{3}}$ dan $g:x\to x^{\frac{2}{3}}$ maka $(g\circ f^{-1})\left(\sqrt{2}\right)$ adalah ....

Jawaban $$\begin{array}{rcl}f\left(x\right)& =& 2x^{\frac{2}{3}}\\ y& =& 2x^{\frac{2}{3}}\\ y^3& =& 2x^2\\ x^2& =& \frac{y^3}{2}\\ x& =& \sqrt{\frac{y^3}{2}}\\ f^{-1}\left(x\right)& =& \sqrt{\frac{x^3}{2}}\end{array}$$ $$\begin{array}{rcl}(g\circ f^{-1})\left(x\right)& =& g\left(f^{-1}\left(x\right)\right)\\ & =& {\left(\sqrt{\frac{x^3}{2}}\right)}^{\frac{2}{3}}\\ & =& {\left({\left(\frac{x^3}{2}\right)}^{\frac{1}{2}}\right)}^{\frac{2}{3}}\\ & =& {\left(\frac{x^3}{2}\right)}^{\frac{1}{3}}\\ (g\circ f^{-1})\left(x\right)& =& \frac{x}{2^{\frac{1}{3}}}\\ (g\circ f^{-1})\left(x\right)& =& x\cdot 2^{-\frac{1}{3}}\\ (g\circ f^{-1})\left(\sqrt{2}\right)& =& 2^{\frac{1}{2}}\cdot 2^{-\frac{1}{3}}\\ & =& 2^{\frac{1}{2}-\frac{1}{3}}\\ (g\circ f^{-1})\left(\sqrt{2}\right)& =& 2^{\frac{1}{6}}\end{array}$$ Sekian dulu yah. Pembahasan selanjutnya saya posting pada postingan mendatang. Selamat belajar