[tex] \sf a. \: Tentukan \: nilai \: dari \: g( - 2) \: jika \: f(x) = 2x + 1 \\ \sf dan \: (f \circ g)(x) = (x - 1)(3x + 5)~~~~~~~~~~~~~~~~ \\ [/tex]
[tex]\sf b. \: Tentukan \: nilai \: dari \: f(5) \: jika \: g(x) = x + 2 \\ \sf dan \: (f \circ g)(x) = (x - 1)(3x + 5)~~~~~~~~~~~ \\ [/tex]
[tex]\sf c. \: Tentukan \: nilai \: dari \: f( - 1) \: jika \: g(x) = 2x + 3 \\ \sf dan \: (g \circ f)(x) = {(2x - 5)}^{2}~~~~~~~~~~~~~~~~~~~~~~~~ \\ [/tex]
[tex] \sf d. \: Tentukan \: nilai \: dari \: g(7) \: jika \: f(x) = 3x - 2 \\ \sf dan \: (g \circ f)(x) = {(2x - 5)}^{2} ~~~~~~~~~~~~~~~~~~~~~~~[/tex]
----------------------
Note :
• jika ragu, lebih baik tidak menjawab
• lengkap (menggunakan cara/langkah²)
• dilarang menggunakan bahasa alien
----------------------
Jawaban:
a. g(–2) = 1
b. f(5) = 28
c. f(–1) = 23
d. g(7) = 1
Pembahasan
Komposisi dan Invers Fungsi
Soal a
[tex]\large\text{$\begin{aligned}&f(x)=2x+1\,,\ (f\circ g)(x)=(x-1)(3x+5)\\\\&\boxed{\ f\left(g(x)\right)=(f\circ g)(x)\ }\\\\{\implies}&f\left(g(-2)\right)=(f\circ g)(-2)\\{\iff} &2g(-2)+1=(-2-1)(3(-2)+5)\\{\iff} &2g(-2)+1=-3(-1)=3\\{\iff} &g(-2)=\frac{3-1}{2}=\frac{2}{2}\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}{\therefore\quad}&g(-2)=\bf1\end{aligned}$}[/tex]
Soal b
[tex]\large\text{$\begin{aligned}&g(x)=x+2\,,\ (f\circ g)(x)=(x-1)(3x+5)\\&g^{-1}(x)=x-2\\\\&\boxed{\ f(x)=\left((f\circ g)\circ g^{-1}\right)(x)\ }\\\\{\implies}&f(5)=\left((f\circ g)\circ g^{-1}\right)(5)\\&{\quad\ \ \;}=(f\circ g)(g^{-1}(5))\\&{\quad\ \ \;}=\left(g^{-1}(5)-1\right)\left(3g^{-1}(5)+5\right)\\&{\quad\ \ \;}=(5-2-1)(3(5-2)+5)\\&{\quad\ \ \;}=(2)(14)\\&{\quad\ \ \;}=28\\\\{\therefore\quad}&f(5)=\bf28\end{aligned}$}[/tex]
Soal c
[tex]\large\text{$\begin{aligned}&g(x)=2x+3\,,\ (g \circ f)(x)={(2x - 5)}^2\\&g^{-1}(x)=\frac{x-3}{2}\\\\&\boxed{\ f(x)=\left(g^{-1}\circ (g \circ f)\right)(x)\ }\\\\{\implies}&f(-1)=\left(g^{-1}\circ (g \circ f)\right)(-1)\\&{\qquad\ \ }=g^{-1}((g \circ f)(-1))\\&{\qquad\ \ }=\frac{(g \circ f)(-1)-3}{2}\\&{\qquad\ \ }=\frac{{(2(-1) - 5)}^2-3}{2}\\&{\qquad\ \ }=\frac{{(-7)}^2-3}{2}\\&{\qquad\ \ }=\frac{49-3}{2}\ =\ \frac{46}{2}\ =\ 23\\\\{\therefore\quad}&f(-1)=\bf23\end{aligned}$}[/tex]
Soal d
[tex]\large\text{$\begin{aligned}&f(x)=3x-2\,,\ (g \circ f)(x)={(2x - 5)}^2\\&f^{-1}(x)=\frac{x+2}{3}\\\\&\boxed{\ g(x)=\left((g \circ f)\circ f^{-1}\right)(x)\ }\\\\{\implies}&g(7)=\left((g \circ f)\circ f^{-1}\right)(7)\\&{\quad\ \;\,}=(g \circ f)\left(f^{-1}(7)\right)\\&{\quad\ \;\,}=\left(2\left(f^{-1}(7)\right)-5\right)^2\\&{\quad\ \;\,}=\left(2\left(\frac{7+2}{3}\right)-5\right)^2\\&{\quad\ \;\,}=\left(2\left(\frac{9}{3}\right)-5\right)^2\\&{\quad\ \;\,}=(6-5)^2\ =\ 1\end{aligned}$}[/tex]
[tex]\large\text{$\begin{aligned}{\therefore\quad}&g(7)=\bf1\end{aligned}$}[/tex]
[answer.2.content]
