Diketahui fungsi \( f(x) = 3x+4 \) dan \( g(x) = \frac{4x-5}{2x+1}, \ x \neq -\frac{1}{2} \). Invers \( (f \circ g)(x) \) adalah….
- \( (f \circ g)^{-1}(x) = \frac{x+11}{-2x+20}, \ x \neq 10 \)
- \( (f \circ g)^{-1}(x) = \frac{x+11}{2x+20}, \ x \neq -10 \)
- \( (f \circ g)^{-1}(x) = \frac{x+11}{2x-20}, \ x \neq 10 \)
- \( (f \circ g)^{-1}(x) = \frac{-x+11}{-2x+20}, \ x \neq 10 \)
- \( (f \circ g)^{-1}(x) = \frac{-x-11}{-2x+20}, \ x \neq 10 \)
Pembahasan:
Pertama, kita cari \( (f \circ g)(x) \) terlebih dahulu.
\begin{aligned} (f \circ g)(x) &= f(g(x))=f \left( \frac{4x-5}{2x+1} \right) \\[8pt] &= 3 \left( \frac{4x-5}{2x+1} \right)+4 \\[8pt] &= \frac{12x-15}{2x+1}+\frac{4(2x+1)}{2x+1} \\[8pt] &= \frac{12x+8x-15+4}{2x+1} \\[8pt] &= \frac{20x-11}{2x+1} \end{aligned}
Sekarang, misalkan \( y = (f \circ g)(x) \) sehingga diperoleh:
\begin{aligned} (f \circ g)(x) = \frac{20x-11}{2x+1} \Leftrightarrow y &= \frac{20x-11}{2x+1} \\[8pt] y(2x+1) &= 20x-11 \\[8pt] 2xy+y &= 20x-11 \\[8pt] 2xy-20x &= -11-y \\[8pt] x(2y-20) &= -11-y \\[8pt] x &= \frac{-11-y}{2y-20} \\[8pt] &= \frac{y+11}{-2y+20} \\[8pt] (f \circ g)^{-1}(x) &= \frac{x+11}{-2x+20}, \ x \neq 10 \end{aligned}
Jawaban A.