1. Diketahui fungsi $f(x-1) = \frac{4x-3}{-2-x}$ dengan $x \neq -2$ dan fungsi $g(x) = \frac{2-x}{-2x+5}$ dengan $x \neq \frac{5}{2}$. Kita diminta menentukan: (i) $f(x)$, (ii) komposisi $(g \circ f)(x)$, dan (iii) komposisi $(f^{-1} \circ g^{-1})(x)$.\n\n2. Menentukan $f(x)$:\nDiketahui $f(x-1) = \frac{4x-3}{-2-x}$. Misalkan $t = x-1$, maka $x = t+1$.\nSehingga $f(t) = \frac{4(t+1)-3}{-2-(t+1)} = \frac{4t + 4 - 3}{-2 - t - 1} = \frac{4t + 1}{-3 - t}$.\nJadi, $f(x) = \frac{4x + 1}{-3 - x}$.\n\n3. Menentukan $(g \circ f)(x) = g(f(x))$:\nSubstitusi $f(x)$ ke $g$:\n$$g(f(x)) = \frac{2 - f(x)}{-2 f(x) + 5} = \frac{2 - \frac{4x + 1}{-3 - x}}{-2 \cdot \frac{4x + 1}{-3 - x} + 5}.$$\nSederhanakan pembilang:\n$$2 - \frac{4x + 1}{-3 - x} = \frac{2(-3 - x) - (4x + 1)}{-3 - x} = \frac{-6 - 2x - 4x - 1}{-3 - x} = \frac{-7 - 6x}{-3 - x}.$$\nSederhanakan penyebut:\n$$-2 \cdot \frac{4x + 1}{-3 - x} + 5 = \frac{-2(4x + 1)}{-3 - x} + 5 = \frac{-8x - 2}{-3 - x} + \frac{5(-3 - x)}{-3 - x} = \frac{-8x - 2 - 15 - 5x}{-3 - x} = \frac{-13x - 17}{-3 - x}.$$\nJadi,\n$$(g \circ f)(x) = \frac{\frac{-7 - 6x}{-3 - x}}{\frac{-13x - 17}{-3 - x}} = \frac{-7 - 6x}{-13x - 17}.$$\n\n4. Menentukan invers $f^{-1}(x)$:\nDari $f(x) = \frac{4x + 1}{-3 - x}$, misalkan $y = f(x)$, maka\n$$y = \frac{4x + 1}{-3 - x}.$$\nKalikan silang:\n$$y(-3 - x) = 4x + 1 \Rightarrow -3y - yx = 4x + 1.$$\nKelompokkan $x$:\n$$- yx - 4x = 1 + 3y \Rightarrow x(-y - 4) = 1 + 3y.$$\nSehingga\n$$x = \frac{1 + 3y}{-y - 4} = \frac{1 + 3y}{-(y + 4)} = -\frac{1 + 3y}{y + 4}.$$\nJadi,\n$$f^{-1}(x) = -\frac{1 + 3x}{x + 4}.$$\n\n5. Menentukan invers $g^{-1}(x)$:\nDari $g(x) = \frac{2 - x}{-2x + 5}$, misalkan $y = g(x)$, maka\n$$y = \frac{2 - x}{-2x + 5}.$$\nKalikan silang:\n$$y(-2x + 5) = 2 - x \Rightarrow -2xy + 5y = 2 - x.$$\nKelompokkan $x$:\n$$-2xy + x = 2 - 5y \Rightarrow x(-2y + 1) = 2 - 5y.$$\nSehingga\n$$x = \frac{2 - 5y}{-2y + 1} = \frac{2 - 5y}{1 - 2y}.$$\nJadi,\n$$g^{-1}(x) = \frac{2 - 5x}{1 - 2x}.$$\n\n6. Menentukan $(f^{-1} \circ g^{-1})(x) = f^{-1}(g^{-1}(x))$:\nSubstitusi $g^{-1}(x)$ ke $f^{-1}$:\n$$f^{-1}\left(\frac{2 - 5x}{1 - 2x}\right) = -\frac{1 + 3 \cdot \frac{2 - 5x}{1 - 2x}}{\frac{2 - 5x}{1 - 2x} + 4}.$$\nSederhanakan pembilang:\n$$1 + 3 \cdot \frac{2 - 5x}{1 - 2x} = \frac{1 - 2x}{1 - 2x} + \frac{3(2 - 5x)}{1 - 2x} = \frac{1 - 2x + 6 - 15x}{1 - 2x} = \frac{7 - 17x}{1 - 2x}.$$\nSederhanakan penyebut:\n$$\frac{2 - 5x}{1 - 2x} + 4 = \frac{2 - 5x}{1 - 2x} + \frac{4(1 - 2x)}{1 - 2x} = \frac{2 - 5x + 4 - 8x}{1 - 2x} = \frac{6 - 13x}{1 - 2x}.$$\nJadi,\n$$(f^{-1} \circ g^{-1})(x) = -\frac{\frac{7 - 17x}{1 - 2x}}{\frac{6 - 13x}{1 - 2x}} = -\frac{7 - 17x}{6 - 13x} = \frac{17x - 7}{6 - 13x}.$$