(c) \( f: x \rightarrow \frac{1}{z-2}, z=2 \) \( g: x \rightarrow 6 x+1 \) Cari/ Find: (ii) fg (iii) \( g f \)

To find \( fg \) and \( gf \), we need to understand the compositions of the functions. Given: \( f(z) = \frac{1}{z - 2} \) For \( g(x) = 6x + 1 \), we need to express \( fg \) and \( gf \): (ii) For \( fg(x) \): First, we substitute \( g(x) \) into \( f(z) \): \[ fg(x) = f(g(x)) = f(6x + 1) = \frac{1}{(6x + 1) - 2} = \frac{1}{6x - 1} \] (iii) For \( gf(x) \): Next, substitute \( f(z) \) into \( g(x) \): \[ gf(z) = g(f(z)) = g\left(\frac{1}{z - 2}\right) = 6\left(\frac{1}{z - 2}\right) + 1 = \frac{6}{z - 2} + 1 = \frac{6 + (z - 2)}{z - 2} = \frac{z + 4}{z - 2} \] Thus, the results are: \[ fg(x) = \frac{1}{6x - 1} \quad \text{and} \quad gf(z) = \frac{z + 4}{z - 2} \]