The functions \( f \) and \( g \) are defined as \( f(x)=8 x+5 \) and \( g(x)=7-9 x \). a) Find the domain of \( f, g, f+g, f-g, f g, f f, \frac{f}{g} \), and \( \frac{g}{f} \). b) Find \( (f+g)(x),(f-g)(x),(f g)(x),(f f)(x),\left(\frac{f}{g}\right)(x) \), and \( \left(\frac{g}{f}\right)(x) \). a) The domain of \( f \) is \( \square \). (Type your answer in interval notation.)

The functions \( f(x) = 8x + 5 \) and \( g(x) = 7 - 9x \) are both linear functions, which means they are defined for all real numbers. Therefore, the domain of \( f \) is \( (-\infty, \infty) \). For the combined functions: - \( f + g \), \( f - g \), and \( f f \) also have a domain of \( (-\infty, \infty) \). - The products \( fg \) and the quotients \( \frac{f}{g} \) and \( \frac{g}{f} \) are also defined for all \( x \) except where \( g(x) = 0 \) and \( f(x) = 0 \), respectively. To find when \( g(x) = 0 \), we solve \( 7 - 9x = 0 \), giving \( x = \frac{7}{9} \). Thus, \( \frac{f}{g} \) is defined everywhere except \( x = \frac{7}{9} \). For \( f(x) = 0 \), solve \( 8x + 5 = 0 \) yielding \( x = -\frac{5}{8} \), so \( \frac{g}{f} \) is undefined there. Summarizing the domains: - Domain of \( f \): \( (-\infty, \infty) \) - Domain of \( g \): \( (-\infty, \infty) \) - Domain of \( f + g \): \( (-\infty, \infty) \) - Domain of \( f - g \): \( (-\infty, \infty) \) - Domain of \( fg \): \( (-\infty, \infty) \) - Domain of \( ff \): \( (-\infty, \infty) \) - Domain of \( \frac{f}{g} \): \( (-\infty, \frac{7}{9}) \cup (\frac{7}{9}, \infty) \) - Domain of \( \frac{g}{f} \): \( (-\infty, -\frac{5}{8}) \cup (-\frac{5}{8}, \infty) \) b) Now let's find the functions: 1. \( (f + g)(x) = (8x + 5) + (7 - 9x) = -x + 12 \) 2. \( (f - g)(x) = (8x + 5) - (7 - 9x) = 17x - 2 \) 3. \( (fg)(x) = (8x + 5)(7 - 9x) = -72x^2 + 43x + 35 \) 4. \( (ff)(x) = f(f(x)) = f(8x + 5) = 8(8x + 5) + 5 = 64x + 45 \) 5. \( \left(\frac{f}{g}\right)(x) = \frac{8x + 5}{7 - 9x} \) 6. \( \left(\frac{g}{f}\right)(x) = \frac{7 - 9x}{8x + 5} \)