Let \( f(x)=4 x+5 \) and \( g(x)=2 x-7 \). Find \( (f+g)(x) \), \( (f-g)(x) \), \( (f g)(x) \), and \( \left(\frac{1}{g}\right)(x) \). Give the domain of each. \( (f+g)(x)=\square \) (Simplify your answer.) \( (f-g)(x)=\square \) (Simplify your answer.) \( \left(\frac{f g}{}\right)(x)=\square \) (Simplify your answer.) \( \left(\frac{f}{g}\right)(x)=\square \) (Simplify your answer.) The domain of \( f+g \) is \( \square \). (Type your answer in interval notation.) The domain of \( f-g \) is \( \square \). (Type your answer in interval notation.) The domain of \( f g \) is \( \square \). (Type your answer in interval notation.)

To find the various operations of functions \( f(x) \) and \( g(x) \): 1. **Addition**: \[ (f+g)(x) = f(x) + g(x) = (4x + 5) + (2x - 7) = 6x - 2 \] 2. **Subtraction**: \[ (f-g)(x) = f(x) - g(x) = (4x + 5) - (2x - 7) = 2x + 12 \] 3. **Multiplication**: \[ (fg)(x) = f(x) \cdot g(x) = (4x + 5)(2x - 7) = 8x^2 - 28x + 10x - 35 = 8x^2 - 18x - 35 \] 4. **Division**: \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{4x + 5}{2x - 7} \] Now, let's determine the domains: - The domain of \( f+g \): All real numbers since there are no restrictions. \[ \text{Domain of } f+g: (-\infty, \infty) \] - The domain of \( f-g \): Also all real numbers, as it has no restrictions. \[ \text{Domain of } f-g: (-\infty, \infty) \] - The domain of \( fg \): Again, all real numbers with no restrictions. \[ \text{Domain of } fg: (-\infty, \infty) \] - The domain of \( \frac{f}{g} \): This requires \( g(x) \neq 0 \), hence \( 2x - 7 \neq 0 \) leads to \( x \neq \frac{7}{2} \). \[ \text{Domain of } \frac{f}{g}: (-\infty, \frac{7}{2}) \cup (\frac{7}{2}, \infty) \] Summing everything up: \[ (f+g)(x) = 6x - 2 \] \[ (f-g)(x) = 2x + 12 \] \[ (fg)(x) = 8x^2 - 18x - 35 \] \[ \left(\frac{f}{g}\right)(x) = \frac{4x + 5}{2x - 7} \] \[ \text{Domain of } f+g: (-\infty, \infty) \] \[ \text{Domain of } f-g: (-\infty, \infty) \] \[ \text{Domain of } fg: (-\infty, \infty) \] \[ \text{Domain of } \frac{f}{g}: (-\infty, \frac{7}{2}) \cup (\frac{7}{2}, \infty) \]