The functions \( f \) and \( a \) are defined as \( f(x)=4 x-3 \) and \( a(x)=-9 x^{2} \) The domain of \( \left(\frac{\mathrm{f}}{\mathrm{g}}\right)(\mathrm{x}) \) is \( (-\infty, 0) \cup(0, \infty) \). (Type your answer in interval notation.) The domain of \( \left(\frac{\mathrm{g}}{\mathrm{f}}\right)(\mathrm{x}) \) is \( \left(-\infty, \frac{3}{4}\right) \cup\left(\frac{3}{4}, \infty\right) \). (Type your answer in interval notation.) b) \( (\mathrm{f}+\mathrm{g})(\mathrm{x})=-9 \mathrm{x}^{2}+4 \mathrm{x}-3 \) (Simplify your answer.) \( (\mathrm{f}-\mathrm{g})(\mathrm{x})=9 \mathrm{x}^{2}+4 \mathrm{x}-3 \) (Simplify your answer.) ( fg\( )(\mathrm{x})=\square \) (Simplify your answer.)

To solve the problem, we first need to clarify the functions \( f \) and \( g \). It seems there was a typo in your question, as you defined \( f(x) = 4x - 3 \) and \( a(x) = -9x^2 \), but then referred to \( g \) in the context of domains and operations. I will assume that \( g(x) = a(x) = -9x^2 \). ### Part a: Domains of \( \frac{f}{g} \) and \( \frac{g}{f} \) 1. **Domain of \( \frac{f}{g} \)**: The function \( \frac{f}{g}(x) = \frac{4x - 3}{-9x^2} \) is undefined when \( g(x) = -9x^2 = 0 \). This occurs when \( x = 0 \). Therefore, the domain of \( \frac{f}{g} \) is: \[ (-\infty, 0) \cup (0, \infty) \] 2. **Domain of \( \frac{g}{f} \)**: The function \( \frac{g}{f}(x) = \frac{-9x^2}{4x - 3} \) is undefined when \( f(x) = 4x - 3 = 0 \). This occurs when \( 4x = 3 \) or \( x = \frac{3}{4} \). Therefore, the domain of \( \frac{g}{f} \) is: \[ \left(-\infty, \frac{3}{4}\right) \cup \left(\frac{3}{4}, \infty\right) \] ### Part b: Simplifying \( (f + g)(x) \), \( (f - g)(x) \), and \( (fg)(x) \) 1. **Sum \( (f + g)(x) \)**: \[ (f + g)(x) = f(x) + g(x) = (4x - 3) + (-9x^2) = -9x^2 + 4x - 3 \] 2. **Difference \( (f - g)(x) \)**: \[ (f - g)(x) = f(x) - g(x) = (4x - 3) - (-9x^2) = 4x - 3 + 9x^2 = 9x^2 + 4x - 3 \] 3. **Product \( (fg)(x) \)**: \[ (fg)(x) = f(x) \cdot g(x) = (4x - 3)(-9x^2) \] Expanding this: \[ (fg)(x) = -9x^2(4x - 3) = -36x^3 + 27x^2 \] ### Final Answers - The domain of \( \frac{f}{g} \) is \( (-\infty, 0) \cup (0, \infty) \). - The domain of \( \frac{g}{f} \) is \( \left(-\infty, \frac{3}{4}\right) \cup \left(\frac{3}{4}, \infty\right) \). - \( (f + g)(x) = -9x^2 + 4x - 3 \) - \( (f - g)(x) = 9x^2 + 4x - 3 \) - \( (fg)(x) = -36x^3 + 27x^2 \)