Find \( f+g, f-g, f g \), and \( \frac{f}{g} \). Determine the domain for each function. \( f(x)=x+1, g(x)=5 x^{2} \) \( (f+g)(x)=(f+g)(x)=5 x^{2}+x+1 \) (Simplify your answer.)

To find \( f+g \), \( f-g \), \( fg \), and \( \frac{f}{g} \): 1. **Sum**: \[ (f+g)(x) = f(x) + g(x) = (x+1) + (5x^2) = 5x^2 + x + 1 \] 2. **Difference**: \[ (f-g)(x) = f(x) - g(x) = (x+1) - (5x^2) = -5x^2 + x + 1 \] 3. **Product**: \[ (fg)(x) = f(x) \cdot g(x) = (x+1) \cdot (5x^2) = 5x^3 + 5x^2 \] 4. **Quotient**: \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x+1}{5x^2} \] Now, let’s determine the domains: - For \( f+g \) and \( f-g \): Both functions \( f \) and \( g \) are polynomials, which are defined for all \( x \in \mathbb{R} \). Thus, the domain is \( (-\infty, \infty) \). - For \( fg \): The domain is again \( (-\infty, \infty) \), since it is a product of polynomials. - For \( \frac{f}{g} \): The function is defined everywhere except where \( g(x) = 0 \). To find where \( g(x) = 5x^2 = 0 \) indicates \( x=0 \) is excluded. So, the domain of \( \frac{f}{g} \) is \( (-\infty, 0) \cup (0, \infty) \). To summarize: - \( f+g = 5x^2 + x + 1 \) - \( f-g = -5x^2 + x + 1 \) - \( fg = 5x^3 + 5x^2 \) - \( \frac{f}{g} = \frac{x+1}{5x^2} \) Domains: - For \( f+g \), \( f-g \), \( fg \): \( (-\infty, \infty) \) - For \( \frac{f}{g} \): \( (-\infty, 0) \cup (0, \infty) \)