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Find \( f+g, f-g, f g \), and \( \frac{f}{g} \). Determine the domain for each function. \( f(x)=x+1, g(x)=5 x^{2} \) \( (f+g)(x)=5 x^{2}+x+1 \) (Simplify your answer.) What is the domain of \( f+g \) ? A. The domain of \( f+g \) is \( (-\infty, \infty) \). (Type your answer in interval notation.) C. The domain of \( f+g \) is \( \{\quad \). (Use a comma to separate answers as needed.) \( (f-g)(x)=\square \) (Simplify your answer.) C \( f+g \) is \( \varnothing \).

Ask by Mccarthy Rose. in the United States
Feb 25,2025

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Answer

\( (f + g)(x) = 5x^2 + x + 1 \) with domain \( (-\infty, \infty) \). \( (f - g)(x) = -5x^2 + x + 1 \) with domain \( (-\infty, \infty) \). \( (f \cdot g)(x) = 5x^3 + 5x^2 \) with domain \( (-\infty, \infty) \). \( \left(\frac{f}{g}\right)(x) = \frac{x + 1}{5x^2} \) with domain \( (-\infty, 0) \cup (0, \infty) \).

Solution

Let's solve the problem step by step. 1. **Extract known conditions:** - \( f(x) = x + 1 \) - \( g(x) = 5x^2 \) 2. **Calculate \( f + g \):** \[ (f + g)(x) = f(x) + g(x) = (x + 1) + (5x^2) = 5x^2 + x + 1 \] This is already simplified. 3. **Determine the domain of \( f + g \):** - Both \( f(x) \) and \( g(x) \) are defined for all real numbers. - Therefore, the domain of \( f + g \) is \( (-\infty, \infty) \). 4. **Calculate \( f - g \):** \[ (f - g)(x) = f(x) - g(x) = (x + 1) - (5x^2) = -5x^2 + x + 1 \] This is also simplified. 5. **Determine the domain of \( f - g \):** - Similar to \( f + g \), both functions are defined for all real numbers. - Therefore, the domain of \( f - g \) is \( (-\infty, \infty) \). 6. **Calculate \( f \cdot g \):** \[ (f \cdot g)(x) = f(x) \cdot g(x) = (x + 1)(5x^2) = 5x^3 + 5x^2 \] This is simplified. 7. **Determine the domain of \( f \cdot g \):** - Again, both functions are defined for all real numbers. - Therefore, the domain of \( f \cdot g \) is \( (-\infty, \infty) \). 8. **Calculate \( \frac{f}{g} \):** \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x + 1}{5x^2} \] This is simplified. 9. **Determine the domain of \( \frac{f}{g} \):** - The function \( g(x) = 5x^2 \) cannot be zero, so we need to find where \( g(x) = 0 \): \[ 5x^2 = 0 \implies x = 0 \] - Therefore, the domain of \( \frac{f}{g} \) is all real numbers except \( x = 0 \), which can be expressed in interval notation as \( (-\infty, 0) \cup (0, \infty) \). ### Summary of Results: - \( (f + g)(x) = 5x^2 + x + 1 \) - Domain: \( (-\infty, \infty) \) - \( (f - g)(x) = -5x^2 + x + 1 \) - Domain: \( (-\infty, \infty) \) - \( (f \cdot g)(x) = 5x^3 + 5x^2 \) - Domain: \( (-\infty, \infty) \) - \( \left(\frac{f}{g}\right)(x) = \frac{x + 1}{5x^2} \) - Domain: \( (-\infty, 0) \cup (0, \infty) \)

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To find \( f-g \), we calculate: \[ (f-g)(x) = f(x) - g(x) = (x + 1) - (5x^2) = -5x^2 + x + 1 \] So, \[ (f-g)(x) = -5x^2 + x + 1. \] Now, moving on to \( f g \): \[ (f g)(x) = f(x) \cdot g(x) = (x + 1) \cdot (5x^2) = 5x^3 + 5x^2. \] Finally, for \( \frac{f}{g} \): \[ \frac{f}{g}(x) = \frac{f(x)}{g(x)} = \frac{x + 1}{5x^2}. \] Now, let's determine the domains: 1. The domain of \( f+g \) is \( (-\infty, \infty) \), as there are no restrictions on \( x \). 2. For \( f-g \), the domain is also \( (-\infty, \infty) \). 3. For \( f g \), the domain remains \( (-\infty, \infty) \). 4. The function \( \frac{f}{g} \) is undefined when \( g(x) = 0 \). Thus, \( 5x^2 = 0 \) implies \( x = 0 \). Therefore, the domain is \( (-\infty, 0) \cup (0, \infty) \). In summary, the domains are: - \( f+g \): \( (-\infty, \infty) \) - \( f-g \): \( (-\infty, \infty) \) - \( fg \): \( (-\infty, \infty) \) - \( \frac{f}{g} \): \( (-\infty, 0) \cup (0, \infty) \)

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