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

Ask by Klein Lambert. in the United States
Feb 25,2025

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Tutor-Verified Answer

Answer

\( (f+g)(x) = 5x² + x + 1 \) with domain all real numbers.

Solution

We are given:   f(x) = x + 1   g(x) = 5x² Below are the operations and the corresponding domains. 1. (f + g)(x):   (f + g)(x) = f(x) + g(x) = (x + 1) + 5x² = 5x² + x + 1   Domain: All real numbers, ℝ. 2. (f − g)(x):   (f − g)(x) = f(x) − g(x) = (x + 1) − 5x² = −5x² + x + 1   Domain: All real numbers, ℝ. 3. (f · g)(x):   (f · g)(x) = f(x) · g(x) = (x + 1)(5x²) = 5x³ + 5x²   Domain: All real numbers, ℝ. 4. (f/g)(x):   (f/g)(x) = f(x) / g(x) = (x + 1) / (5x²)   Domain: Since the denominator cannot be zero, we must have 5x² ≠ 0 ⟹ x ≠ 0.   Thus, Domain: ℝ \ {0}. For the specific question about (f+g)(x), the answer is:   (f+g)(x) = 5x² + x + 1  with domain ℝ.

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To find the functions \( f+g, f-g, fg, \) and \( \frac{f}{g} \): 1. **Finding \( f+g \)**: \[ (f+g)(x) = f(x) + g(x) = (x + 1) + (5x^2) = 5x^2 + x + 1 \] 2. **Finding \( f-g \)**: \[ (f-g)(x) = f(x) - g(x) = (x + 1) - (5x^2) = -5x^2 + x + 1 \] 3. **Finding \( fg \)**: \[ (fg)(x) = f(x) \cdot g(x) = (x + 1)(5x^2) = 5x^3 + 5x^2 \] 4. **Finding \( \frac{f}{g} \)**: \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x + 1}{5x^2} \] Note that this fraction is defined for all \( x \) except \( x = 0 \) (since that would make the denominator zero). Now for the domains: - For \( f+g \) and \( f-g \): Both are polynomial functions, which are defined for all real numbers, so the domain is \( \mathbb{R} \). - For \( fg \): This is also a polynomial function, so its domain is \( \mathbb{R} \). - For \( \frac{f}{g} \): The function is defined for all \( x \) except \( x = 0 \), so the domain is \( \mathbb{R} \setminus \{0\} \).

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