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Suppose that the functions \( f \) and \( g \) are defined as follows. \[ \begin{array}{l} f(x)=x-1 \\ g(x)=\sqrt{2 x-3} \end{array} \] Find \( f+g \) and \( f \cdot g \). Then, give their domains using interval notation \[ (f+g)(x)= \] \( \square \) Domain of \( f+g \) : \( \square \) \[ (f \cdot g)(x)= \] \( \square \) Domain of \( f \cdot g \) : \( \square \)

Ask by Carrillo Craig. in the United States
Feb 25,2025

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Answer

\[ (f+g)(x) = x - 1 + \sqrt{2x - 3} \] Domain of \( f+g \): \([ \frac{3}{2}, \infty )\) \[ (f \cdot g)(x) = (x - 1) \cdot \sqrt{2x - 3} \] Domain of \( f \cdot g \): \([ \frac{3}{2}, \infty )\)

Solution

We are given:   f(x) = x – 1   g(x) = √(2x – 3) Step 1. Find (f + g)(x): (f + g)(x) = f(x) + g(x) = (x – 1) + √(2x – 3) Step 2. Find the domain for (f + g): Since g(x) = √(2x – 3) is only defined when the radicand is non-negative, set:   2x – 3 ≥ 0 ⟹ x ≥ 3/2 Thus, the domain of (f + g) is: [3/2, ∞). Step 3. Find (f · g)(x): (f · g)(x) = f(x) · g(x) = (x – 1) · √(2x – 3) Step 4. Find the domain for (f · g): Once again, √(2x – 3) requires that 2x – 3 ≥ 0 ⟹ x ≥ 3/2. There are no additional restrictions from (x – 1), so the domain of (f · g) is also: [3/2, ∞). Final Answers:   (f + g)(x) = (x – 1) + √(2x – 3)   Domain of (f + g): [3/2, ∞)   (f · g)(x) = (x – 1) · √(2x – 3)   Domain of (f · g): [3/2, ∞)

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The Deep Dive

To find \( f + g \) and \( f \cdot g \), we'll start by combining the two functions mathematically. 1. The sum of the functions: \[ (f+g)(x) = f(x) + g(x) = (x - 1) + \sqrt{2x - 3} \] 2. The product of the functions: \[ (f \cdot g)(x) = f(x) \cdot g(x) = (x - 1) \cdot \sqrt{2x - 3} \] Next, let's find the domains of each function before concluding. **Domain of \( f + g \)**: - The function \( f(x) = x - 1 \) is defined for all real numbers. - The function \( g(x) = \sqrt{2x - 3} \) requires that \( 2x - 3 \geq 0 \), which simplifies to \( x \geq \frac{3}{2} \). Thus, the domain of \( (f + g)(x) \) is \( [\frac{3}{2}, \infty) \). **Domain of \( f \cdot g \)**: - Similar to before, \( f(x) \) is defined for all \( x \), while \( g(x) \) also requires \( x \geq \frac{3}{2} \). Therefore, the domain of \( (f \cdot g)(x) \) is also \( [\frac{3}{2}, \infty) \). In conclusion, we have: \[ (f+g)(x) = x - 1 + \sqrt{2x - 3} \] Domain of \( f+g \) : \( [\frac{3}{2}, \infty) \) \[ (f \cdot g)(x) = (x - 1) \sqrt{2x - 3} \] Domain of \( f \cdot g \) : \( [\frac{3}{2}, \infty) \)

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