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Find \( f+g, f-g, f g \), and \( \frac{-}{g} \). Determine the domain for each function. \[ f(x)=\sqrt{x-3} ; g(x)=\sqrt{x-8} \] \( (f+g)(x)=\square \)

Ask by Bradley French. in the United States
Feb 25,2025

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\( (f+g)(x) = \sqrt{x - 3} + \sqrt{x - 8} \) with domain \( x \geq 8 \).

Solución

We are given   f(x) = √(x – 3)  and  g(x) = √(x – 8). Remember that for a square root function √(h(x)) to be defined (in ℝ), the expression h(x) must be ≥ 0. • For f(x) = √(x – 3): x – 3 ≥ 0 ⟹ x ≥ 3. • For g(x) = √(x – 8): x – 8 ≥ 0 ⟹ x ≥ 8. Thus, the domain of f(x) is [3, ∞) and the domain of g(x) is [8, ∞). When combining functions, we must take the intersection of their domains. Below are the four operations with their expressions and domains. ──────────────────────── 1. (f + g)(x) ──────────────────────── We have:   (f + g)(x) = f(x) + g(x) = √(x – 3) + √(x – 8). The domain is the intersection of the domains of f and g, that is, x must satisfy both x ≥ 3 and x ≥ 8. Thus, the domain is:   x ≥ 8  (i.e. [8, ∞)). ──────────────────────── 2. (f – g)(x) ──────────────────────── We have:   (f – g)(x) = f(x) – g(x) = √(x – 3) – √(x – 8). Again, the domain is x ≥ 8 (i.e. [8, ∞)). ──────────────────────── 3. (f · g)(x) ──────────────────────── We have:   (f · g)(x) = f(x) · g(x) = √(x – 3) · √(x – 8) = √[(x – 3)(x – 8)]. The function is defined only when both square roots exist, i.e. when x is in the intersection of the domains of f and g (x ≥ 8). Thus, the domain is:   x ≥ 8 (i.e. [8, ∞)). ──────────────────────── 4. (f / g)(x) ──────────────────────── (Note: The given notation “\( \frac{-}{g} \)” appears to be a typo. We interpret it as the quotient f/g.) We have:   (f / g)(x) = f(x) / g(x) = √(x – 3) / √(x – 8). For this quotient to be defined, in addition to needing x to be in the domain of both f and g (x ≥ 8), we must also have the denominator nonzero:   √(x – 8) ≠ 0 ⟹ x – 8 > 0 ⟹ x > 8. Thus, the domain of (f / g)(x) is:   x > 8 (i.e. (8, ∞)). ──────────────────────── Summary of Answers ──────────────────────── 1. (f + g)(x) = √(x – 3) + √(x – 8),  Domain: [8, ∞) 2. (f – g)(x) = √(x – 3) – √(x – 8),  Domain: [8, ∞) 3. (f · g)(x) = √[(x – 3)(x – 8)],  Domain: [8, ∞) 4. (f / g)(x) = √(x – 3) / √(x – 8),  Domain: (8, ∞) ──────────────────────── Answer for (f+g)(x):   (f+g)(x)= √(x – 3) + √(x – 8)  with Domain: [8, ∞). These are your final answers for the functions and their respective domains.

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Beyond the Answer

To find \( (f+g)(x) \), we simply add the two functions together: \[ (f + g)(x) = f(x) + g(x) = \sqrt{x - 3} + \sqrt{x - 8} \] Next, let's determine the domain for \( (f + g)(x) \). The square root functions require the expressions under the square roots to be non-negative: 1. For \( f(x) = \sqrt{x - 3} \), we need: \( x - 3 \geq 0 \) \( x \geq 3 \) 2. For \( g(x) = \sqrt{x - 8} \), we need: \( x - 8 \geq 0 \) \( x \geq 8 \) The domain of \( (f + g)(x) \) is determined by the intersection of these two conditions, so we need \( x \geq 8 \). Thus, the domain is \( [8, \infty) \). Now, let's calculate \( (f - g)(x) \): \[ (f - g)(x) = \sqrt{x - 3} - \sqrt{x - 8} \] The domain for \( (f - g)(x) \) is the same as that of \( (f + g)(x) \), which is \( [8, \infty) \). Next, we compute \( (f \cdot g)(x) \): \[ (f \cdot g)(x) = f(x) \cdot g(x) = \sqrt{x - 3} \cdot \sqrt{x - 8} = \sqrt{(x - 3)(x - 8)} \] The domain for \( (f \cdot g)(x) \) is also \( [8, \infty) \) since both component functions must be defined. Finally, let's define \( \frac{f}{g}(x) \): \[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{x - 3}}{\sqrt{x - 8}} \] For this function to be defined, \( g(x) \) must not be zero, which means \( \sqrt{x - 8} \neq 0 \) or \( x \neq 8 \). Thus, the domain is \( (8, \infty) \). To summarize: \[ (f+g)(x) = \sqrt{x-3} + \sqrt{x-8} \quad \text{with domain } [8, \infty) \] \[ (f-g)(x) = \sqrt{x-3} - \sqrt{x-8} \quad \text{with domain } [8, \infty) \] \[ (f \cdot g)(x) = \sqrt{(x-3)(x-8)} \quad \text{with domain } [8, \infty) \] \[ \left(\frac{f}{g}\right)(x) = \frac{\sqrt{x-3}}{\sqrt{x-8}} \quad \text{with domain } (8, \infty) \]

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