Find \( f+g, f-g, f g \), and \( \frac{-}{g} \). Determine the domain for each function. \[ f(x)=\sqrt{x-3} ; g(x)=\sqrt{x-8} \] \( (f+g)(x)=\square \)
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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: \[ (f+g)(x) = \sqrt{x-3} + \sqrt{x-8} \quad \text{with domain } [8, \infty) \] \[ (f-g)(x) = \sqrt{x-3} - \sqrt{x-8} \quad \text{with domain } [8, \infty) \] \[ (f \cdot g)(x) = \sqrt{(x-3)(x-8)} \quad \text{with domain } [8, \infty) \] \[ \left(\frac{f}{g}\right)(x) = \frac{\sqrt{x-3}}{\sqrt{x-8}} \quad \text{with domain } (8, \infty) \]
