Let \( f(x)=\sqrt{5 x-2} \) and \( g(x)=\frac{1}{x} \). Find \( (f+g)(x),(f-g)(x),(f g)(x) \), and \( \left(\frac{f}{g}\right)(x) \). Give the domain of each. \( (f+g)(x)=\square \) (Simplify your answer.)

We are given   f(x) = √(5x – 2)  and  g(x) = 1/x. To find (f+g)(x), we add the functions:   (f+g)(x) = f(x) + g(x) = √(5x – 2) + 1/x. Now, we need to determine the domain. Step 1. Find the domain of f(x):   For √(5x – 2) to be defined (with a real result), we need    5x – 2 ≥ 0  ⟹  x ≥ 2/5. Step 2. Find the domain of g(x):   g(x) = 1/x is defined when x ≠ 0. Step 3. Find the intersection of the domains:   Since f requires x ≥ 2/5 and 2/5 > 0, the x-values in [2/5, ∞) already satisfy x ≠ 0.   Therefore, the domain of (f+g)(x) is x ∈ [2/5, ∞). Thus, the simplified answer for (f+g)(x) is:   (f+g)(x) = √(5x – 2) + 1/x  with domain x ∈ [2/5, ∞). For completeness, here are the other operations: 1. (f – g)(x):   (f – g)(x) = √(5x – 2) – 1/x,   with domain x ∈ [2/5, ∞). 2. (fg)(x):   (fg)(x) = f(x) ⋅ g(x) = (√(5x – 2))/x,   with domain x ∈ [2/5, ∞) (since x cannot be 0 and x must be ≥ 2/5). 3. (f/g)(x):   (f/g)(x) = f(x)/g(x) = (√(5x – 2))/(1/x) = x√(5x – 2),   with domain x ∈ [2/5, ∞) (again, the restriction x ≠ 0 is automatically satisfied). Thus, the answer to your problem is: (f+g)(x)= √(5x – 2) + 1/x,  Domain: [2/5, ∞).