Questions: Let f(x) = sqrt(2x - 1) and g(x) = 1/x. Find (f+g)(x), (f-g)(x), (fg)(x), and (f/g)(x). Give the domain of each. (f+g)(x) = (f-g)(x) = (fg)(x) = (f/g)(x) = The domain of f+g is The domain of f-g is The domain of fg is

Solution Steps

To solve the given problem, we need to perform the following steps:

1. Define the functions \( f(x) = \sqrt{2x - 1} \) and \( g(x) = \frac{1}{x} \).
2. Compute the sum, difference, product, and quotient of the functions \( f \) and \( g \).
3. Determine the domain of each resulting function.

1. Sum of functions: \( (f+g)(x) = f(x) + g(x) \)
2. Difference of functions: \( (f-g)(x) = f(x) - g(x) \)
3. Product of functions: \( (fg)(x) = f(x) \cdot g(x) \)
4. Quotient of functions: \( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \)

For the domain:

• \( f(x) \) is defined when \( 2x - 1 \geq 1 \Rightarrow x \geq \frac{1}{2} \).
• \( g(x) \) is defined when \( x \neq 0 \).
• The domain of \( f+g \), \( f-g \), \( fg \), and \( \frac{f}{g} \) will be the intersection of the domains of \( f \) and \( g \).

Given: \[ f(x) = \sqrt{2x - 1} \] \[ g(x) = \frac{1}{x} \]

\[ (f+g)(x) = f(x) + g(x) = \sqrt{2x - 1} + \frac{1}{x} \]

\[ (f-g)(x) = f(x) - g(x) = \sqrt{2x - 1} - \frac{1}{x} \]

\[ (fg)(x) = f(x) \cdot g(x) = \sqrt{2x - 1} \cdot \frac{1}{x} = \frac{\sqrt{2x - 1}}{x} \]

\[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{\sqrt{2x - 1}}{\frac{1}{x}} = x \sqrt{2x - 1} \]

• The domain of \( f(x) \) is \( 2x - 1 \geq 0 \Rightarrow x \geq \frac{1}{2} \).
• The domain of \( g(x) \) is \( x \neq 0 \).

The intersection of these domains is: \[ x \geq \frac{1}{2} \]

Final Answer