Question

section 2.7 - the algebra of functions

in the following exercise, for the functions \\(f\\) and \\(g\\), find each of the following and identify its domain.

a. \\((f + g)(x) = f(x) + g(x)\\)
b. \\((f - g)(x) = f(x) - g(x)\\)
c. \\((fg)(x) = f(x) * g(x)\\)
d. \\((f/g)(x) = f(x) / g(x)\\)

1. \\(f(x) = 1/x\\); \\(g(x) = 1/(2x - 1)\\)

Explanation:

Determine individual domains

Find domain of \(f\) and \(g\).
For \(f(x) = \frac{1}{x}\), \(x
eq 0\).
For \(g(x) = \frac{1}{2x - 1}\), \(2x - 1
eq 0 \implies x
eq \frac{1}{2}\).
Intersection domain is \(D = \{x \mid x
eq 0 \text{ and } x
eq \frac{1}{2}\}\).

Find sum and its domain

Compute \((f+g)(x)\).

$$ (f+g)(x) = \frac{1}{x} + \frac{1}{2x-1} = \frac{2x-1+x}{x(2x-1)} = \frac{3x-1}{x(2x-1)} $$

Domain is \(D = \{x \mid x
eq 0, x
eq \frac{1}{2}\}\).

Find difference and its domain

Compute \((f-g)(x)\).

$$ (f-g)(x) = \frac{1}{x} - \frac{1}{2x-1} = \frac{2x-1-x}{x(2x-1)} = \frac{x-1}{x(2x-1)} $$

Domain is \(D = \{x \mid x
eq 0, x
eq \frac{1}{2}\}\).

Find product and its domain

Compute \((fg)(x)\).

$$ (fg)(x) = \frac{1}{x} \cdot \frac{1}{2x-1} = \frac{1}{x(2x-1)} $$

Domain is \(D = \{x \mid x
eq 0, x
eq \frac{1}{2}\}\).

Find quotient and its domain

Compute \((f/g)(x)\).

$$ (f/g)(x) = \frac{1/x}{1/(2x-1)} = \frac{2x-1}{x} $$

For quotient, we also require \(g(x)
eq 0\). Since \(g(x) = \frac{1}{2x-1}
eq 0\) for all \(x\) in its domain, no additional values are excluded.
Domain is \(D = \{x \mid x
eq 0, x
eq \frac{1}{2}\}\).

Answer:

a. \((f+g)(x) = \frac{3x-1}{x(2x-1)}\), Domain: \(\{x \mid x
eq 0, x
eq \frac{1}{2}\}\)

b. \((f-g)(x) = \frac{x-1}{x(2x-1)}\), Domain: \(\{x \mid x
eq 0, x
eq \frac{1}{2}\}\)

c. \((fg)(x) = \frac{1}{x(2x-1)}\), Domain: \(\{x \mid x
eq 0, x
eq \frac{1}{2}\}\)

d. \((f/g)(x) = \frac{2x-1}{x}\), Domain: \(\{x \mid x
eq 0, x
eq \frac{1}{2}\}\)