Sure! Let's dive into the problem step-by-step:
### Part (a): Finding the Domain of Each Function and Combinations
1. Domain of [tex]\( f(x) = x - 2 \)[/tex]:
- The function [tex]\( f(x) = x - 2 \)[/tex] is a linear function.
- Linear functions are defined for all real numbers.
- Hence, domain of [tex]\( f \)[/tex] is [tex]\(\mathbb{R}\)[/tex] (all real numbers).
2. Domain of [tex]\( g(x) = \sqrt{x + 1} \)[/tex]:
- The function [tex]\( g(x) = \sqrt{x + 1} \)[/tex] is defined for the values inside the square root being non-negative.
- So, [tex]\( x + 1 \geq 0 \implies x \geq -1 \)[/tex].
- Hence, domain of [tex]\( g \)[/tex] is [tex]\([-1, \infty)\)[/tex].
3. Domain of [tex]\( f+g \)[/tex]:
- [tex]\( (f+g)(x) = f(x) + g(x) = (x - 2) + \sqrt{x + 1} \)[/tex].
- The domain is the intersection of the domains of [tex]\( f \)[/tex] and [tex]\( g \)[/tex].
- Intersection of [tex]\(\mathbb{R}\)[/tex] and [tex]\([-1, \infty)\)[/tex] is [tex]\([-1, \infty)\)[/tex].
- Hence, domain of [tex]\( f+g \)[/tex] is [tex]\([-1, \infty)\)[/tex].
4. Domain of [tex]\( f-g \)[/tex]:
- [tex]\( (f-g)(x) = f(x) - g(x) = (x - 2) - \sqrt{x + 1} \)[/tex].
- Domain is the same as for [tex]\( f+g \)[/tex] (intersection of [tex]\(\mathbb{R}\)[/tex] and [tex]\([-1, \infty)\)[/tex]).
- Hence, domain of [tex]\( f-g \)[/tex] is [tex]\([-1, \infty)\)[/tex].
5. Domain of [tex]\( f \cdot g \)[/tex]:
- [tex]\( (fg)(x) = f(x) \cdot g(x) = (x - 2) \cdot \sqrt{x + 1} \)[/tex].
- Domain is the same as for [tex]\( f+g \)[/tex] (intersection of [tex]\(\mathbb{R}\)[/tex] and [tex]\([-1, \infty)\)[/tex]).
- Hence, domain of [tex]\( f \cdot g \)[/tex] is [tex]\([-1, \infty)\)[/tex].
6. Domain of [tex]\( \frac{f}{g} \)[/tex]:
- [tex]\( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x - 2}{\sqrt{x + 1}} \)[/tex].
- [tex]\( g(x) \neq 0 \)[/tex], so [tex]\( \sqrt{x + 1} \neq 0 \implies x \neq -1 \)[/tex].
- Intersection of [tex]\(\mathbb{R}\)[/tex] and [tex]\([-1, \infty)\)[/tex] excluding [tex]\( x = -1 \)[/tex].
- Hence, domain of [tex]\( \frac{f}{g} \)[/tex] is [tex]\((-1, \infty)\)[/tex].
7. Domain of [tex]\( \frac{g}{f} \)[/tex]:
- [tex]\( \left(\frac{g}{f}\right)(x) = \frac{g(x)}{f(x)} = \frac{\sqrt{x + 1}}{x - 2} \)[/tex].
- [tex]\( f(x) \neq 0 \)[/tex], so [tex]\( x - 2 \neq 0 \implies x \neq 2 \)[/tex].
- Intersection of [tex]\(\mathbb{R}\)[/tex] and [tex]\([-1, \infty)\)[/tex] excluding [tex]\( x = 2 \)[/tex].
- Hence, domain of [tex]\( \frac{g}{f} \)[/tex] is [tex]\(\left[-1, 2\right) \cup \left(2, \infty\right)\)[/tex].
### Part (b): Calculating the Expressions
1. [tex]\( (f+g)(x) \)[/tex]:
[tex]\[
(f+g)(x) = f(x) + g(x) = (x - 2) + \sqrt{x + 1} = x + \sqrt{x + 1} - 2
\][/tex]
2. [tex]\( (f-g)(x) \)[/tex]:
[tex]\[
(f-g)(x) = f(x) - g(x) = (x - 2) - \sqrt{x + 1} = x - \sqrt{x + 1} - 2
\][/tex]
3. [tex]\( (fg)(x) \)[/tex]:
[tex]\[
(fg)(x) = f(x) \cdot g(x) = (x - 2) \cdot \sqrt{x + 1}
\][/tex]
4. [tex]\( (ff)(x) \)[/tex]:
[tex]\[
(ff)(x) = f(f(x)) = f(x - 2) = (x - 2) - 2 = x - 4
\][/tex]
5. [tex]\( \left(\frac{f}{g}\right)(x) \)[/tex]:
[tex]\[
\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x - 2}{\sqrt{x + 1}}
\][/tex]
6. [tex]\( \left(\frac{g}{f}\right)(x) \)[/tex]:
[tex]\[
\left(\frac{g}{f}\right)(x) = \frac{g(x)}{f(x)} = \frac{\sqrt{x + 1}}{x - 2}
\][/tex]
This concludes the detailed step-by-step solution for the given functions and their combinations along with their domains.
