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The functions [tex]$f$[/tex] and [tex]$g$[/tex] are defined as [tex]$f(x) = x - 8$[/tex] and [tex]$g(x) = \sqrt{x + 4}$[/tex].

a) Find the domain of [tex]$f$[/tex], [tex]$g$[/tex], [tex]$f+g$[/tex], [tex]$f-g$[/tex], [tex]$f \cdot g$[/tex], [tex]$f \cdot f$[/tex], [tex]$\frac{f}{g}$[/tex], and [tex]$\frac{g}{f}$[/tex].

b) Find [tex]$(f+g)(x)$[/tex], [tex]$(f-g)(x)$[/tex], [tex]$(f \cdot g)(x)$[/tex], [tex]$(f \cdot f)(x)$[/tex], [tex]$\left(\frac{f}{g}\right)(x)$[/tex], and [tex]$\left(\frac{g}{f}\right)(x)$[/tex].

Sagot :

Alright, let's tackle each part step-by-step.

### Part (a): Finding the Domains

1. Domain of [tex]\( f(x) = x - 8 \)[/tex]:
The function [tex]\( f \)[/tex] is a linear function defined for all real numbers. Thus, the domain of [tex]\( f \)[/tex] is:
[tex]\[ \text{Domain of } f: \mathbb{R} \][/tex]

2. Domain of [tex]\( g(x) = \sqrt{x + 4} \)[/tex]:
The square root function is defined where its argument is non-negative, meaning:
[tex]\[ x + 4 \geq 0 \implies x \geq -4 \][/tex]
Thus, the domain of [tex]\( g \)[/tex] is:
[tex]\[ \text{Domain of } g: [-4, \infty) \][/tex]

3. Domain of [tex]\( (f+g)(x) = f(x) + g(x) \)[/tex]:
The domain of [tex]\( f+g \)[/tex] is the intersection of the domains of [tex]\( f \)[/tex] and [tex]\( g \)[/tex]:
[tex]\[ \text{Domain of } (f+g): [-4, \infty) \][/tex]

4. Domain of [tex]\( (f-g)(x) = f(x) - g(x) \)[/tex]:
The domain of [tex]\( f-g \)[/tex] is also the intersection of the domains of [tex]\( f \)[/tex] and [tex]\( g \)[/tex]:
[tex]\[ \text{Domain of } (f-g): [-4, \infty) \][/tex]

5. Domain of [tex]\( (fg)(x) = f(x) \cdot g(x) \)[/tex]:
The domain of [tex]\( fg \)[/tex] is likewise the intersection of the domains of [tex]\( f \)[/tex] and [tex]\( g \)[/tex]:
[tex]\[ \text{Domain of } (fg): [-4, \infty) \][/tex]

6. Domain of [tex]\( (ff)(x) = f(x) \cdot f(x) \)[/tex]:
Since [tex]\( f \)[/tex] is defined for all real numbers:
[tex]\[ \text{Domain of } (ff): \mathbb{R} \][/tex]

7. Domain of [tex]\( \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} \)[/tex]:
Here, we need [tex]\( g(x) \neq 0 \)[/tex] and [tex]\( g \)[/tex] itself must be defined. Thus, the domain is:
[tex]\[ g(x) \neq 0 \implies \sqrt{x+4} \neq 0 \implies x \neq -4 \][/tex]
Therefore, the domain is:
[tex]\[ \text{Domain of } \left(\frac{f}{g}\right): (-4, \infty) \setminus \{-4\} \][/tex]
Which can be written more compactly as:
[tex]\[ \text{Domain of } \left(\frac{f}{g}\right): (-4, \infty) \][/tex]

8. Domain of [tex]\( \left(\frac{g}{f}\right)(x) = \frac{g(x)}{f(x)} \)[/tex]:
Here, we need [tex]\( f(x) \neq 0 \)[/tex] and [tex]\( f \)[/tex] and [tex]\( g \)[/tex] must both be defined. Thus, the domain is:
[tex]\[ f(x) \neq 0 \implies x - 8 \neq 0 \implies x \neq 8 \][/tex]
So the intersection of the domains and the exclusion gives us:
[tex]\[ \text{Domain of } \left(\frac{g}{f}\right): [-4, 8) \cup (8, \infty) \][/tex]

### Part (b): Finding the Functions

1. [tex]\( (f+g)(x) \)[/tex]:
[tex]\[ (f+g)(x) = f(x) + g(x) = (x - 8) + \sqrt{x+4} = x + \sqrt{x+4} - 8 \][/tex]

2. [tex]\( (f-g)(x) \)[/tex]:
[tex]\[ (f-g)(x) = f(x) - g(x) = (x - 8) - \sqrt{x+4} = x - \sqrt{x+4} - 8 \][/tex]

3. [tex]\( (fg)(x) \)[/tex]:
[tex]\[ (fg)(x) = f(x) \cdot g(x) = (x - 8) \cdot \sqrt{x + 4} \][/tex]

4. [tex]\( (ff)(x) \)[/tex]:
[tex]\[ (ff)(x) = f(x) \cdot f(x) = (x - 8) \cdot (x - 8) = (x - 8)^2 \][/tex]

5. [tex]\( \left(\frac{f}{g}\right)(x) \)[/tex]:
[tex]\[ \left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)} = \frac{x - 8}{\sqrt{x + 4}} \][/tex]

6. [tex]\( \left(\frac{g}{f}\right)(x) \)[/tex]:
[tex]\[ \left(\frac{g}{f}\right)(x) = \frac{g(x)}{f(x)} = \frac{\sqrt{x + 4}}{x - 8} \][/tex]

This covers both parts (a) and (b) in detail.