Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
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.
### 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.
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
