At Westonci.ca, we connect you with the best answers from a community of experienced and knowledgeable individuals. Get immediate and reliable solutions to your questions from a community of experienced experts on our Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Let's consider the functions [tex]\( f(x) = \frac{1}{8} \sqrt{x} \)[/tex] and [tex]\( g(x) = 8 \sqrt{x} \)[/tex] and analyze their domains and ranges.
### Domains:
1. [tex]\( f(x) = \frac{1}{8} \sqrt{x} \)[/tex]
- The function [tex]\( f(x) \)[/tex] involves the square root of [tex]\( x \)[/tex]. Since the square root function [tex]\( \sqrt{x} \)[/tex] is only defined for [tex]\( x \geq 0 \)[/tex], the domain of [tex]\( f(x) \)[/tex] is [tex]\( x \geq 0 \)[/tex], or in interval notation, [tex]\([0, \infty)\)[/tex].
2. [tex]\( g(x) = 8 \sqrt{x} \)[/tex]
- Similarly, the function [tex]\( g(x) \)[/tex] also involves the square root of [tex]\( x \)[/tex]. Hence, the domain of [tex]\( g(x) \)[/tex] is also [tex]\( x \geq 0 \)[/tex], or in interval notation, [tex]\([0, \infty)\)[/tex].
### Ranges:
1. [tex]\( f(x) = \frac{1}{8} \sqrt{x} \)[/tex]
- For [tex]\( f(x) = \frac{1}{8} \sqrt{x} \)[/tex], as [tex]\( x \)[/tex] goes from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex], [tex]\( \sqrt{x} \)[/tex] also goes from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex]. Multiplying by [tex]\( \frac{1}{8} \)[/tex], the output [tex]\( f(x) \)[/tex] ranges from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex] multiplied by [tex]\( \frac{1}{8} \)[/tex], which results in:
[tex]\[ f(x) \in \left[0, \infty \right) \][/tex]
2. [tex]\( g(x) = 8 \sqrt{x} \)[/tex]
- For [tex]\( g(x) = 8 \sqrt{x} \)[/tex], as [tex]\( x \)[/tex] goes from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex], [tex]\( \sqrt{x} \)[/tex] also goes from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex]. Multiplying by [tex]\( 8 \)[/tex], the output [tex]\( g(x) \)[/tex] ranges from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex] multiplied by [tex]\( 8 \)[/tex], which results in:
[tex]\[ g(x) \in \left[0, \infty \right) \][/tex]
Although both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same domain [tex]\([0, \infty)\)[/tex], their ranges are different due to the different scaling factors involved ([tex]\(\frac{1}{8}\)[/tex] for [tex]\( f(x) \)[/tex] and [tex]\( 8 \)[/tex] for [tex]\( g(x)\)[/tex]). Thus, if we compare the ranges:
- [tex]\( f(x) = \frac{1}{8} \sqrt{x} \)[/tex] yields a smaller range scaled down by [tex]\( \frac{1}{8} \)[/tex].
- [tex]\( g(x) = 8 \sqrt{x} \)[/tex] yields a larger range scaled up by [tex]\( 8 \)[/tex].
Therefore, the correct statement is:
"They have the same domain but different ranges."
So, the answer is 4.
### Domains:
1. [tex]\( f(x) = \frac{1}{8} \sqrt{x} \)[/tex]
- The function [tex]\( f(x) \)[/tex] involves the square root of [tex]\( x \)[/tex]. Since the square root function [tex]\( \sqrt{x} \)[/tex] is only defined for [tex]\( x \geq 0 \)[/tex], the domain of [tex]\( f(x) \)[/tex] is [tex]\( x \geq 0 \)[/tex], or in interval notation, [tex]\([0, \infty)\)[/tex].
2. [tex]\( g(x) = 8 \sqrt{x} \)[/tex]
- Similarly, the function [tex]\( g(x) \)[/tex] also involves the square root of [tex]\( x \)[/tex]. Hence, the domain of [tex]\( g(x) \)[/tex] is also [tex]\( x \geq 0 \)[/tex], or in interval notation, [tex]\([0, \infty)\)[/tex].
### Ranges:
1. [tex]\( f(x) = \frac{1}{8} \sqrt{x} \)[/tex]
- For [tex]\( f(x) = \frac{1}{8} \sqrt{x} \)[/tex], as [tex]\( x \)[/tex] goes from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex], [tex]\( \sqrt{x} \)[/tex] also goes from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex]. Multiplying by [tex]\( \frac{1}{8} \)[/tex], the output [tex]\( f(x) \)[/tex] ranges from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex] multiplied by [tex]\( \frac{1}{8} \)[/tex], which results in:
[tex]\[ f(x) \in \left[0, \infty \right) \][/tex]
2. [tex]\( g(x) = 8 \sqrt{x} \)[/tex]
- For [tex]\( g(x) = 8 \sqrt{x} \)[/tex], as [tex]\( x \)[/tex] goes from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex], [tex]\( \sqrt{x} \)[/tex] also goes from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex]. Multiplying by [tex]\( 8 \)[/tex], the output [tex]\( g(x) \)[/tex] ranges from [tex]\( 0 \)[/tex] to [tex]\( \infty \)[/tex] multiplied by [tex]\( 8 \)[/tex], which results in:
[tex]\[ g(x) \in \left[0, \infty \right) \][/tex]
Although both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] have the same domain [tex]\([0, \infty)\)[/tex], their ranges are different due to the different scaling factors involved ([tex]\(\frac{1}{8}\)[/tex] for [tex]\( f(x) \)[/tex] and [tex]\( 8 \)[/tex] for [tex]\( g(x)\)[/tex]). Thus, if we compare the ranges:
- [tex]\( f(x) = \frac{1}{8} \sqrt{x} \)[/tex] yields a smaller range scaled down by [tex]\( \frac{1}{8} \)[/tex].
- [tex]\( g(x) = 8 \sqrt{x} \)[/tex] yields a larger range scaled up by [tex]\( 8 \)[/tex].
Therefore, the correct statement is:
"They have the same domain but different ranges."
So, the answer is 4.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Stay curious and keep coming back to Westonci.ca for answers to all your burning questions.