Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To find the domain of the composition function [tex]\((C \circ d)(x)\)[/tex], where [tex]\(C(x) = \frac{5}{x-2}\)[/tex] and [tex]\(d(x) = x+3\)[/tex], follow these steps:
1. Domain of [tex]\(d(x)\)[/tex]:
- The function [tex]\(d(x) = x + 3\)[/tex] is defined for all real numbers. Thus, the domain of [tex]\(d(x)\)[/tex] is all real values of [tex]\(x\)[/tex].
2. Range of [tex]\(d(x)\)[/tex]:
- Since [tex]\(d(x)\)[/tex] maps all real numbers to all real numbers, the range of [tex]\(d(x)\)[/tex] is also all real values.
3. Domain of [tex]\(C(x)\)[/tex]:
- The function [tex]\(C(x) = \frac{5}{x-2}\)[/tex] is defined for all [tex]\(x\)[/tex] except [tex]\(x = 2\)[/tex] because [tex]\(x = 2\)[/tex] would make the denominator zero, which is undefined.
- Therefore, the domain of [tex]\(C(x)\)[/tex] is all real values except [tex]\(x = 2\)[/tex].
4. Determine [tex]\((C \circ d)(x)\)[/tex]:
- The composition [tex]\((C \circ d)(x)\)[/tex] means we apply [tex]\(d(x)\)[/tex] first and then apply [tex]\(C\)[/tex] to the result. In other words, [tex]\((C \circ d)(x) = C(d(x))\)[/tex].
5. Expression for [tex]\((C \circ d)(x)\)[/tex]:
- First, find [tex]\(d(x)\)[/tex]:
[tex]\[ d(x) = x + 3 \][/tex]
- Then, substitute [tex]\(d(x)\)[/tex] into [tex]\(C(x)\)[/tex]:
[tex]\[ (C \circ d)(x) = C(d(x)) = C(x + 3) = \frac{5}{(x + 3) - 2} = \frac{5}{x + 1} \][/tex]
6. Domain of [tex]\((C \circ d)(x)\)[/tex]:
- For [tex]\((C \circ d)(x)\)[/tex] to be defined, the denominator [tex]\(x + 1\)[/tex] must not be zero. Therefore:
[tex]\[ x + 1 \neq 0 \implies x \neq -1 \][/tex]
- Additionally, since [tex]\(d(x) = x + 3\)[/tex], the input to [tex]\(C(x)\)[/tex] should not be 2 for [tex]\(C(x)\)[/tex] to be defined:
[tex]\[ x + 3 \neq 2 \implies x \neq -1 \][/tex]
- Therefore, [tex]\(x \neq -1\)[/tex] covers both conditions automatically by ensuring that [tex]\(d(x) \neq 2\)[/tex] and that the denominator of the final expression is non-zero.
Thus, the domain of [tex]\((C \circ d)(x)\)[/tex] is all real numbers except [tex]\(x = -1\)[/tex]. In the original multiple-choice responses given:
- All real values of [tex]\(x\)[/tex] except [tex]\(x = 2\)[/tex] and [tex]\(x = -3\)[/tex].
The correct answer is all real values of [tex]\(x\)[/tex] except [tex]\(x = -1\)[/tex]. However, none of the given choices directly match this exact answer. If we are confined to the given choices, the closest conceptually would be "all real values of \$ except [tex]\(x = 2\)[/tex] and [tex]\(x = -3\)[/tex]," but this is not strictly correct for the function analyzed here.
1. Domain of [tex]\(d(x)\)[/tex]:
- The function [tex]\(d(x) = x + 3\)[/tex] is defined for all real numbers. Thus, the domain of [tex]\(d(x)\)[/tex] is all real values of [tex]\(x\)[/tex].
2. Range of [tex]\(d(x)\)[/tex]:
- Since [tex]\(d(x)\)[/tex] maps all real numbers to all real numbers, the range of [tex]\(d(x)\)[/tex] is also all real values.
3. Domain of [tex]\(C(x)\)[/tex]:
- The function [tex]\(C(x) = \frac{5}{x-2}\)[/tex] is defined for all [tex]\(x\)[/tex] except [tex]\(x = 2\)[/tex] because [tex]\(x = 2\)[/tex] would make the denominator zero, which is undefined.
- Therefore, the domain of [tex]\(C(x)\)[/tex] is all real values except [tex]\(x = 2\)[/tex].
4. Determine [tex]\((C \circ d)(x)\)[/tex]:
- The composition [tex]\((C \circ d)(x)\)[/tex] means we apply [tex]\(d(x)\)[/tex] first and then apply [tex]\(C\)[/tex] to the result. In other words, [tex]\((C \circ d)(x) = C(d(x))\)[/tex].
5. Expression for [tex]\((C \circ d)(x)\)[/tex]:
- First, find [tex]\(d(x)\)[/tex]:
[tex]\[ d(x) = x + 3 \][/tex]
- Then, substitute [tex]\(d(x)\)[/tex] into [tex]\(C(x)\)[/tex]:
[tex]\[ (C \circ d)(x) = C(d(x)) = C(x + 3) = \frac{5}{(x + 3) - 2} = \frac{5}{x + 1} \][/tex]
6. Domain of [tex]\((C \circ d)(x)\)[/tex]:
- For [tex]\((C \circ d)(x)\)[/tex] to be defined, the denominator [tex]\(x + 1\)[/tex] must not be zero. Therefore:
[tex]\[ x + 1 \neq 0 \implies x \neq -1 \][/tex]
- Additionally, since [tex]\(d(x) = x + 3\)[/tex], the input to [tex]\(C(x)\)[/tex] should not be 2 for [tex]\(C(x)\)[/tex] to be defined:
[tex]\[ x + 3 \neq 2 \implies x \neq -1 \][/tex]
- Therefore, [tex]\(x \neq -1\)[/tex] covers both conditions automatically by ensuring that [tex]\(d(x) \neq 2\)[/tex] and that the denominator of the final expression is non-zero.
Thus, the domain of [tex]\((C \circ d)(x)\)[/tex] is all real numbers except [tex]\(x = -1\)[/tex]. In the original multiple-choice responses given:
- All real values of [tex]\(x\)[/tex] except [tex]\(x = 2\)[/tex] and [tex]\(x = -3\)[/tex].
The correct answer is all real values of [tex]\(x\)[/tex] except [tex]\(x = -1\)[/tex]. However, none of the given choices directly match this exact answer. If we are confined to the given choices, the closest conceptually would be "all real values of \$ except [tex]\(x = 2\)[/tex] and [tex]\(x = -3\)[/tex]," but this is not strictly correct for the function analyzed here.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We're dedicated to helping you find the answers you need at Westonci.ca. Don't hesitate to return for more.