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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.
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