Answer :
To determine the domain of the function [tex]\((c \cdot d)(x)\)[/tex], where [tex]\(c(x) = \frac{5}{x-2}\)[/tex] and [tex]\(d(x) = x+3\)[/tex], let's analyze each function and how they combine.
1. Domain of [tex]\(c(x)\)[/tex]:
- The function [tex]\(c(x) = \frac{5}{x-2}\)[/tex] is a rational function. Rational functions are defined for all real numbers except where the denominator is zero.
- Therefore, for [tex]\(c(x)\)[/tex] to be defined, [tex]\(x - 2 \neq 0\)[/tex].
- Solving [tex]\(x - 2 = 0\)[/tex] gives [tex]\(x = 2\)[/tex].
Hence, the domain of [tex]\(c(x)\)[/tex] is all real numbers except [tex]\(x = 2\)[/tex].
2. Domain of [tex]\(d(x)\)[/tex]:
- The function [tex]\(d(x) = x + 3\)[/tex] is a linear function. Linear functions are defined for all real numbers.
- Therefore, the domain of [tex]\(d(x)\)[/tex] is all real numbers.
3. Domain of the product [tex]\((c \cdot d)(x) = c(x) \cdot d(x)\)[/tex]:
- The product [tex]\((c \cdot d)(x) = \left(\frac{5}{x-2} \right) \cdot (x + 3)\)[/tex].
- To find where [tex]\((c \cdot d)(x)\)[/tex] is defined, we need both [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex] to be defined.
Analyzing:
- From [tex]\(c(x)\)[/tex], we know [tex]\(x \neq 2\)[/tex] because [tex]\(\frac{5}{x-2}\)[/tex] is undefined at [tex]\(x = 2\)[/tex].
- Additionally, consider what happens when we directly substitute [tex]\((c \cdot d)(x)\)[/tex]: [tex]\(\left(\frac{5}{x-2}\right) \cdot (x + 3)\)[/tex].
- If [tex]\(x = -3\)[/tex], then [tex]\(x + 3 = 0\)[/tex], which would make the product zero with no undefined issues.
- However, [tex]\(\left(\frac{5}{x-2}\right)\)[/tex] remains undefined solely at [tex]\(x = 2\)[/tex], no new exclusions are introduced from the product directly unless denominator again trickles back to 0.
Therefore, the domain of [tex]\( (c \cdot d)(x) \)[/tex] is all real values of [tex]\( x \)[/tex] except [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex], as these points lead to division by zero in either the original functions or product.
So, the correct answer is:
[tex]\[ \text{all real values of } x \text{ except } x=2 \text{ and } x=-3 \][/tex]
1. Domain of [tex]\(c(x)\)[/tex]:
- The function [tex]\(c(x) = \frac{5}{x-2}\)[/tex] is a rational function. Rational functions are defined for all real numbers except where the denominator is zero.
- Therefore, for [tex]\(c(x)\)[/tex] to be defined, [tex]\(x - 2 \neq 0\)[/tex].
- Solving [tex]\(x - 2 = 0\)[/tex] gives [tex]\(x = 2\)[/tex].
Hence, the domain of [tex]\(c(x)\)[/tex] is all real numbers except [tex]\(x = 2\)[/tex].
2. Domain of [tex]\(d(x)\)[/tex]:
- The function [tex]\(d(x) = x + 3\)[/tex] is a linear function. Linear functions are defined for all real numbers.
- Therefore, the domain of [tex]\(d(x)\)[/tex] is all real numbers.
3. Domain of the product [tex]\((c \cdot d)(x) = c(x) \cdot d(x)\)[/tex]:
- The product [tex]\((c \cdot d)(x) = \left(\frac{5}{x-2} \right) \cdot (x + 3)\)[/tex].
- To find where [tex]\((c \cdot d)(x)\)[/tex] is defined, we need both [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex] to be defined.
Analyzing:
- From [tex]\(c(x)\)[/tex], we know [tex]\(x \neq 2\)[/tex] because [tex]\(\frac{5}{x-2}\)[/tex] is undefined at [tex]\(x = 2\)[/tex].
- Additionally, consider what happens when we directly substitute [tex]\((c \cdot d)(x)\)[/tex]: [tex]\(\left(\frac{5}{x-2}\right) \cdot (x + 3)\)[/tex].
- If [tex]\(x = -3\)[/tex], then [tex]\(x + 3 = 0\)[/tex], which would make the product zero with no undefined issues.
- However, [tex]\(\left(\frac{5}{x-2}\right)\)[/tex] remains undefined solely at [tex]\(x = 2\)[/tex], no new exclusions are introduced from the product directly unless denominator again trickles back to 0.
Therefore, the domain of [tex]\( (c \cdot d)(x) \)[/tex] is all real values of [tex]\( x \)[/tex] except [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex], as these points lead to division by zero in either the original functions or product.
So, the correct answer is:
[tex]\[ \text{all real values of } x \text{ except } x=2 \text{ and } x=-3 \][/tex]