To determine the domain of [tex]\((c \cdot d)(x)\)[/tex], where [tex]\(c(x) = \frac{5}{x-2}\)[/tex] and [tex]\(d(x) = x + 3\)[/tex], we need to consider the domains of both [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex] individually and then determine the combined effect of these domains.
1. Domain of [tex]\(c(x) = \frac{5}{x-2}\)[/tex]:
- The function [tex]\(c(x)\)[/tex] involves a division by [tex]\(x-2\)[/tex]. Division by zero is undefined, so [tex]\(x-2 \neq 0\)[/tex].
- This means that [tex]\(x \neq 2\)[/tex].
- Therefore, [tex]\(c(x)\)[/tex] is undefined at [tex]\(x = 2\)[/tex]. The domain of [tex]\(c(x)\)[/tex] is all real numbers except [tex]\(x=2\)[/tex].
2. Domain of [tex]\(d(x) = x + 3\)[/tex]:
- The function [tex]\(d(x)\)[/tex] is a linear polynomial and is defined for all real numbers.
- Therefore, there are no restrictions on the domain of [tex]\(d(x)\)[/tex]. The domain of [tex]\(d(x)\)[/tex] is all real numbers.
3. Domain of [tex]\((c \cdot d)(x)\)[/tex]:
- [tex]\((c \cdot d)(x)\)[/tex] denotes the product of [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex].
- For [tex]\((c \cdot d)(x)\)[/tex] to be defined, both [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex] have to be defined.
- From the domain of [tex]\(c(x)\)[/tex], we know [tex]\(x\)[/tex] cannot be [tex]\(2\)[/tex].
- Since [tex]\(d(x)\)[/tex] is defined for all [tex]\(x\)[/tex], it does not introduce any additional restrictions.
- Therefore, the domain of [tex]\((c \cdot d)(x)\)[/tex] is all real numbers except [tex]\(x=2\)[/tex].
Putting this all together, the domain of [tex]\((c \cdot d)(x)\)[/tex] is:
[tex]\[
\text{all real values of } x \text{ except } x=2
\][/tex]
Hence, the domain of [tex]\((c \cdot d)(x)\)[/tex] is:
[tex]\[
\boxed{\text{all real values of } x \text{ except } x=2}
\][/tex]
