To determine the domain of the function [tex]\((c \cdot d)(x)\)[/tex], defined as the product of the functions [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex], where [tex]\(c(x) = \frac{5}{x-2}\)[/tex] and [tex]\(d(x) = x+3\)[/tex], we need to look at the domains of the individual functions and find their intersection.
1. Domain of [tex]\(c(x)\)[/tex]:
- [tex]\(c(x) = \frac{5}{x-2}\)[/tex]
- The expression [tex]\(\frac{5}{x-2}\)[/tex] is defined for all [tex]\(x\)[/tex] except [tex]\(x = 2\)[/tex], because the denominator cannot be zero.
- Therefore, the domain of [tex]\(c(x)\)[/tex] is all real numbers except [tex]\(x = 2\)[/tex]. Symbolically, this can be expressed as [tex]\(x \in \mathbb{R} \setminus \{2\}\)[/tex].
2. Domain of [tex]\(d(x)\)[/tex]:
- [tex]\(d(x) = x + 3\)[/tex]
- This is a linear function which is defined for all real numbers.
- Hence, the domain of [tex]\(d(x)\)[/tex] is all real numbers, [tex]\(x \in \mathbb{R}\)[/tex].
3. Domain of [tex]\((c \cdot d)(x)\)[/tex]:
- The domain of the product function [tex]\((c \cdot d)(x) = c(x) \cdot d(x)\)[/tex] is the intersection of the domains of [tex]\(c(x)\)[/tex] and [tex]\(d(x)\)[/tex].
- The intersection of the domains of [tex]\(c(x) \)[/tex] and [tex]\(d(x)\)[/tex] would be all real numbers except [tex]\(x = 2\)[/tex], since this is the only value at which [tex]\(c(x)\)[/tex] is undefined.
Thus, the domain of [tex]\((c \cdot d)(x)\)[/tex] is all real numbers except [tex]\(x = 2\)[/tex]. Therefore, the domain of [tex]\((c \cdot d)(x)\)[/tex] can be expressed as:
[tex]\[ x \in \mathbb{R} \setminus \{2\}. \][/tex]
This concludes that the domain of [tex]\((c \cdot d)(x)\)[/tex] is all real numbers except [tex]\(x = 2\)[/tex].
